[Colectiv] Inequalities Methods and Olympiad Probl(BookZZ.org)

Inequalities Methods and Olympiad Problems

Contents 1 Part I 1.1 Squares are positive . . . . . . . . . . . . . . . . . . . . . 1.2 Some special inequalities and identities for two numbers . 1.3 Some special inequalities and identities for more numbers 1.4 The mathematical induction . . . . . . . . . . . . . . . . . 1.5 The AM-GM inequality . . . . . . . . . . . . . . . . . . . 1.6 The quadratic trinomial . . . . . . . . . . . . . . . . . . . 1.7 Cauchy-Schwartz Inequality . . . . . . . . . . . . . . . . . 1.8 Young Inequality . . . . . . . . . . . . . . . . . . . . . . . 1.9 Advanced techniques with Cauchy-Buniakowski-Schwarz Holder Inequalities . . . . . . . . . . . . . . . . . . . . . . 1.10 The principle of extremality and monotonicity . . . . . . . 1.11 Breaking the inequality . . . . . . . . . . . . . . . . . . . 1.12 Separating the squares . . . . . . . . . . . . . . . . . . . . 1.13 The Dual Principle . . . . . . . . . . . . . . . . . . . . . . 1.14 Substitutions . . . . . . . . . . . . . . . . . . . . . . . . . 1.15 Homogenization and dehomogenization . . . . . . . . . . . 1.16 Unimonotonic sequences . . . . . . . . . . . . . . . . . . . 1.17 Working backwards . . . . . . . . . . . . . . . . . . . . . . 1.18 Mixing variables . . . . . . . . . . . . . . . . . . . . . . . 1.19 Limits in inequalities . . . . . . . . . . . . . . . . . . . . . 1.20 Derivatives in Inequalities . . . . . . . . . . . . . . . . . . 1.21 Convexity . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.22 Jensen’s Inequality . . . . . . . . . . . . . . . . . . . . . . 1.23 The shrinking principle and Karamata’s Inequality . . . . 1.24 Schur’s Inequality . . . . . . . . . . . . . . . . . . . . . . . 1.25 The generalized Means . . . . . . . . . . . . . . . . . . . . 1.26 Inequalities between the symmetric sums . . . . . . . . . . 1.27 The pqr technique . . . . . . . . . . . . . . . . . . . . . . 1.28 The tangent line technique and its extensions . . . . . . . 1.29 Using identities to prove inequalities . . . . . . . . . . . .

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3 3 10 13 17 25 29 41 48 50 84 86 92 101 104 109 111 128 129 133 135 139 141 144 148 149 150 151 159 178

2 Problems

185

3 Solutions

225 1

4 Anexa 1 4.1 Metoda 4.2 Metoda 4.3 Metoda 4.4 Metoda 4.5 Metoda

compar˘ arii . . . . . . . . . major˘ arii ¸si minor˘arii . . . coeficient¸ilor nedeterminat¸i normaliz˘ arii . . . . . . . . omogeniz˘ arii . . . . . . . .

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489 489 499 504 510 515

Chapter 1

Part I 1.1

Squares are positive

Perhaps the first inequality one learns is ”The square of any real number is nonnegative”. It’s a very simple one, but despite its simplicity it is one of the most important inequalities, used frequently by beginners and advanced problem solvers alike. It lies √ at the base of virtually every inequality: for √ 2 a+b 1 √ example, the inequality 2 ≥ ab can be rewritten as 2 ( a − b) ≥ 0. But writing an expression as a square, or as a sum of squares, is usually far from obvious. It requires a certain level of intuition and creativity, but perhaps more importantly, experience. We thus start with some introductory problems. Example 1. If a, b, c are arbitrary real numbers, then a2 +b2 +c2 ≥ ab+bc+ca. Solution: Let us first see what we have here: squares of numbers versus their pairwise products. Next, we ask ourselves: ”Have I ever seen things like these before?”. Well, yes. Namely, we’ve just seen that (x − y)2 = x2 − 2xy + y 2 , where x, y are two (randomly chosen) real numbers. Motivated by this identity, we multiply by two our inequality, and see that it is equivalent with 2(a2 + b2 + c2 ) ≥ 2(ab + bc + ca). We now transfer everything to the left side. In this case, 2a2 + b2 + c2 − 2(ab + bc + ca) = = (a2 − 2ab + b2 ) + (b2 − 2bc + c2 ) + (c2 − 2ca + a2 )   = (a − b)2 + (b − c)2 + (c − a)2 , which is clearly nonnegative, and thus our inequality is proven. Example 2. If a, b are real numbers such that a + b = 2, then a4 + b4 ≥ 2. Solution: Since 2(a4 + b4 ) − (a2 + b2 )2 = a4 − 2a2 b2 + b4 = (a2 − b2 )2 ≥ 0, we (a2 + b2 )2 get that a4 + b4 ≥ . Similarly, 2(a2 + b2 ) − (a + b)2 = a2 − 2ab + b2 = 2 (a + b)2 (a2 + b2 )2 (a − b)2 ≥ 0, and therefore a2 + b2 ≥ = 2.Thus a4 + b4 ≥ ≥ 2 2 2 2 2 = 2. 3

Example 3. Let a, b, c > 0. Prove that a3 + b3 + c3 + ab2 + bc2 + ca2 ≥ 2(a2 b + b2 c + c2 a). Solution: Note that a3 + ab2 − 2a2 b = a(a2 − 2ab + b2 )2 = a(a − b)2 ≥ 0. In this case, we analogously establish that b3 +bc2 −2b2 c ≥ 0 and c3 +ca2 −2c2 a ≥ 0. Summing up all these three inequalities yields our result. Example 4. Let a, b, x, y be arbitrary real numbers. Show that (a2 + b2 )(x2 + y 2 ) ≥ (ax + by)2 . Solution:Transferring the terms to the left side, the inequality rewrites as (a2 + b2 )(x2 + y 2 ) − (ax + by)2 ≥ 0. Denoting the quantity from the left hand side by P , we get = (a2 + b2 )(x2 + y 2 ) − (ax + by)2

P

= a2 x2 + a2 y 2 + b2 x2 + b2 y 2 − (a2 x2 + b2 y 2 + 2abxy) = a2 y 2 + b2 x2 − 2abxy = (ay − bx)2 ≥ 0. Remarks. Notice that this inequality is a particular case of the following: (a21 + a22 + . . . + a2n )(b21 + b22 + . . . + b2n ) ≥ (a1 b1 + a2 b2 + . . . + an bn )2 , where a1 , a2 , . . . , an and b1 , b2 , . . . , bn are arbitrary real numbers. This last one is known in literature as the Cauchy-Schwartz inequality, to which we have dedicated the next chapter. Anyway, before this, let us see some other (. . .)2 ≥ 0 problems. Exercises 1. Show that a2 + b2 + c2 + 3 ≥ 2(a + b + c). Solution: Moving everything to the left side, we get a2 + b2 + c2 + 3 − 2(a + b + c) = (a2 − 2a + 1) + (b2 − 2b + 1) + (c2 − 2c + 1) = (a − 1)2 + (b − 1)2 + (c − 1)2 ≥ 0, which proves our inequality. 2. If x, y are real numbers, show that x2 + y 2 ≥ Solution: This is equivalent to

x2 +y 2 +2xy 2

(x−y)2 2

≥ 0 i.e.

(x+y)2 2

≥ 0.

3. Show that x4 + y 4 + z 2 + 1 ≥ 2x(xy 2 − x + z + 1). Solution: Moving everything to the left side, and denoting by P the quantity from the left hand side, we get P

= x4 + y 4 + z 2 + 1 − 2x(xy 2 − x + z + 1) = (x4 − 2x2 y 2 + y 4 ) + (x2 − 2xz + z 2 ) + (x2 − 2x + 1) = (x2 − y 2 )2 + (x − z)2 + (x − 1)2 ≥ 0. 4

4. Show that for any real x, we have (x − 1)(x − 3)(x − 4)(x − 6) + 10 > 0. Solution: Factoring our expression, we obtain that (x − 1)(x − 3)(x − 4)(x − 6) + 10 = (x2 − 7x + 6)(x2 − 7x + 12) + 10 = (x2 − 7x + 6)2 + 6(x2 − 7x + 6) + 10 = (x2 − 7x + 9)2 + 1 > 0. a 5. If a and b are strictly (nonzero) positive real numbers, prove that √ + b √ √ b √ ≥ a + b. a Solution: Moving everything to the left side, it follows that     √ √ √ √ a b a b √ + √ − a− b = √ − b + √ − a a a b b   1 1 a−b b−a = √ + √ = (a − b) √ − √ a a b b √ √ 2 √ √  a− b a+ b √ = ≥ 0. ab 6. Show that for any positive real number x, we have x5 − x2 − 3x + 5 > 0. Solution: Note that the following chain of equalities holds: x5 − x2 − 3x + 5 = x2 (x3 − 1) − 3(x − 1) + 2   = (x − 1) x2 (x2 + x + 1) − 3 + 2   = (x − 1) (x4 − 1) + (x3 − 1) + (x2 − 1) + 2 = (x − 1)2 (x3 + 2x2 + 3x + 3) + 2. Since this last term is nonnegative, we deduce our inequality. 7. If x, y are two real numbers, prove that x4 + y 4 ≥ x3 y + y 3 x. Solution: Since x2 + xy + y 2 = 41 (x − y)2 + 34 (x + y)2 ≥ 0, we have that x4 + y 4 − x3 y − y 3 x = (x4 − x3 y) + (y 4 − y 3 x) = x3 (x − y) + y 3 (y − x) = (x − y)(x3 − y 3 ) = (x − y)2 (x2 + xy + y 2 ) ≥ 0. 8. Let a, b, c, m, n, p be real numbers (a, b, c 6= 0), such that ap−2bn+cm = 0 and ac = b2 . Show that n2 ≥ mp. Solution: For any two real numbers x and y, we have that (x+y)2 −4xy = x2 − 2xy + y 2 = (x − y)2 ≥ 0, or equivalently, (x + y)2 ≥ 4xy. In this case, 4b2 n2 = (ap + cm)2 ≥ 4ap · cm = 4acpm = 4b2 mp, and thus n2 ≥ mp. 5

9. Show that a2 + b2 + c2 + d2 + e2 ≥ a(b + c + d + e). Solution: Moving everything to the left side, it follows that A = a2 + b2 + c2 + d2 + e2 − a(b + c + d + e)  2   2   2   2  a a a a 2 2 2 2 − ab + b + − ac + c + − ad + d + − ae + e = 4 4 4 4 a 2  a 2  a 2  a 2 = −b + −c + −d + − e ≥ 0. 2 2 2 2 10. If a, b, c are nonnegative real numbers, show that a3 b3 c3 a+b+c + + ≥ . 2 2 2 2 2 2 a + ab + b b + bc + c c + ca + a 3

Solution: Since

a3 2a − b (a − b)2 (a + b) − = ≥ 0, we get a2 + ab + b2 3 3(a2 + ab + b2 )

a3 b3 c3 + + a2 + ab + b2 b2 + bc + c2 c2 + ca + a2

≥ =

2a − b 2b − c 2c − a + + 3 3 3 a+b+c . 3

11. Let a, b, c be three arbitrary real numbers. Prove that p p p √ a2 + ab + b2 + b2 + bc + c2 + c2 + ca + a2 ≥ 3(a + b + c). Solution: Since a2 + ab + b2 = deduce that

1 4 (a

− b)2 + 34 (a + b)2 ≥

3 4 (a

+ b)2 , we

√ p p p a2 + ab + b2 + b2 + bc + c2 + c2 + ca + a2 ≥

3 (|a + b| + |b + c| + |c + a|) . 2

Now, for any real number x we have |x| ≥ x, and therefore √ √ 3 3 (|a + b| + |b + c| + |c + a|) ≥ (a + b + b + c + c + a) 2 √2 = 3(a + b + c), which yields our result. 12. Prove that 19x2 + 54y 2 + 16z 2 + 36xy − 24yz − 16zx ≥ 0. Solution: The inequality follows from the chain of equalities written below: P

= 19x2 + 54y 2 + 16z 2 + 36xy − 24yz − 16zx = (9x2 + 36xy + 36y 2 ) + (18y 2 − 24yz + 8z 2 ) + (8x2 − 16xz + 8z 2 ) + 2x2 = 9(x + 2y)2 + 2(3y − 2z)2 + 8(x − z)2 + 2x2 ≥ 0. 6

13. Let x, y be real numbers such that xy ≥ 1. Show that

x2

1 1 + 2 ≥ +1 y +1

2 . xy + 1 Solution: Note that 1 1 2 + 2 − 2 x + 1 y + 1 xy + 1

   1 1 1 1 + − − x2 + 1 xy + 1 y 2 + 1 xy + 1 x(y − x) y(x − y) + (xy + 1)(x2 + 1) (xy + 1)(y 2 + 1)   x−y y x − xy + 1 y 2 + 1 x2 + 1 (x − y)2 (xy − 1) , (xy + 1)(x2 + 1)(y 2 + 1)

 = = = =

which is clearly nonnegative. This proves the inequality in question. 14. If a, b, c are positive real numbers, prove that

a b c + + ≥ 32 . b+c c+a a+b

Solution: We have P

= = = = =

b c 3 a + + − b + c c +a a + b 2     a 1 b 1 c 1 − + − + − b+c 2 c+a 2 a+b 2 (a − b) + (a − c) (b − c) + (b − a) (c − a) + (c − b) + + 2(b + c) 2(c + a) 2(a + b)       a−b 1 1 b−c 1 1 c−a 1 1 − + − + − 2 b+c c+a 2 c+a a+b 2 a+b b+c (a − b)2 (b − c)2 (c − a)2 + + , 2(b + c)(a + c) 2(a + b)(a + c) 2(a + b)(b + c)

which is clearly nonnegative. 15. Let a, b, c be nonnegative real numbers such that a + b + c = 1. Prove that r r r (b − c)2 (c − a)2 (a − b)2 a+ + b+ + c+ ≤ 2. 4 4 4 Solution: Squaring both sides, the inequality to prove rewrites as s    X  X (c − a)2 (a − b)2 (b − c)2 +2 b+ c+ ≤ 4. a+ 4 4 4 cyc cyc Now, notice that (b − c)2 (1 + a − b − c) = 2a(b − c)2 ≥ 0, which gives us that s   (c − a)2 (a − b)2 (a − b)(a − c) 2 b+ c+ ≤b+c+ . 4 4 2 7

Thus it suffices to show that  X  X (a − b)(a − c) (b − c)2 b+c+ a+ + ≤ 4, 4 2 cyc cyc which since

P

(a − b)(a − c) = a2 + b2 + c2 − ab − bc − ca, reduces to

cyc

3(a + b + c) + a2 + b2 + c2 − ab − bc − ca ≤ 4, or equivalently, a2 + b2 + c2 − ab − bc − ca ≤ 1. This is obviously true, because 1 = (a + b + c)2 = (a2 + b2 + c2 − ab − bc − ca) + 3(ab + bc + ca) ≥ a2 + b2 + c2 − ab − bc − ca. 16. Let x, y be two strictly (nonzero) positive real numbers such that x2 + y 3 ≥ x3 + y 4 . Prove that x3 + y 3 ≤ x2 + y 2 ≤ x + y ≤ 2. Solution: Since y 2 − y 3 − (y 3 − y 4 ) = y 2 (y − 1)2 ≥ 0, we have that x2 − x3 + y 2 − y 3 ≥ x2 − x3 + y 3 − y 4 = (x2 + y 3 ) − (x3 + y 4 ) ≥ 0, and therefore x3 + y 3 ≤ x2 + y 2 . Now, since x − x2 − (x2 − x3 ) = x(x − 1)2 ≥ 0, and y − y 2 − (y 3 − y 4 ) = y(y − 1)2 (y + 1) ≥ 0, we get x2 + y 2 ≤ x + y, and thus the problem reduces to proving that x + y ≤ 2. Now, note that x − (1 − x2 + x3 ) = −(x − 1)2 (x + 1) ≤ 0, y − (1 − y 3 + y 4 ) = −(y − 1)2 (y 2 + y + 1) ≤ 0, and therefore, it follows that x + y ≤ 1 − x2 + x3 + 1 − y 3 + y 4 = 2 + (x3 + y 4 ) − (x2 + y 3 ) ≤ 2.  17. Let a, b, c > 0. Show that ab

 c  a  b − 1 + bc − 1 + ca − 1 ≥ 0. c a b

Solution: Moving everything to the left side, it follows that   c  a  b P = ab − 1 + bc − 1 + ca −1 c a b 2 2 2 ab bc ca + + − ab − bc − ca = c a b  2   2   2  ab bc ca = + ca − 2ab + + ab − 2bc + + bc − 2ca c a b 2 2 2 a(b − c) b(c − a) c(a − b) = + + ≥ 0. c a b 18. Consider three positive real numbers a, b, c such that a > c and b > c. Prove that p p √ c(a − c) + c(b − c) ≤ ab. Solution: According to Example 4, we get hp i2 √ √ p √ √ 2 c(a − c) + c(b − c) = c· a−c+ b−c· c ≤ (c + b − c) (a − c + c) = ab. 8

19. Let a, b, c be nonnegative real numbers, from which at least two are nonzero. Prove that r r r a(b + c) b(c + a) c(a + b) + + ≥ 2. b2 + c2 c2 + a2 a2 + b2 Solution: Notice that for any two nonnegative real numbers x, y, we have √ √ √
2 √ x + y − 2 xy = x − y ≥ 0, and thus x + y ≥ 2 xy. Now, assume that a ≥ b ≥ c; then r r r a(b + c) a(b + c) a ≥ = , 2 2 2 b +c b + bc b r r r b(c + a) b(c + a) b ≥ = , 2 2 2 c +a ca + a a and therefore sr r r r r r r a(b + c) b(c + a) c(a + b) a b a b + + ≥ + ≥2 · = 2. b2 + c2 c2 + a2 a2 + b2 b a b a 20. Let a, b, c be nonnegative real numbers, from which at least two are nonzero. Prove that r r r a(b + c) b(c + a) c(a + b) + + ≥ 2. 2 2 a + bc b + ca c2 + ab Solution: Assume that a ≥ b ≥ c. Since (x+y)2 = x2 +y 2 +2xy ≥ x2 +y 2 , for all x, y ≥ 0, we have r r r a(b + c) c(a + b) a(b + c) c(a + b) + ≥ + 2 , a2 + bc c2 + ab a2 + bc c + ab a(b + c) c(a + b) b2 + ca + 2 ≥ , or equiv2 a + bc c + ab b(c + a) c(a + b) b2 + ca a(b + c) c(a + b) alently, 2 ≥ − 2 . This rewrites as 2 ≥ c + ab b(c + a) a + bc c + ab c(a + b)(a − b)2 , or b(c + a)(a2 + bc) ≥ (a − b)2 (ab + c2 ) which is obvib(c + a)(a2 + bc) ously true since b(c + a) = ab + bc ≥ ab + c2 , and a2 + bc ≥ a2 ≥ (a − b)2 . Hence s r r a(b + c) c(a + b) b2 + ca + ≥ , a2 + bc c2 + ab b(c + a) s r b(c + a) b2 + ca and therefore the problem reduces to + ≥ 2. In b2 + ca b(c + a) √ the proof of Exercise 19, we have showed that x + y ≥ 2 xy for any nonnegative real numbers x, y. Therefore v s s ur r u b(c + a) 2 b(c + a) b + ca b2 + ca t + ≥ 2 · = 2, b2 + ca b(c + a) b2 + ca b(c + a)

and we will now show that

which completes our proof.

9

1.2

Some special inequalities and identities for two numbers

We have already met the arithmetic mean of two numbers: AM =

a+b 2

and the geometric mean of two numbers: √ GM = ab We can also consider the harmonic mean HM =

2 2ab = 1 1 a+b + a b

and the quadratic mean of two numbers r a2 + b2 QM = 2 For positive a, b, these means satisfy the following chain of inequalities: QM ≥ AM ≥ GM ≥ HM. Proof : From the transitivity of ≥ it suffices to prove QM ≥ AM, AM ≥ GM, GM ≥ HM. Let’s take them step by step: a) QM ≥ AM . This is equivalent to a2 + b2 QM ≥ AM or ≥ 2 2

2



a+b 2

2 =

a2 + 2ab + b2 . 4

Multiplying by 4 we must prove 2(a2 +b2 ) ≥ a2 +b2 +2ab or a2 +b2 ≥ 2ab which is just the inequality AM ≥ GM for a2 , b2 . b) AM ≥ GM was proven by us in the previous topic. 1 1 + 1 1 1 c) GM ≥ HM . This is equivalent to ≤ , or to √ ≤ a b GM HM 2 ab 1 1 which is just AM ≥ GM for , . a b [Remark: This chain of inequalities can be generalized in the following form: Let a1 , a2 , . . . , an be positive real numbers, and for every k ∈ R consider the generalized means s mk =

k

ak1 + ak2 + . . . + akn n 10

if k 6= 0 and m0 =

√ n

a1 a2 . . . an

Then mk is increasing as a function in k: that is mα ≥ mβ if α ≥ β Our inequality chain QM ≥ AM ≥ GM ≥ HM is just a special case of this inequality for n − 2 and k = 3, 2, 1, 0. The proof of this inequality requires more advanced methods, and we will return to it in the further sections.] We now move to identities. There is one basic identity for two numbers: the Newton Binomial Theorem: (a + b)n =

n   X n i n−i ab i i=0

where


n i

=

n! (recall that n! = 1 × 2 × . . . × n). i!(n − i)!

Particularly (a + b)2 = a2 + 2ab + b2 (a + b)3 = a3 + 3a2 b + 3ab2 + b3 (a + b)4 = a4 + 4a3 b + 6a2 b2 + 4ab3 + b4 (a − b)2 = a2 − 2ab + b2 (a − b)3 = a3 − 3a2 b + 3ab2 − b3 (a − b)4 = a4 − 4a3 b + 6a2 b2 − 4ab3 + b4 etc. Another well-known identity is an − bn = (a − b)(an−1 + an−2 b + . . . + bn−2 a + bn−1 ), and for n odd, an + bn = (a + b)(an−1 − an−2 b + . . . − abn−2 + bn−1 ). Note: the above identities are often used for b = 1. This is because homogeneous inequalities in two numbers a, b can be reduced to inequalities in just one variable x by letting x = ab (but more on that in the section about homogeneity). Finally, a very simple but sometimes crucial inequality is the triangle inequality |x| + |y| ≥ |x + y|. Exercises 1. Let a > b > 0. Show that 2(a2 − b2 )2 ≤ a2 (a − b)2 + a2 (a − b)2 2

2

2

2

Proof: This is equivalent to 4(a−b)4a(a+b) which is just the ≤ (a−b) +(a−b) 2 2 inequality QM ≥ HM squared for the numbers a + b, a − b. Of course, 11

this inequality can also be shown by opening the brackets and using the AM-GM inequality or turning it into a sum of squares. That’s not surprising since all the inequalities AM ≥ GM, GM ≥ HM, QM ≥ AM were proven using the same method, which is essentially that the square of the difference of two numbers is non-negative. xmn − 1 xn − 1 ≥ . m x Solution: We can extract the common factor xn − 1 to get

2. If m, n are natural numbers and x > 0 then

xmn − 1 xn − 1 − m x

= =

h i 1 n (x − 1) x(xn(m−1) + . . . + 1) − m nx h i 1 (x − 1)(xn−1 + . . . + 1) x(xn(m−1) + . . . + 1) − m . nx

Now as the right parenthesis vanishes for x = 1, extracting x − 1 out of it produces

=

1 n (x − 1)(xn(m−1) − 1 + xn(m−2) − 1 + . . . + 1 − 1) = nx h i 1 (x − 1)2 (xn−1 + . . . + 1) (xn(m−1) + . . . + 1) + (xn(m−2) + . . . + 1) + . . . , nx

and this is clearly positive. 3. For a1 , a2 , a3 , . . . , an ∈ R show that 12 min (ak − ai )2 ≤ 1≤i x + y and (a − b)2 + (b − c)2 + (c − a)2 ≥ x2 + y 2 and so the inequality follows from 2(x + y)(x2 + y 2 ) ≥ (x + y)3 which by cancelling common terms turns to x3 + y 3 ≥ x2 y + xy 2 which is true. 6. Let x, y, z be positive numbers such that x2 + y 2 + z 2 = 1. Determine xy xz yz the minimum value of + + . z y x Solution: We have

xy xz yz x2 y 2 + y 2 z 2 + z 2 x2 + + = . z y x xyz

Now we use the famous inequality (a+b+c)2 ≥ 3(ab+bc+ca) for a = uv, b = vw, c = uw, deducing that (uv + uw + vw)2 ≥ 3(u + v + w)uvw. Applying this for u = x2 , v = y 2 , w = z 2 , we deduce that (x2 y 2 + z 2 x2 + y 2 z 2 )2 ≥ 3x2 y 2 z 2 (x2 + y 2 + z 2 ). Taking square root and keeping in mind that x2 + y 2 + z 2 = 1 we have √ x2 y 2 + y 2 z 2 + z 2 x2 ≥ 3xyz. The value

√

1 3 can be achieved, when x = y = z = √ . 3

7. Let 0 < x < y < z < t such that x + t = y + z. Show that for any natural number n ≥ 2, xn + tn > y n + z n . Solution: Let a = y − x = t − z. We rewrite the conclusion as (x + a)n − xn < (z + a)n − z n , 15

which by Newton Binomial Theorem is equivalent to nxn−1 a +

n(n − 1) n−2 n(n − 1) n−2 x a + . . . + an < nz n−1 a + z a + . . . + an 2 2

which is clearly true from z > x. 8. Let a, b, c be length of the sides of triangle; and let t ≥ 1 be real number. Prove that:  2  a b2 c2 b2 c2 a2 (t + 1) + + ≥ + + + t(a + b + c). c a b c a b Solution: It suffices to prove it for t = 1 since a2 b2 c2 + + ≥ a + b + c. c a b This latter inequality can be proven in many ways. A simple way is to 2 note that a2 −c(2a−b) = (a−c)2 ≥ 0 hence c(2a−c) ≤ a2 so ac ≥ 2a−c 2 2 and similarly ba ≤ 2b − a, cb ≥ 2c − b and we sum these three inequalities to obtain the desired result. 2

2

2

2

2

2

So it suffices to show 2( ac + ba + cb ) ≥ cb + bc + ca + a + b + c. We multiply this latter inequality by abc to turn it into the equivalent form 2(a3 b + b3 c + c3 a) ≥ ab3 + bc3 + ca3 + abc(a + b + c). This can be rewritten as ab3 + bc3 + ca3 − a3 b − b3 c − c3 a ≤ a3 b + b3 c + c3 a − abc(a + b + c) i.e.

(a − b)(b − c)(c − a)(a + b + c) ≤ a3 b + b3 c + c3 a − abc(a + b + c). 2

2

The right-hand side is non-negative since we’ve just shown that ac + ba + c2 b ≥ a+b+c, thus is suffices to consider the case (a−b)(b−c)(c−a) > 0. In this case we can suppose a < b < c (check!) and so the RHS is b(b2 − a2 )(c − a) + a(c − b)(c2 − b2 ) By the AM-GM inequality this is greater than or equal to p 2(c − b) ab(c − b)(b − a)(c + b)(b + a) and so by squaring both sides and canceling reducing the common factors it remains to prove that 4ab(c + b)(b + a) ≥ (c − b)(b − a)(a + b + c)2 Indeed, a ≥ c − b, b ≥ b − a, 4(c + b)(b + a) − (a + b + c)2 = 4c(b + a) − (a + b + c)2 + 4b(b + a) = 4b(b + a) − (a + b − c)2 ≥ 4b(b + a) − a2 > 0 and by multiplying we get the result. 16

1.4

The mathematical induction

The Principle of Mathematical Induction is formulated as follows: Suppose we have to prove an affirmation A(n) depending on n ∈ N. If we can prove it for some k (the basis), and then we can prove that A(n + 1) (the induction step) follows from (the induction assumption) A(n) then A(n) is true for any n. Another variation is to show that A(n+1) follows from A(k), A(k +1), . . . A(n) - this is called ”strong induction” as opposed to the ”weak induction” method above. Of course, strong induction become weak induction one we replace the propositions A(n) with the equivalent set of propositions B(n) that state that A(k), A(k + 1), . . . , A(n) are all true. Consider for example Bernoulli’s Inequality If α ≥ 0 then (1 + α)n ≥ 1 + nα. For n = 1 this is actually an identity (the basis). Now if (1 + α)n ≥ 1 + nα (the induction assumption) then (1+α)n+1 = (1+α)n (1+α) ≥ (1+nα)(1+α) = 1+(n+1)α+α2 ≥ 1+(n+1)α (The induction step). Hence the affirmation is true for all n. We may have noted that the inequality a3 + b3 + c3 ≥ 3abc resembles the AM-GM inequality, generalized for three numbers. Indeed, if we put a3 = x+y+z √ x, b3 = y, c3 = z we transform it to ≥ 3 xyz. 3 A natural question appears: is this inequality true for any number of variables, x1 + x2 + . . . + xn ≥ i.e., is it true that for positive x1 , x2 , . . . , xn we have n √ n x1 x2 . . . xn ? We know it’s true for n = 4, and we know how to deduce it from the twovariables case: q √ √ √ √ √ 4 a + b + c + d ≥ 2 ab + 2 cd ≥ 4 ab cd = 4 abcd. We can prove the inequality for n = 2m the very same way using induction: x1 + . . . + x2m = (x1 + . . . + x2m−1 ) + (x2m−1 +1 + . . . + x2m ) ≥ ≥ 2m−1

√

2m−1

x1 x2 . . . x2m−1 + 2m−1

√

2m−1

x2m−1 +1 . . . x2m .

Applying the AM-GM Inequality for two numbers, we deduce this quantity is √ at greater or equal than 2m 2m x1 x2 . . . x2m . So, we can prove the inequality for any power of 2. To pass to the general case, pick up such a m with 2m > n and √ consider the system b1 , b2 , . . . , b2m with bi = ai if i ≤ n and bi = n a1 a2 . . . an √ otherwise. The geometric mean of the system is still n a1 a2 . . . an and so we deduce √ b1 + b2 + . . . + b2m ≥ 2m n a1 a2 . . . an so

√ √ a1 + a2 + . . . + an + (2m − n) n a1 a2 . . . an ≥ 2m n a1 a2 . . . an 17

so

√ a1 + a2 + . . . + an ≥ n a1 a2 . . . an . n

This inequality is called the general Arithmetic Mean-Geometric Mean (AMGM) Inequality. The method we used is called Cauchy Induction, which is based on proving the inequality for an infinite class of numbers (in our case powers of two) an then showing that if m < n and for n the proposition is true, then for m it also must be true. In a simpler way, if for n the assertion holds, then it holds for n − 1. The mathematical induction is a tool that can be used in inequalities with many variables. Sometimes the induction step is quite clear, but it can not be proven, In this case it’s good to sharpen the induction assumption. Look at the following problem: 1 3 2n − 1 1 · · ... · −2. an an−1 =

so the induction step is fulfilled. For n = 1 the condition clearly holds (with equality). 7. Show that for any n ≥ 4, x1 , x2 , . . . , xn ∈ R+ we have (x1 + x2 + . . . + xn )2 ≥ 4(x1 x2 + x2 x3 + . . . + xn−1 xn + xn x1 ). Solution: For n = 4 this is (x1 + x3 − x2 − x4 )2 ≥ 0. Now let n ≥ 5 and denote f (x1 , x2 , . . . , xn ) = (x1 + x2 + . . . + xn )2 − 4(x1 x2 + . . . + xn−1 xn + xn x1 ). Then f (x1 + x2 , x3 , x4 , . . . , xn ) = f (x1 , x2 , . . . , xn ) + x3 x1 + x2 xn − x1 x2 . Now if x3 ≥ x2 (which we may assume as the inequality is cyclic) we get f (x1 + x2 , x3 , x4 , . . . , xn ) ≥ f (x1 , x2 , . . . , xn ) and so we are done by induction. 21

8. Let a1 , a2 , . . . , an ∈ (0, 1) with a1 a2 . . . an = An . Show that 1 n 1 + ... + ≤ . 1 + a1 1 + an 1+A Solution: If n = 2 the inequality is equivalent to

1 1 + ≤ 2 1+x 1 + y2

2 or by clearing denominators to (2+x2 +y 2 )(1+xy)−2(1+x2 )(1+ 1 + xy an an+1 . y 2 ) ≤ 0 or (x − y)2 (xy − 1) ≤ 0, which is true. Now set b = A n Then a1 a2 . . . an−1 b = A so by induction assumption we have 1 1 1 n + ... + + ≤ 1 + a1 1 + an−1 1 + b 1+A and we are left to prove that 1 1 1 1 1 1 1 1 +. . .+ + + ≥ +. . .+ + + , 1 + a1 1 + an−1 1 + b 1 + A 1 + a1 1 + an−1 1 + an 1 + an+1 1 1 A 1 + ≤ + . 1 + an 1 + an+1 1 + A A + an an+1 By clearing denominators this reduce to (an −A)(an+1 −A)(1−an an+1 ) ≤ 0. As an , an+1 < 1 this inequality will be fulfilled iff one of an , an+1 is greater than A ad the other is smaller (or equal). But we can clearly assume this as the inequality is cyclic and we can’t have all ai greater than A or all ai smaller than A.

or

1 1 1 > + +. . .. nn S(n + 1) S(n + 2) Solution: The quickest idea that comes to our mind is to use ”induction”, 1 1 1 < n− . (Actually this is not induction i.e. prove that S(n) n (n + 1)n+1 but telescoping sums, however it resembles induction). Surprisingly it works! (on difficult exams, it won’t). We need to prove that

9. Let S(n) = 11 +22 +. . .+nn . Show that

(n + 1)n+1 − nn 1 > 1 . n n+1 2 n × (n + 1) 1 + 2 + . . . + nn (n + 1)n+1 − nn 1 + 22 + . . . + nn · > 1. (n + 1)n+1 nn     n+1 n n + 1 n+1 n n+1 (n + 1)n+1 nn 22

   1 1 1 1− 1+ + + ... . ne ne n(n − 1)e2

Since    1 1 1 1 1 1 1− 1+ = 1− 2 2 + − 2 + 2 2 ne ne n(n − 1)e n e n(n − 1)e n (n − 1)e3 1 1 = 1+ 2 − > 1. n (n − 1)e2 n2 (n − 1)e3 And we are done. 10. If a1 ≥ a2 ≥ . . . ≥ an ≥ 0 then a21 − a22 + a23 ± . . . ± a2n ≥ (a1 − a2 + a3 ± . . . ± an )2 Solution: Again we want to use induction by deleting some of the variables. Since we don’t really know the sign of an , it’s probably better to start from a1 . However deleting it produces an expression of opposite sign and applying induction to it would give an inequality also of opposite sign, not helpful to us. So we should delete the first terms, a1 , a2 . Indeed, by the induction assumption for n − 2 we can deduce that a21 − a22 + a23 ± . . . ± a2n ≥ a21 − a22 + (a3 − a4 + a5 ± . . . ± an )2 . If we denote by t = a2 − a3 ± . . . ± an then we must prove that a21 − a22 + t2 ≥ (a1 − a2 + a3 )2 . It resembles our inequality for n = 3 and it is indeed our inequality for n = 3 because t = (a3 −a4 )+. . . = a3 −(a4 −a5 )−. . . so 0 ≤ t ≤ a3 ≤ a2 . Therefore it suffices to prove the basis, which consists of two separate cases: n = 2 and n = 3 (this is because we did an unusual induction step of 2 - if you want, we do induction on [ n2 ]). For n = 2 we must prove (a1 − a2 )(a1 + a2 ) ≥ (a1 − a2 )2 which is clear. For n = 3 we must prove a21 − a22 + a23 − (a1 − a2 + a3 )2 ≥ 0 which is equivalent to 2(a1 − a2 )(a2 − a3 ) ≥ 0, true! xn n + , for n ≥ 1. Prove that the sequence (xn ) is n xn increasing and that [x2n ] = n for n ≥ 4

11. Let x1 = 1, xn+1 =

Solution: To perform the induction step, we will prove a stronger hy√ n pothesis, that for n ≥ 3 we have n ≤ xn ≤ √ . n−1 x n We may note that f (x) = + is decreasing for x ≤ n so it suffices to n x  √ √ n+1 n prove that f ( n) ≤ √ and f √ ≥ n + 1. We have n n−1   √ √ n 1 n xn+1 ≥ f √ = n−1+ √ =√ > n + 1. n−1 n−1 n−1 n−1 Replacing n + 1 by n we get xn ≥ √ . Therefore n−2   n−1 n2 (n − 2) + (n − 1)2 √ xn+1 ≤ f √ = . n−2 n(n − 1) n − 2 23

√ n2 (n − 2) + (n − 1)2 √ < n + 2 which is n(n − 1) n − 2 3 2 n − n − 2n + 1 p 2 equivalent to < n − 4 which after squaring is equivn2 − n alent to 2n2 (n − 3) + 4n − 1 ≥ 0, true for n ≥ 4. We thus deduce n < x2n < n + 1 so [x2n ] = n. So we are left to prove that

12. Let n > 2. Find the least constant k such that for any a1 , . . . , an > 0 with product 1 we have (a21

a1 a2 a2 a3 an a1 + 2 +...+ 2 ≤ k. 2 2 + a2 )(a2 + a1 ) (a2 + a3 )(a3 + a2 ) (an + a1 )(a21 + an )

Solution: We prove the answer is n − 2. If we put a1 = . . . = an−1 = x tending to 0 we get close to n − 2. Now to prove another part: Firstly (a21 + a2 )(a22 + a1 ) ≥ (a21 + a1 )(a22 + a2 ) (this is easily deduced by opening the brackets). So we replace to prove n X i=1

1 ≤ n − 2. (ai + 1)(ai+1 + 1)

For n = 3 it’s simple replacing a1 =

<

x y z , a2 = , a3 = : y z x

yz zx xy + + < (x + y)(x + z) (y + z)(y + z) (z + x)(z + y) xy yz xz + + = 1. xy + yz + zx xy + yz + zx xy + yz + zx

For n > 3 we prove by induction: If a1 a2 ≤ 1 and a2 a3 ≥ 1 (we may assume this since the inequality is cyclic) we can replace a2 and a3 with a2 a3 and prove the value decreases by at most 1 so we can use induction step. Indeed, the value decreases by 1 1 1 + + − (a1 + 1)(a2 + 1) (a2 + 1)(a3 + 1) (a3 + 1)(a4 + 1) − =

1 1 − = (a1 + 1)(a2 a3 + 1) (a2 a3 + 1)(a4 + 1)

a2 (a3 − 1) a3 (a2 − 1) a2 + a3 + a2 a3 + +1− . (a1 + 1)(a2 + 1)(a2 a3 + 1) (a4 + 1)(a3 + 1)(a2 a3 + 1) (a2 + 1)(a3 + 1) + 1

Now if a2 ≤ 1, a3 ≥ 1 then this quantity is at most (a2 + 1)(a3 + 1) − 1 a2 a3 − ≤ a2 a3 + 1 (a2 + 1)(a3 + 1) 1 1 ≤ 1+1− −1+ ≤ 1, a2 a3 + 1 a2 a3 + a2 + a3 + 1 1+

24

and analogously if a2 ≥ 1, a3 ≤ 1. Finally if a2 , a3 > 1 then the quantity is at most a2 a3 a2 a3 a2 a3 + a2 + a3 + − = 2(a2 a3 + 1) 2(a2 a3 + 1) (a2 + 1)(a3 ) + 1 a2 a3 a2 a3 + a2 + a3 = 1+ − ≤1 a2 a3 + 1 a2 a3 + a2 + a3 + 1 1+

again.

1.5

The AM-GM inequality

The inequality proven in the previous chapter: √ a1 + a2 + . . . + an ≥ n a1 a2 . . . an n is the most important of all the inequalities used, and has enormously many applications. Here’s one of them: If ai , bi > 0 then s n Q n ai n X i=1 ai ≥n· s ai + bi n Q i=1 n (ai + bi ) i=1

and also

s n

n X

bi ≥n· s ai + bi n Q i=1 n

n Q

bi

i=1

(ai + bi )

i=1

Summing these two inequalities and using the fact that we deduce

s n

n≥n·

n Q

s ai +

n

i=1

s n

n Q

bi

i=1

n Q

ai bi + =1 ai + bi ai + bi

or

(ai + bi )

i=1

v u n uY n t (ai + bi ) ≥

v u n uY n t ai +

i=1

i=1

v u n uY n t bi i=1

This is called Huygens Inequality. If we have the sequence a1 , a1 , . . . , a1 , a2 , . . . , a2 , . . . , an , . . . , an , where ai repeats pi times then according to AM-GM we have q p1 a1 + p2 a2 + . . . + pn an ≥ (p1 + p2 + . . . + pn ) p1 +p2 +...+pn ap11 . . . apnn . 25

If we denote αi =

pi , we obtain the following generalization p1 + p2 + . . . + pn

of AM-GM: If α1 , . . . , αn are positive numbers that sum to 1 then for any positive x1 , x2 , . . . , xn we have α1 x1 + . . . + αn xn ≥ xα1 1 . . . xαnn This is called Weighted AM-GM. Of course, in this setting αi need to be rational - but in fact this condition is unnecessary. Exercises  1. Show that for any a1 , a2 , . . . , an > 0 we have (a1 +. . .+an )

1 1 + ... + a1 an

 ≥

n2 . √ Solution: We have a1 + a2 + . . . + an ≥ n n a1 a2 . . . an and r 1 1 1 1 + + ... + ≥nn a1 a2 an a1 a2 . . . an and by multiplying these two relations we get the result. 2. Let a1 , a2 , . . . , an > 0 with a1 a2 . . . an = 1. Let sk = ak1 + ak2 + . . . + akn . a) Show that if k ≥ l ≥ 0 then sk ≥ sl ; b) Find all k > 0 for which sk ≥ s−1 . k−l k Solution: a) We have aki + ≥ ali by weighted AM-GM. The l l inequality now follows by summing, as sk ≥ n = s0 by AM-GM b) We notice that sn−1 ≥ s−1 because + . . . + xnn−1 ≥ (n − 1)x2 x3 . . . xn = + xn−1 xn−1 3 2

n−1 x1

and the inequality follows by summing. So using a) we get sk ≥ s−1 for any k ≥ n − 1. If k < n − 1 then taking x1 = x2 = . . . = xn−1 = 1 1 n−1 x, xn = n−1 we have sk = (n − 1)xk + (n−1)k , s−1 = + xn−1 and x x x s−1 ≥ sk for all sufficiently big x. So k ∈ [n − 1, +∞). p √ 3. Show that n (n + 1)! > n n! + 1 Solution: By Huygens Inequality p p √ √ n n n (n + 1)! = n (1 + 1)(2 + 1)(3 + 1) . . . (n + 1) ≥ 1 · 2 . . . · n+1 = n!+1. The equality cannot take place.   n+1 1 1 n < 1+ ; 4. a) Show that 1 + n n+1    n+2 1 n+1 1 b) Show that 1 + > 1+ . n n+1 

26

    1 1 + 1+ + Solution: a) By AM-GM we have n + 2 = 1 + 1 + n s n     1 1 1 n n+1 . . .+ 1 + (1 + meets n times) is bigger than n + 1 1+ , n n n which rising to power n + 1 gives us exactly the conclusion. n n b) By AM-GM we have n + 1 = 1 + + ... + ≥ (n + n + 1 n + 1 s n+1 n n+2 which after rising to power n + 2 yields the con2) n+1 clusion. 5. 2(a4 + b4 ) + 17 > 16ab. Solution: 2(a4 + b4 ) + 17 ≥ 4a2 b2 + 17 > 4a2 b2 + 16 ≥ 16ab by AM-GM. √ √ 6. n n a1 a2 . . . an − (n − 1) n−1 a1 a2 . . . an−1 ≤ an , for ai > 0. Solution: The conclusion from AM-GM applied to the n numbers an , √ √ √ n−1 a a . . . a 1 2 n−1 , n−1 a1 a2 . . . an−1 , . . ., n−1 a1 a2 . . . an−1 .   √
1 1 1 7. Show that n 1 − √ > n n n + 1 − 1 , if + 1 > 1 + + ... + n 2 n n n ∈ N∗ . n−1 1 1 2 ≥n√ Solution: By AM-GM we have 1 + + + . . . + , or n 2 3 n n   1 1 1 1 n− + + ... + ≥n√ , n 2 3 n n 1 1 1 1 thus n(1 − √ ) ≥ + + . . . + , from where left inequality follows. n 2 3 n n √ 3 4 n+1 For the right inequality, we see that 2 + + + . . . + ≥ n n n + 1, 2 3 n   √ 1 1 so n + 1 + + . . . + ≥ n n n + 1, and the rest is clear. 2 n 8. Let x1 , x2 , . . . , x2n be positive reals that sum to 1. Let S = x21 x22 . . . x2n + x22 x23 . . . x2n+1 + . . . + x22n x21 . . . x2n−1 . 1 . n2n Solution: We can notice that Show that S <

S ≤ (x21 + x2n+1 ) . . . (x2n + x22n ) ≤ (x1 + xn+1 )2 . . . (xn + x2n )2   x1 + xn+1 + . . . + xn + x2n 2 1 = 2n . ≤ n n n The inequality cannot hold in all cases. r r r a b c 9. If a, b, c > 0 then + + > 2. b+c a+c a+b 27

Solution: This time wershall use r the inequality between the geometric a a 2 2a and harmonic means: = ≥ , = b + c b+c (b + c) · 1 a+b+c +1 a and the inequality now follows by summation. 10. For a, b, c > 0 we have b c 3 a p +p +p ≤ . 2 (a + b)(a + c) (b + a)(b + c) (c + a)(c + b) Solution: By AM-GM we have r   a a a a 1 a p = ≤ + a+ba+c 2 a+b a+c (a + b)(a + c) and summing with the analogous results we get the conclusion. 11. For x, y, z are positive numbers such that x + x2 + y + y 2 + z + z 2 = 6, find the maximum of x2 y + y 2 z + z 2 x Solution: We see that x3 + y 3 + z 3 + xy 2 + yz 2 + zx2 = (x3 + xy 2 ) + (y 3 + yz 2 ) + (z 3 + zx2 ) ≥ 2(x2 y + y 2 z + z 2 x). 2

2

2

≤ 3. The This means x2 y + y 2 z + z 2 x ≤ (x+y+z)(x3 +y +z ) = a(6−a) 3 equality can hold for x = y = z = 1. So the maximum is 3. 12. If a1 , a2 , . . . , an > 0 then 1 2 n + + ... + 0 then √ √ √ a1 + a1 a2 + 3 a1 a2 a3 + . . . + n a1 a2 . . . an < 3(a1 + a2 + . . . + an ). Solution: As in the above exercise, we assign ”weights” to ai . We have √ k

1 p 1 k (a1 +2a2 +. . .+kak ). a1 a2 . . . ak = √ a1 × 2a2 × . . . × kak ≤ √ k k k! k k!

After summation, the term alongside am will be " # 1 1 1 √ p + m+1 . + ... + √ m m n m!m n!n (m + 1)!(m + 1) Now we prove that "

√ k

k! ≥

k+1 which will imply 3

# 1 p + ... + √ ≤ m n m+1 n!n (m + 1)!(m + 1)   3 3 3 + + ... + ≤ m m(m + 1) (m + 1)(m + 2) n(n + 1)   3 3 3 3 3 3 = m − + − + ... + − m m+1 m+1 m+2 n n+1   3 3 3m = m − =3− 0, we have ax2 + bx + c ≥ 0 for any x if and only if b2 ≤ 4ac. b) For a < 0, we have ax2 + bx + c ≤ 0 for any x if and only if b2 ≤ 4ac. 29

We can also make the inequalities strict: a) For a > 0, we have ax2 + bx + c > 0 for any x if and only if b2 < 4ac. b) For a < 0, we have ax2 + bx + c < 0 for any x if and only if b2 < 4ac. This principle can be used for solving many problems or deducing theorems. n X Consider the quadratic trinomial f (x) = (ai x + bi )2 . It is always positive, i=1

and hence must have non-positive discriminant. However, by opening the brackets we see that ! n n n X X X 2 2 f (x) = ai x + 2 ai bi x + b2i i=1

i=1

i=1

and so the discriminant is 4

n X

!2 ai bi

−4

i=1

n X

! a2i

i=1

n X

! b2i

.

i=1

Therefore we have the following inequality (Cauchy-Buniakovsky-Schwartz): For any reals a1 , a2 , . . . , an , b1 , b2 , . . . , bn we have ! n ! !2 n n X X X a2i b2i ≥ ai bi . i=1

i=1

i=1

If we take the equation f (x) = (a1 x − b1 )2 − (a2 x − b2 )2 − . . . − (an x − bn )2 then does takes non-positive values. Therefore if the leading coefficient is positive, it must have roots and so the discriminant must be positive, which leads to Aczel’s Inequality: If a21 > a22 + . . . + a2n then (a1 b1 − a2 b2 − . . . − an bn )2 ≥ (a21 − a22 − . . . − a2n )(b21 − b22 − . . . − b2n ). If the dominant coefficient of the trinomial f (x) = ax2 + bx + c is positive,   b 2 D + 2 , it has its minimal value at then as f (x) can be written as a x + 2a a b x = − , called the vertex of the trinomial (or parabola). We can also see 2a b that f is increasing for x > − and decreasing otherwise. It can be therefore 2a deduced that a quadratic trinomial with positive dominant coefficient attains its maximum at the extremities of an interval. For a negative, all properties above change sign. The last principle deserves more a more detailed discussion. Let us firstly consider the case a > 0, then for all x ∈ [t1 , t2 ] , we have f (x) − f (t1 ) = a(x2 − t21 ) + b(x − t1 ) = (x − t1 )(ax + at1 + b), f (x) − f (t2 ) = a(x2 − t22 ) + b(x − t2 ) = (x − t2 )(ax + at2 + b). From this, we have that: if ax + at1 + b ≤ 0 then f (x) ≤ f (t1 ), and in the case ax + at1 + b ≥ 0, we have ax + at2 + b ≥ ax + at1 + b ≥ 0, and we get that 30

f (x) ≤ f (t2 ). Or in the other words, in any cases, the following inequality is valid: f (x) ≤ max {f (t1 ), f (t2 )} ∀x ∈ [t1 , t2 ] . In the case a < 0, we proceed by the same way as the above for g(x) = −f (x) = −ax2 − bx − c, and we also have that g(x) ≤ max {f (t1 ), f (t2 )} ∀x ∈ [t1 , t2 ] , or −f (x) ≤ max {−f (t1 ), −f (t2 )} ∀x ∈ [t1 , t2 ] , which is equivalent to f (x) ≥ min {f (t1 ), f (t2 )} ∀x ∈ [t1 , t2 ] . Finally, let us consider the case a = 0, then if b ≥ 0, it is trivial that bx1 + c ≤ bx + c ≤ bx2 + c, or f (t1 ) ≤ f (x) ≤ f (t2 ). And if b ≤ 0, it is also trivial that bt2 + c ≤ bx + c ≤ bt1 + c, or f (t2 ) ≤ f (x) ≤ f (t1 ).Therefore, in any cases, we have min {f (t1 ), f (t2 )} ≤ f (x) ≤ max {f (t1 ), f (t2 )} ∀x ∈ [t1 , t2 ] . All these results can be summarized in the following proposition: Proposition. Consider the following polynomial f (x) = ax2 + bx + c with a, b, c ∈ R and x ∈ [t1 , t2 ] , then 1. If a > 0, then f (x) ≤ max {f (t1 ), f (t2 )} ∀x ∈ [t1 , t2 ] . 2. If a < 0, then f (x) ≥ min {f (t1 ), f (t2 )} ∀x ∈ [t1 , t2 ] . 3. If a = 0, then min {f (x1 ), f (x2 )} ≤ f (x) ≤ max {f (x1 ), f (x2 )} ∀x ∈ [t1 , t2 ] . This theorem provides us a very interesting way to prove inequalities. For example, we need to prove that f (x) = ax2 + bx + c ≤ m for all x ∈ [t1 , t2 ] and we know that a ≥ 0, then from the above theorem, it suffices to prove that max {f (t1 ), f (t2 )} ≤ m. It means that we only need to consider the original inequality in the cases x = t1 and x = t2 which is not hard to prove. For more clearly, see the following examples

Example 1. Let a, b, c ∈ [1, 2] , show that a2 + b2 + c2 ≤ 3abc. Solution: We need to prove that P (a, b, c) = a2 − 3abc + b2 + c2 ≤ 0. This is a quadratic polynomial in term of a with the highest coefficient it positive, therefore, from the above theorem, we only have to prove that max {P (1, b, c), P (2, b, c)} ≤ 0. Now, we see that P (1, b, c), P (2, b, c) are also quadratic polynomials in terms of b, hence it suffices to prove the inequality max{max {P (1, 1, c), P (1, 2, c)} , max {P (2, 1, c), P (2, 2, c)}} 0, which is equivalent to max {P (1, 1, c), P (1, 2, c), P (2, 1, c), P (2, 2, c)} ≤ 0. Again, we see that P (1, 1, c), P (1, 2, c), P (2, 1, c), P (2, 2, c) are also quadratic polynomials in terms of c, hence it suffices to prove the inequality max{P (1, 1, 1), P (1, 1, 2), P (1, 2, 1), P (1, 2, 2), P (2, 1, 1), P (2, 1, 2), P (2, 2, 1), P (2, 2, 2)} ≤ 0. But our inequality is symmetric, therefore we only need to consider the following cases i) If a = b = c = 1, then a2 + b2 + c2 − 3abc = 0. ii) If a = b = 1, c = 2, then a2 + b2 + c2 − 3abc = 0. iii) If a = 1, b = c = 2, then a2 + b2 + c2 − 3abc = −3 < 0. iv) If a = b = c = 2, then a2 + b2 + c2 − 3abc = −12 < 0. Our proof is complete. Example 2. If a, b, c ≥ 0 and a+b+c = 1. Prove that 1−4(ab+bc+ca)+9abc ≥ 0. 31

(a + b)2 (1 − c)2 = . And our in4 4 equality becomes (9c − 4)ab + 1 − 4c(a + b) ≥ 0, or equivalently, f (x) = (9c − 4)x + 1 − 4c + 4c2 ≥ 0. This is a quadratic polynomial with zero quadraticcoefficient,  from the above theorem, it suffices to prove  therefore (1 − c)2 ≥ 0. But it is true since f (0) = 1 − 4c + 4c2 = that min f (0), f 4 (1 − 2c)2 ≥ 0, and   (1 − c)2 (1 − c)2 1 f = (9c − 4) · + 1 − 4c + 4c2 = c(1 − 3c)2 ≥ 0. 4 4 4 Solution: Put x = ab, then 0 ≤ x ≤

Remark. This is Schur’s inequality for third degree. In the Example 2, if we put P (a, b, c) = 1 − 4(ab  + bc + ca) then we  + 9abc,2  (a + b) ≥ 0. have showed that we only need to prove that min f (0), f 4 But it is easy to see that     (a + b)2 a+b a+b f (0) = P (a + b, 0, c), f =P , ,c . 4 2 2    a+b a+b , ,c ≥ Therefore, our statement becomes min P (a + b, 0, c), P 2 2 0. From this, we can obtain the following interesting idea: in order to prove the original inequality, it suffices to consider the cases there is one variable equal to 0 or there are two variables equal. This is a very helpful idea in solving inequalities. Example 3. Let a, b, c, d be nonnegative real numbers such that a+b+c+d = 4. Show that a2 b2 c2 + b2 c2 d2 + c2 d2 a2 + d2 a2 b2 + abc + bcd + cda + dab ≤ 8. Solution: Put x = ab, then 0 ≤ x ≤

(a + b)2 , our inequality becomes 4

f (x) = (c2 + d2 )x2 + (c + d − 2c2 d2 )x + cd(a + b) + (a + b)2 c2 d2 − 8 ≤ 0, This is a quadratic polynomial of x with  the highest  coefficient  is nonnegative, (a + b)2 therefore, it suffices to prove that max f (0), f ≤ 0. It shows 4 that it suffices to consider the cases a = b or ab = 0. Similarly, we also obtain that it suffices to consider the cases c = d or cd = 0. Combining both these statements, we obtain that, in order to prove the original inequality, it enough to prove it in the following cases Case 1. b = 0, d = 0, then the inequality is trivial since a2 b2 c2 + b2 c2 d2 + c2 d2 a2 + d2 a2 b2 + abc + bcd + cda + dab = 0. Case 2. a = b, d = 0, then c = 4 − 2a ≥ 0, and our inequality becomes a4 (4 − 2a)2 + a2 (4 − 2a) ≤ 8, which is true since by AM-GM inequality, we 32

 a + a + 4 − 2a 3 64 have ≤ = , hence a4 (4−2a)2 +a2 (4−2a)−8 ≤ 3 27 8 642 64 + −8=− < 0. 272 27 729 Case 3. a = b, c = d, then c = 2 − a ≥ 0, and our inequality becomes 2a4 (2 − a)2 + 2(2 − a)4 a2 + 2a2 (2 − a) + 2(2 − a)2 a ≤ 8, or equivalent to a6 −6a5 +14a4 −16a3 +7a2 +2a−2 ≤ 0, which is (a2 −2a−1)(a2 −2a+2)(a−1)2 ≤ 0. The last inequality is valid since a ≤ 2. a2 (4−2a)



The property of always attaining the maximal (respectively minimal) value at the extremities of the interval is not limited to quadratic polynomials - it is characteristic for convex (respectively) concave functions, but we will deal with these concepts later. Exercises 1. If x, y, z ∈ R then 3(x2 y 2 + 1)(z 2 − y 2 + 1) ≥ 2(x2 y 2 z 2 + xyz + 1) Solution: Grouping by the powers of z we get a quadratic trinomial in z with positive leading coefficient: (x2 y 2 + 3)z 2 − (3x2 y 2 + 2xy + 3)z + 3x2 y 2 + 1 ≥ 0. Its discriminant can be computed to be −3(xy − 1)4 ≤ 0 and from here we deduce the result. 2. If x, y, z ∈ [0, 1] then x2 + y 2 + z 2 ≤ x2 y + y 2 z + z 2 x + 1. Solution: Let f (t) = t2 (1 − y) − z 2 t + y 2 + z 2 − y 2 z − 1. We must prove that f (x) ≤ 0. If y = 1 then f (t) = −z 2 t+z 2 +1−z −1 = z 2 (1−t)−z ≤ z 2 − z ≤ 9. If y < 1 then f is a quadratic in t with positive dominant coefficient and attains its maximum either at t = 0 or at t = 1. But f (1) = 1 − y − z 2 + y 2 + z 2 − y 2 z − 1 = y 2 (1 − z) − y ≥ y 2 − y ≤ 0 and f (0) = y 2 + z 2 − y 2 z − 1 = (1 − z)(y 2 − 1) ≤ 0. 3. If a < b < c < d then (a + b + c + d)2 > 8(ac + bd) Solution: This time, we can try two different approaches. The first is to look at (a + b + c + d)2 − 8(ac + bd) as at a quadratic expression in, say, a. It is written as a2 − 2(3c − b − d)a + (b + c + d)2 − 8bd. It’s discriminant is D = 32(c − b)(c − d) < 0 so the expression is always positive (moreover we could restrict the condition to just b < c < d or just c between b and d). The second is to write (a + b + c + d)2 − 8(ac + bd) as D = B 2 − 4AC. It’s convenient to set B = a + b + c + d, A = 2, C = ac + bd. Then the quadratic function f (x) = Ax2 + Bx + C = (x + a)(x + c) + (x + b)(x + d) satisfies f (−a) = (b − a)(d − a) > 0 and f (−b) = (a − b)(c − b) < 0 so f must have a root and hence its discriminant is positive. 3 4. Show that a2 + b2 + c2 − ab − bc − ca ≥ (a − b)2 . 4 Solution: Regarding this as a quadratic trinomial in c, we write it as   (a + b)2 a+b 2 2 c − (a + b)c + ≥ 0 or c − ≥ 0. 4 2 33

5. Let 0 < a < b and xi ∈ [a, b]. Show that  (x1 + x2 + . . . + xn )

1 1 1 + + ... + x1 x2 xn

 ≤

(a + b)2 2 n . 4ab

Solution: If xi ∈ [a, b] then (xi − a)(xi − b) ≤ 0 so x2i + ab ≤ (a + b)xi or n n X X 1 ab ≤ a + b. We conclude that ≤ (a + b)n and we xi + ab xi + xi x i=1 i=1 i use AM-GM to conclude the proof. 6. If ai ∈ [a, A], bi ∈ [b, B] where 0 < a < A, 0 < b < B then (a21 + . . . + a2n )(b21 + . . . + b2n ) 1 ≤ 2 (a1 b1 + . . . + an bn ) 4

r

AB + ab

r

ab AB

! .

Solution: As this inequality has a form similar to the Cauchy-Schwartz, but has a inverse sign, we might try to find a quadratic function with a zero and therefore positive discriminant. We can take ! ! r r n X AB ab f (x) = a2i x2 − + x + b2i ab AB i=1 ! ! r r n X AB ab b b i i = a2i x − x− . ai ab ai AB i=1

r r ab bi AB bi ≤t≤ , so To have f (t) ≤ 0 it suffices to find a t s.t. ai AB ai ab " r # r bi ab bi AB the intervals ; must have non-empty intersection, or ai AB ai ab bi ai

r

bj ab ≤ AB aj

for any i, j. This is equivalent to

r

AB ab

aj bi AB ≤ , and is true from the ai bj ab

condition. 7. If a1 , a2 , . . . , an ∈ [1, n − 1] then n(a1 + a2 + . . . + an )2 ≥ 4(n − 1)(a21 + a22 + . . . + a2n ). Solution: We have (ai − 1) [ai − (n − 1)] ≤ 0 so a2i + (n − 1) ≤ nai . Thus (a21 + a22 + . . . + a2n ) + n(n − 1) ≤ n(a1 + a2 + . . . + an ). Thus 4(n − 1)(a21 + a22 + . . . + a2n ) ≤ 4n(n − 1)(a1 + a2 + . . . + an ) − 4n(n − 1)2 . But 4n(n − 1)(a1 + a2 + . . . + an ) − 4n(n − 1)2 ≤ n(a1 + a2 + . . . + an )2 as this is equivalent to n [a1 + a2 + . . . + an − 2(n − 1)]2 ≥ 0. 34

8. Show that a2 + b2 + c2 ≥

p 3(a3 b + b3 c + c3 a), where a, b, c ≥ 0.

Solution: If all of a, b, c are equal, equality holds. If not, since the inequality is cyclic, we may assume that b − a > 0. As the inequality is obvious we can set b − a = 1. Let x = c − a. Then f (a) = (a2 + b2 + c2 )2 − 3(a3 b + b3 c + c3 a) can be written as (x2 − x + 1)a2 + (x3 − 5x2 + 4x + 1)a + x4 − 3x3 + 2x2 + 1. Its discriminant can be evaluated as −3(x3 −x2 −2x−1)2 so is non-positive and thus f is always non-negative, QED. 9. Let a, b, c ∈ [0, 1] . Determine the greatest value of P = a + b + c − ab − bc − ca. Solution: We have that P (a, b, c) = a + b + c − ab − bc − ca is a quadratic polynomial of a with the highest coefficient is 0, therefore, to find the maximum of P , we only need to consider the cases a = 0 or a = 1. Similarly, P is also a quadratic polynomial of b and c, therefore it suffices to consider the cases b = 0 or b = 1, and c = 0 or c = 1. Combining all cases with notice that the expression P is symmetric, we see that it suffices to consider the following cases Case 1. If a = b = c = 0, then P = 0. Case 2. If a = b = 0, c = 1, then P = 1. Case 3. If a = 0, b = c = 1, then P = 1. Case 4. If a = b = c = 1, then P = 0. Therefore, we obtain that max P = 1.with equality holds when a = b = 0, c = 1 or a = 0, b = c = 1 and their cyclic permutations.   1 a b c 7 10. Let a, b, c ∈ , 3 , then show that + + ≥ . 3 a+b b+c c+a 5 Solution: Since the inequality is cyclic, we may assume that c = max {a, b, c} ,  1 c then since a, b, c ∈ , 3 , we have that ≤ a ≤ c. We will change the 3 9 inequailty into a quadratic polynomial of a in order to use our theorem. 7 b a c Put x = − , then our inequality becomes + ≥ x, 5 b+c a+b c+a which is equivalent to a2 + ac + ac + bc ≥ x(a2 + ac + ab + bc), or (1 − x)a2 + [(2 − x)c − xb] a + (1 − x)bc ≥ 0. 3 If 1 − x ≤ 0 or eqivalently c ≥ b, we have that the last inequality is a 2 quadratic polynomial of a with the highest coefficient is not greater than c 0, therefore, it suffices to consider the cases a = or a = c to prove the 9 original inequality. If a = 9c , then we have that a b c c b 9 7 (3b − c)2 7 + + = + + = + ≥ , a+b b+c c+a 9b + c b + c 10 5 2(9b + c)(b + c) 5 b c c b 1 a + + = + + = a+b b+c c+a c+b b+c 2 3 7 > . So our inequality is proved in this case. 2 5

and if a = c, then we have that

35

2 c, then we have that 3 c (2 − x)c − xb ≥ (2 − x)c − xc = 2(1 − x)c ≥ 0. Therefore, since a ≥ , 9 we have Now, let us consider the case 1 − x ≥ 0 or b ≥

(1 − x)a2 + [(2 − x)c − xb] a + (1 − x)bc ≥  c 2 c ≥ (1 − x) + [(2 − x)c − xb] + (1 − x)bc. 9 9 c  c 2 + [(2 − x)c − xb] + (1 − And it suffices to prove that (1 − x) 9 9 x)bc ≥ 0. Or in another word, it means that we only need to prove the c original inequality in the case a = , but this case has been already 9 proved in case 1. Our proof is now complete.   1 1 1 ≤ 10. 11. Let a, b, c ∈ [1, 2] , then show that (a + b + c) + + a b c Solution: We can rewrite the original inequality as P (a, b, c) = (a + b + c)(ab + bc + ca) − 10abc ≤ 0 which is a quadratic polynomial of a with the highest coefficient is (b + c) > 0, therefore, it suffices to consider the cases a = 1 or a = 2. Similarly, by the same manner, we also see that P is also a quadratic polynomial of b and c, and thus, it suffices to consider the cases b = 1 or b = 2, and c = 1 or c = 2. Combining all cases with notice that the inequality is symmetric, we obtain that it suffices to consider the following cases Case 1. If a = b = c = 1, then P (1, 1, 1) = −1 < 0. Case 2. If a = b = 1, c = 2, then P (1, 1, 2) = 0. Case 3. If a = 1, b = c = 2, then P (1, 2, 2) = 0. Case 4. If a = b = c = 2, then P (2, 2, 2) = −1 < 0. Therefore, our inequality is proved. 12. Let a, b, c be nonnegative real numbers such that a + b + c = 3, then a2 + b2 + c2 + abc ≥ 4. (a + b)2 , and our inequality becomes 4 f (x) = (c − 2)x + 2c2 − 6c + 5 ≥ 0 which is a quadratic polynomial of x with coefficient is 0, therefore, it suffices to prove that  the highest   (a + b)2 min f (0), f ≥ 0. Now, putting P (a, b, c) = a2 + b2 + c2 + 4 abc, then we see that     (a + b)2 a+b a+b f (0) = P (a + b, 0, c), f =P , ,c . 4 2 2    a+b a+b which means that we have to prove min P (a + b, 0, c), P , ,c ≥ 2 2 0, or in the other words, we only need to consider the original inequality in the cases there is one variable equal to 0 and there are 2 variables

Solution: Put x = ab, then 0 ≤ x ≤

36

equal. Since the inequality is symmetric, therefore from the above statements, we have 2 cases Case 1. If a = 0, then b = 3−c, and our inequality becomes (3−c)2 +c2 ≥ 4, or equivalently 2c2 − 6c + 5√ ≥ 0, which is true since by AM-GM inequality, we have 2c2 + 5 ≥ 2 10c ≥ 6c. Case 2. If a = b, then c = 3 − 2a ≥ 0, and our inequality becomes 2a2 + (3 − 2a)2 + a2 (3 − 2a) ≥ 4, or equivalently (5 − 2a)(a − 1)2 ≥ 0 which is true since a ≤ 23 .This ends our proof. 13. Let a, b, c, d ∈ [0, 1] . Prove that a2 b+b2 c+c2 d+d2 a−ab2 −bc2 −cd2 −da2 ≤ 8 . 27 Solution: Since the inequality is cyclic, we may assume that a = max {a, b, c, d} , then from a, b, c, d ∈ [0, 1] , we have that 0 ≤ d ≤ a. Now, notice that we can rewrite our inequality as f (d) = (a − c)d2 + (c2 − a2 )d + a2 b + b2 c − ab2 − bc2 −

8 ≤0 27

which is a quadratic polynomial of d with the highest coefficient is nonnegative, therefore, it suffices to prove that max {f (0), f (a)} ≤ 0. We have 8 8 f (0) = a2 b + b2 c − ab2 − bc2 − = b(a − c)(a − b + c) − 27 27   b + (a − c) + (a − b + c) 3 8 8 ≤ − = (a3 − 1) ≤ 0, 3 27 27 and f (a) = (a − c)a2 + (c2 − a2 )a + a2 b + b2 c − ab2 − bc2 − = (a − b)c2 + (b2 − a2 )c + a2 b − ab2 −

8 27

8 27

8 8 ≤ b(a − c)(a − b) − 27 27   8 8 b + (1 − b) 2 8 5 ≤ ab(a − b) − ≤ b(1 − b) − ≤ − =− . 27 27 2 27 108 = (b − c)(a − c)(a − b) −

Therefore, our inequality is proved. 14. If a, b, c are the side-lengths of a triangle, then     b c a a b c 3 + + ≥2 + + + 1. b c a a b c Solution: Without loss of generality, we may assume that a ≥ b ≥ c, then b ≤ a ≤ b + c. The inequality is equivalent to f (a) = (3c − 2b)a2 + (3b2 − 3bc − 2c2 )a + 3bc2 − 2b2 c ≥ 0. If 3c ≤ 2b, then f (a) is a quadratic polynomial of a with the highest coefficient is 3c − 2b ≤ 0, therefore according to our theorem, it suffices to prove that min {f (b), f (b + c)} ≥ 37

0. But this inequality is true because f (b) = b(b − c)2 ≥ 0, and f (b + c) = c(b − c)2 ≥ 0. Now, let us consider the case 3c ≥ 2b, then by AM-GM inequality, we have that a2 + b2 ≥ 2ab, therefore
f (a) ≥ (3c − 2b) 2ab − b2 + (3b2 − 3bc − 2c2 )a + 3bc2 − 2b2 c = a(b − c)(2c − b) + b(2b − 3c)(b − c) ≥ b(b − c)(2c − b) + b(2b − 3c)(b − c) = b(b − c)2 ≥ 0. Remark. We still have to justify why we can assume that a ≥ b ≥ c because the inequality it is only cyclic. Put P (a, b, c) =   is not symmetric,   a b c b c a −2 − 3, then since the inequality is cyclic, 3 + + + + b c a a b c we may assume that a = max {a, b, c} . If b ≥ c, then the statement is trivial. If c ≥ b, then we have that   a b c b c a P (a, b, c) − P (a, c, b) = 5 + + − − − b c a a b c 5(a − b)(a − c)(c − b) = ≥ 0. abc Hence, it suffices to prove that P (a, c, b) ≥ 0 which is equivalent to P (a, b0 , c0 ) ≥ 0 with a ≥ b0 ≥ c0 . Therefore, we only need to consider the case a ≥ b ≥ c in order to prove the original inequality. We may use similar methods to explain the reason why we can assume that a ≥ b ≥ c or c ≥ b ≥ a in proving cyclic inequalities. 15. If a, b, c are the side-lengths of a triangle, then  2   
1 a b2 c2 1 1 2 2 2 3 2 + 2 + 2 ≥ a +b +c + + . b c a a2 b2 c2 Solution: Put x = a2 , y = b2 , z = c2 and assume that x ≥ y ≥ z (we can argue this like in the previous problem), then we have that y ≤ x ≤ √
2 √ y + z sincea, b, c are theside-lengths of atriangle. Our  inequality x y z 1 1 1 is equivalent to 3 + + ≥ (x + y + z) + + ,which can y z x x y z be rewritten as f (x) = (2z − y)x2 + (2y 2 − 3yz − z 2 )x + 2yz 2 − y 2 z ≥ 0. If y ≥ 2z, then f (x) is a quadratic polynomial of x with the n highest  coeffi√
2 o √ y+ z ≥ cient is 2z−y ≤ 0, therefore, it suffices to prove that min f (y), f 0 which is true because f (y) = y(y − z)2 ≥ 0, and  √ √
2  f y+ z = √
4 √
2 √ √ = (2z − y) y + z + (2y 2 − 3yz − z 2 ) y + z + 2yz 2 − y 2 z = (2c2 − b2 )(b + c)4 + (2b4 − 3b2 c2 − c4 )(b + c)2 + 2b2 c4 − b4 c2 = b6 − 6b4 c2 − 2b3 c3 + 9b2 c4 + 6bc5 + c6 = (b3 − 3bc2 − c3 )2 ≥ 0. 38

Now, let use consider the case 2z ≥ y, then by AM-GM inequality, we have that x2 + y 2 ≥ 2xy. Therefore f (x) = (2z − y)x2 + (2y 2 − 3yz − z 2 )x + 2yz 2 − y 2 z ≥ (2z − y)(2xy − y 2 ) + (2y 2 − 3yz − z 2 )x + 2yz 2 − y 2 z = xz(y − z) + y(y − z)(y − 2z) ≥ yz(y − z) + y(y − z)(y − 2z) = y(y − z)2 ≥ 0. Our proof is complete. 16. If a, b, c, d are nonnegative real numbers such that a + b + c + d = 1, then abc + bcd + cda + dab ≤

1 176 + abcd. 27 27 2

Solution: Put x = ab, then 0 ≤ x ≤ (a+b) 4 , our inequality becomes 1 176 cd x + cd(a + b) − ≤ 0. This is a quadratic f (x) = c + d − 27 27 polynomial of x  with thehighest coefficient is 0, therefore, it suffices to  (a + b)2 prove that max f (0), f ≤ 0. It shows that it suffices to 4 consider the cases a = b or ab = 0. Similarly, we also obtain that it suffices to consider the cases c = d or cd = 0. Combining both these statements, we obtain that, in order to prove the orginal inequality, it enough to prove it in the following cases Case 1. If b = d = 0, then the inequality is trivial since abc + bcd + cda + 176 dab − abcd = 0. 27 Case 2. If a = b, d = 0, then c = 1 − 2a ≥ 0, the inequality becomes 1 a2 (1 − 3a) ≤ which is true since by AM-GM inequality, we have  27  a + a + (1 − 2a) 3 1 2 a (1 − 3a) ≤ = . 3 27 2 Case 3. If a = b, c = d, then c = 1−2a 2 , the inequality becomes a (1 − 1 1 44 2 2 2 2a) + 2 a(1 − 2a) ≤ 27 + 27 a (1 − 2a) , which is equivalent to (4a − 1)2 (22a2 − 11a + 2) ≥ 0. The last √ inequality is valid since by AM-GM inequality, we have 22a2 + 2 ≥ 4 11a ≥ 11a. This ends our proof.

17. If a, b, c, d are nonnegative real numbers such that a + b + c + d = 4, then (1 + 3a)(1 + 3b)(1 + 3c)(1 + 3d) ≤ 125 + 131abcd. Solution: Proceeding similar to the previous problem, we see that it enough to prove it in the following cases Case 1. If b = d = 0, then the inequality is trivial since 1 (1 + 3a)(1 + 3b)(1 + 3c)(1 + 3d) = (1 + 3a)(1 + 3c) ≤ (2 + 3a + 3c)2 4 = 49 < 125 = 125 + 131abcd. 39

Case 2. If a = b, d = 0, then c = 4 − 2a ≥ 0, and the inequality becomes (1 + 3a)2 (13 − 6a) ≤ 125 which is true since by AM-GM inequality, we have that   (1 + 3a) + (1 + 3a) + (13 − 6a) 3 2 (1 + 3a) (13 − 6a) ≤ = 125. 3 Case 3. If a = b, c = d, then c = 2 − a ≥ 0, and the inequality becomes (1 + 3a)2 (7 − 3a)2 ≤ 125 + 131a2 (2 − a)2 , which is equivalent to (25a2 − 50a + 38)(a − 1)2 ≥ 0 which is true. Our proof is complete. 18. Let a, b, c, d be nonnegative real numbers with sum 4, prove that 3(a2 + b2 + c2 + d2 ) + 4abcd ≥ 16. Solution: Again, proceeding similar to the previous problem, we see that it enough to prove it in the following cases Case 1. If b = d = 0, then the inequality is trivial since 3 3(a2 + b2 + c2 + d2 ) + 4abcd = 3(a2 + c2 ) ≥ (a + c)2 = 24 > 16. 2 Case  2. If a = b,d = 0, then c = 4 − 2a and the inequality becomes 3 2a2 + (4 − 2a)2 ≥ 16, which is true since by Cauchy Schwartz inequality (see at the next chapter), we have that 2a2 + (4 − 2a)2 ≥

1 16 [a + a + (4 − 2a)]2 = . 3 3

Case 3. If a = b, c = d, then c = 2 − a ≥ 0 and the inequality becomes 6a2 +6(2−a)2 +4a2 (2−a)2 ≥ 16, or equivalently (a2 −2a+2)(a−1)2 ≥ 0 which is trivial. The proof is complete. 19. Let a, b, c be nonnegative real numbers such that a + b + c = 3. Prove that 1 1 1 3 + + ≤ . 6 − ab 6 − bc 6 − ca 5 Solution: By expanding, the inequality is equivalent to 108 + 13abc(a + b + c) − 48(ab + bc + ca) − 3a2 b2 c2 ≥ 0, or 36 + 13abc − a2 b2 c2 − 16(ab + (a + b)2 bc + ca) ≥ 0. Put x = ab, then 0 ≤ x ≤ , our inequality becomes 4 2 2 f (x) = −c x + (13c − 16)x + 36 − 16c(3 − c) ≥ 0. This is a quadratic 2 polynomial of x with then highest coefficient o is −c ≤ 0, therefore, it (a+b)2 ≥ 0. It shows that it suffices suffices to prove that min f (0), f 4 to consider the cases a = b or ab = 0. Case 1. If b = 0, then a = 3 − c, and the inequality becomes 36 − 16c(3 − c) ≥ 0, or equivalently 4(3 − 2c)2 ≥ 0. Case 2. If a = b, then c = 3 − 2a ≥ 0, and the inequality becomes 36 + 13a2 (3 − 2a) − a4 (3 − 2a)2 − 16a2 − 32a(3 − 2a) ≥ 0, or equivalently −4a6 + 12a5 − 9a4 − 26a3 + 87a2 − 96a + 36 ≥ 0, or (3 − 2a)(a + 2)(2a2 − 3a + 6)(a − 1)2 ≥ 0. Our proof is complete. 40

20. Let a, b, c, d be nonnegative real numbers, no three of which are zero. Show that r r r r a b c d 4 + + + ≤√ . a+b+c b+c+d c+d+a d+a+b 3 Solution: By Cauchy Schwarz inequality, we have !2 Xr a ≤ a + b + c cyc # " #" X X a ≤ (a + b + d)(a + c + d) (a + b + c)(a + b + d)(a + c + d) cyc cyc =

2[2(a + b + c + d)2 + (a + c)(b + d)][(a + c)(b + d) + ac + bd] . (a + b + c)(b + c + d)(c + d + a)(d + a + b)

We need to prove that P (a, b, c, d) = 8(a + b + c)(b + c + d)(c + d + a)(d + a + b) −[2(a + b + c + d)2 + (a + c)(b + d)][(a + c)(b + d) + ac + bd] ≥ 0. It is easy to see that P (a, b, c, d) is a quadratic polynomial of x = ac with the highest coefficient is 0, therefore, in order to prove P (a, b, c, d) ≥ 0, it suffices to prove it in the case a = c or ac = 0. Similarly, P (a, b, c, d) is also a quadratic polynomial of t = bd with the highest coefficient is 0, therefore, in order to prove P (a, b, c, d) ≥ 0, it suffices to prove it in the case b = d or bd = 0. Combining all cases, we see that it is enough to prove the inequality in the following cases Case 1. c = d = 0, then the inequality becomes 8ab(a + b)2 ≥ 3ab(2a2 + 5ab + 2b2 ), or equivalently ab(2a2 + ab + 2b2 ) ≥ 0. Case 2. a = c, d = 0, then the inequality becomes 16a(2a + b)(a + b)2 ≥ 6a(a + b)(a + 2b)(4a + b), or equivalently 2a(a + b)(4a2 − 3ab + 2b2 ) ≥ 0. Case 3. a = c, b = d, then the inequality becomes 8(2a + b)2 (a + 2b)2 ≥ 12(2a2 +5ab+2b2 )(a2 +4ab+b2 ), or equivalently 4(a+2b)(2a+b)(a−b)2 ≥ 0.Our proof is now complete. Equality holds if and only if a = b = c = d.

1.7

Cauchy-Schwartz Inequality

The inequality (a21 + a22 + . . . + a2n )(b21 + b22 + . . . + b2n ) ≥ (a1 b1 + a2 b2 + . . . + an bn )2 has incredibly many applications. Especially useful is the following form of the inequality: CBS Lemma: For yi > 0, x21 x2 (x1 + x2 + . . . + xn )2 + ... + n ≥ y1 yn y1 + y2 + . . . + yn 41

It can be easily noticed that it’s equivalent to CBS inequality, and it can be proven also by induction on n. With this lemma, inequalities like the following become straightforward: Example 1. If a, b, c > 0 then a b c + + ≥ 1. a + 2b b + 2c c + 2a Indeed, multiplying the denominator and numerator of the first fraction with a, of the second with b, and of the third with c, we get a2 b2 c2 (a + b + c)2 + + ≥ = 1. a2 + 2ab b2 + 2bc c2 + 2ca a2 + 2ab + b2 + 2bc + c2 + 2ca The importance of this inequality is revealed by the multitude of exercises it helps solve. Exercises 1. If x, y, z ≥ 0 show that

p

(x + y)(x + z) ≥ x +

√

yz. √ √ √ √ Solution: Immediate from Cauchy-Schwartz for x, y and x, z.

2. Show that n(x21 + x22 + . . . + x2n ) ≥ (x1 + x2 + . . . + xn )2 Solution: In the form (1 + 1 + . . . + 1)(x21 + x22 + x23 + . . . + x2n ) ≥ (x1 + . . . + xn )2 , Cauchy-Schwartz is clearly seen. a b c 3 + + ≥ . a + 2b + c b + 2c + a c + 2a + b 4 Solution: Amplify the first fraction by a, the second by b, the third by c and apply CBS Lemma.

3. For a, b, c > 0, show that

4. For a, b, c, d, e, f > 0 show that

a b c d e + + + + + b+c c+d d+e e+f f +a

f ≥ 3. a+b Solution: We again amplify by the corresponding numerators and apply CBS Lemma, obtaining

≥

b c d e f a + + + + + ≥ b+c c+d d+e e+f f +a a+b (a + b + c + d + e + f )2 . a(b + c) + b(c + d) + c(d + e) + d(e + f ) + e(f + a) + f (a + b)

It remains to show that a(b+c)+b(c+d)+c(d+e)+d(e+f )+e(f +a)+f (a+b) ≤

(a + b + c + d + e + f )2 . 3

However this can be written as (a+d)(b+e)+(b+e)(c+f )+(c+f )(a+d) (u + v + w)2 and we apply the inequality uv + vw + wu ≤ . 3 42

5. If

√ √ √ √ 1 1 1 + + = 2 then x + y + z ≥ x − 1 + y − 1 + z − 1. x y z

Solution: By CBS, √

x−1+ 

≤ (x + y + z)

2 p √ y−1+ z−1 ≤

x−1 y−1 z−1 + + x y z

 = x + y + z.

a b c + + ≥ 1. 1+b+c 1+c+a 1+a+b Solution: Amplifying the fractions by a, b, c we get

6. If abc = 1, a, b, c > 0 then

a2 b2 c2 + + a + ab + ac b + ab + bc c + ac + bc

≥ ≥

(a + b + c)2 a + b + c + 2(ab + bc + ca) (a + b + c)2 = 1. (a + b + c)2

The last inequality follows from the fact that a2 + b2 + c2 ≥ a + b + c as a2 + 1 + b2 + 1 + c2 + 1 ≥ 2a + 2b + 2c and a + b + c ≥ 3. 7. Let n ≥ 3, a1 , a2 , . . . , an ≥ 0 with a21 + a22 + . . . + a2n = 1. Prove that √ √ √ a1 a2 an 4 + + ... + 2 ≥ (a1 a1 + a2 a2 + . . . + an an )2 . 5 a22 + 1 a23 + 1 a1 + 1 Solution: Amplifying the fractions by a2i and applying the CBS Lemma we get √ √ √
2 a1 a1 + a2 a2 + . . . + an an a1 a2 an + + ... + 2 ≥ . a22 + 1 a23 + 1 a1 + 1 a21 a22 + a21 + . . . + a2n a21 + a2n 1 So we are left to show that a21 a22 + . . . + a2n a21 ≤ which we have already 4 proven in exercise 7 at induction. 8. Show that x y z p p p + + ≤ 1. x + (x + y)(x + z) y + (y + z)(y + x) z + (z + x)(z + y) Solution: We have hence

p p √ √ (x + y)(x + z) = (x + y)(z + x) ≥ xz + xy

√ x x p √ =√ √ . ≤ √ √ x + xy + xz x+ y+ z x + (x + y)(x + z) x

Summing with the analogously deduced inequalities we get the result. 43

a b c 1 + + + ≥ 1. 1+a 1+b 1+c 1+a+b+c

9. If a, b, c ≥ 0 then Solution:

≥

b2 c2 1 a2 + + + ≥ 2 2 2 a +a b +b c +c a+b+c+1 (a + b + c + 1)2 = 1. a2 + a + b2 + b + c2 + c + a + b + c + 1

10. If m ≥ n ≥ p ≥ q are integers numbers with m + q = n + p and a1 , a2 , . . . , ak are positive reals then ! k ! ! k ! k k X X q X X p m n ai ai ≥ ai ai . i=1

Solution: Let sr =

i=1

i=1

i=1

k X ari . We must prove that sm sq ≥ sn sp . For i=1

n = p = n − 1 = q + 1 we have sn+1 sn−1 ≥ s2n , which is true in virtue of sn sn+1 sn+1 CBS inequality. Rewriting this as ≥ . Therefore rn = sn sn−1 sn is increasing in n. Hence rm−1 rm−2 . . . rn ≥ rp−1 rp−2 . . . rq (recall that sp sm m − n = p − q) which means that ≥ or sm sq ≥ sn sp . sn sq r r k k l l k a1 + . . . + an l a1 + . . . + an 11. If n, k > l ∈ N, a1 , a2 , . . . , an ≥ 0 then ≥ . n n Solution: We shall use induction on k − l. If k = l this is obvious. Rephrasing the problem using the notation in the previous exercise we need to show that slk nk−l ≥ skl . Now remember that we proved that si+1 si ri = ≥ ri−1 = . From here we deduce that si si−1     sk+1 l sk+1 l sl l rk ≥ rl−1 rl−2 . . . r0 or ≥ or n ≥ sl , sk n sk and this relation performs the induction step on k − l: multiplying it slk nk−l ≥ skl we get slk+1 nk−l+1 ≥ sk+1 . l 12. (Minkowski) If ai , bi are reals then v u n !2 n q u X X a2i + b2i ≥ t ai + i=1

i=1

n X

!2 bi

.

i=1

Solution: If n = 2 the inequality is equivalent to p p p a2 + b2 + c2 + d2 ≥ (a + c)2 + (b + d)2 , which after squaring and cancelling common terms becomes p (a2 + b2 )(c2 + d2 ) ≥ ac + bd, which is CBS. The general case follows by induction on n. 44

13. If a, b, c > 0 add up to 1, then

a+1 b+1 c+1 36 + + ≥ . c(2 − b) a(2 − c) b(2 − a) 5

Solution:   1 1 1 2 2 1 1 , = = ≤ 2 + + 2−a 2 (a + b + c) − a 2b + 2c + a 5 b c a   2 2 1 where the last inequality is because (2b + 2c + a) ≥ 52 + + b c a (Cauchy-Schwartz). Similarly,     1 1 1 2 2 1 1 2 2 1 , . ≤ 2 + + ≤ 2 + + 2−b 5 c a b 2−c 5 a b c Summing up results in

≤ =

1 1 1 + + ≤ 2 −a 2 − b 2 − c     1 2 2 1 1 2 2 1 1 2 2 1 + + + + + + + + 52 b c a 52 c a b 52 a b c     1 5 5 5 1 1 1 1 + + = + + . 2 5 a b c 5 a b c

Thus, using a + 1 = 2 − (1 − a) = 2 − (b + c) = (2 − b) − c and similarly b + 1 = (2 − c) − a and c + 1 = (2 − a) − b, we get b+1 c+1 a+1 + + = c (2 − b) a (2 − c) b (2 − a) (2 − b) − c (2 − c) − a (2 − a) − b = + + c (2 − b) a (2 − c) b (2 − a)       1 1 1 1 1 1 = + + − − − c 2−b a 2−c b 2−a     1 1 1 1 1 1 = + + − + + a b c 2−a 2−b 2−c       1 1 1 1 1 1 1 4 1 1 1 ≥ + + − + + = + + . a b c 5 a b c 5 a b c   1 1 1 36 1 1 1 So it remains to show that 45 + + ≥ , i. e. that + + ≥ 9. a b c 5 a b c   1 1 1 But this is clear, since Cauchy-Schwartz gives (a + b + c) + + ≥ a b c 9, and a + b + c = 1. This completes the proof. 14. Let a, b, c, x, y, z be reals such that (a + b + c)(x + y + z) = 3 and (a2 + b2 + c2 )(x2 + y 2 + z 2 ) = 4. Show that ax + by + cz ≥ 0 Solution: Set u = ax + by + xz. As (a, b, c) and (x, y, z) are unrelated in the condition, we can also take its ”brothers”: v = ay + bz + cx, w = az + bx + cy. Then u + v + w = (a + b + c)(x + y + z) = 3. u2 + v 2 + w2 = (a2 + b2 + c2 )(x2 + y 2 + z 2 ) + 2(ab + bc + ca)(xy + yz + zx) = 4 + 2(ab + bc + ca)(xy + yz + xz). 45

Assume now that u < 0, then v+w ≥ 3 so a(y+z)+b(x+z)+c(x+y) > 3. Applying CBS we get   9 (a2 +b2 +c2 ) 2(x2 + y 2 + z 2 ) + 2(xy + yz + zx) ≥ (a2 +b2 +c2 )(x2 +y 2 +z 2 ). 4 This means that xy + yz + zx ≥

x2 + y 2 + z 2 and analogously ab + bc + 8

a2 + b2 + c2 ca ≥ . This estimates are too rough, however they tell us 8 that ab + bc + ca, xy + yz + zx are positive, and so we can apply CBS in the following form:  2   a + b2 + c2 + 2(ab + bc + ca) x2 + y 2 + z 2 + 2(xy + yz + zx) ≥ hp i2 p ≥ (a2 + b2 + c2 )(x2 + y 2 + z 2 ) + 2 (ab + bc + ca)(xy + yz + zx) h i2 p or 9 ≥ 2 + 2 (ab + bc + ca)(xy + yz + zx) which implies 1 (ab + bc + ca)(xy + yz + zx) ≤ . 4 It suffices to derive a contradiction now: we have 9 u2 + v 2 + w2 = 4 + 2(ab + bc + ca)(xy + yz + xz) ≤ , 2 however from the other side u2 + v 2 + w2 > v 2 + w2 ≥

(v + w)2 9 > . 2 2

15. If a, b, c, x, y, z > 0 then b c xy + yz + zx a (y + z) + (x + z) + (x + y) ≥ 3 . b+c a+c a+b x+y+z Solution: We add (y + z) + (x + z) + (x + y) to both sides to get a more comfortable form of the LHS:   y+z z+x x+y 2(x + y + z)2 + 3(xy + yz + zx) (a+b+c) + + = . b+c c+a a+b x+y+z Now from CBS we have   y+z z+x x+y (a + b + c) + + = b+c c+a a+b   1 y+z z+x x+y [(b + c) + (c + a) + (a + b)] + + = 2 b+c c+a a+b
√ √ √ 1 2 ≥ y+z+ z+x+ x+y . 2 So it suffices to prove that
2 √ √ √ xy + yz + zx y + z + z + x + x + y ≥ 4(x + y + z) + 6 , x+y+z 46

or 2(x + y + z) + 2

Xp xy + yz + zx (x + y)(x + z) ≥ 4(x + y + z) + 6 x+y+z cyc

or p p p xy + yz + zx (x + y)(x + z)+ (y + z)(y + x)+ (z + x)(z + y) ≥ (x+y+z)+3 . x+y+z However p p p (x + y)(x + z) + (y + z)(y + x) + (z + x)(z + y) = p p p = x2 + (xy + yz + zx) + y 2 + (xy + yz + zx) + z 2 + (xy + yz + zx) p (x + y + z)2 + 9(xy + yz + zx) ≥ from Minkowski. The fact that p xy + yz + zx (x + y + z)2 + 9(xy + yz + zx) ≥ (x + y + z) + 3 x+y+z follows by squaring. 16. If a1 , a2 , . . . , an are positive numbers that sum to 1, then   a1 a2 an n . + 2 + ... + 2 (a1 a2 +a2 a3 +. . .+an a1 ) ≥ 2 n + 1 a2 + a2 a3 + a3 a1 + a1 Solution: We may note that

ai ai ai − , thus = ai+1 ai+1 + 1 a2i+1 + ai+1

a1 a2 an + + ... + 2 = a22 + a2 a23 + a3 a1 + a1     a1 a2 an an a1 = + + ... + + ... + . − a2 a3 a1 a2 + 1 a1 + 1   a1 a2 an But + + ... + (a1 a2 + . . . + an a1 ) ≥ 1 so it suffices to prove a2 a3 a1 that   a1 a2 an a1 an + + ... + ≥ (n + 1) + ... + . a2 a3 a1 a2 + 1 a1 + 1 a1 an + ... + ≥ n but as a2 a1     1 n+1 2 (a2 + 1) +1 ≥ (CBS) n2 a2 n

However we may note that

a1 1 a1 n2 ≤ + a1 . Summing with the anala2 + 1 (n + 1)2 a2 (n + 1)2 ogous relations we get the result.

we deduce

47

1.8

Young Inequality

In the AM-GM inequality put m terms equal a and n terms equal b. We deduce m n ma + nb ≥ (m + n)a m+n b m+n . m n m+n m+n , q = , x = a m+n , y = b m+n we rewrote the Now setting p = m n xp y q inequality as + ≥ pq for any a, b > 0, p, q ∈ Q with p1 + 1q = 1. We can p q extend this inequality to any positive real p, q: taking sequences of rationals 1 1 + = 1 and pn tending to p, qn tending to q we deduce pn , qn with pn qn p q xn yn + ≥ xy. Passing to limit we have pn qn

Theorem. If p, q > 0 with

1 p

+

1 q

> 0 then for any positive reals x, y > 0 xp y q + ≥ xy. p q

This is called Young’s inequality. Exercise: a) Prove that Young Inequality changes sign when one of p, q is q

negative. (Hint: if, for example, q < 0, use normal Young Inequality for b 1−q , q 1−q (ab) q−1 and weights 1 − q, ). q b) Prove the generalization of Young Inequality: If p1 , p2 , . . . , pn , x1 , x2 , . . . , xn > 0 and p1 + p2 + . . . + pn = 1 then p1 x1 + . . . + pn xn ≥ xp11 . . . xpnn (hint: use induction). Take now some A, B, a1 , a2 , . . . , an , b1 , b2 , . . . , bn positive reals. Then   1  ai p 1 bi q ai bi ≤ + . AB p A q B Summing this over all i we have 

n X

p p ai 

 a1 b1 + . . . + an bn 1 i=1 ≤  AB p  A

Taking now A =

n X i=1

!1 api

p

,B =

n X



n X

q

bpi 

    + 1  i=1  q   B

  .  

!1 nbqi

q

we deduce

i=1

Theorem. If p, q > 0 and p1 + 1q = 1 then for any positive reals a1 , . . . , an , b1 , . . . , bn the following inequality holds: 48

n X

n X

ai bi ≤

i=1

!1 api

p

i=1

!1 X

nbqi

q

i=1

This is Holder’s Inequality. Like Young’s Inequality, when one of p, q is negative it changes sign. [Note: for p = q = 2 this is Cauchy-Schwartz]. Take p > 1 and apply Holder for the numbers ai , (ai + bi )p−1 with weights p p, . We deduce p−1 n X

ai (ai + bi )p−1 ≤

n X

i=1

# p−1 !1 " n p X p p p . (ai + bi ) ai i=1

i=1

Writing the analogous inequality for bi , (ai + bi )p−1 and summing we obtain Minkovski’s Inequality: n X

!1

p

(ai + bi )p

≤

n X

!1 api

p

+

i=1

i=1

n X

!1 bpi

p

.

i=1

When 0 < p < 1 it changes sign. Exercises 1. Show that for any x ≥ y ≥ 0, y −y −x y −y (ax1 + . . . + axn )(a−x 1 + . . . + an ) ≥ (a1 + . . . + an )(a1 + . . . + an ).

Solution: By Holder, y

x−y x

y

x−y x

(ax1 + . . . + axn ) x (1 + 1 + . . . + 1)

≥ (ay1 + . . . + ayn ),

and analogously −x x (a−x 1 + . . . + an ) (1 + 1 + . . . + 1)

−y ≥ (a−y 1 + . . . + an ).

The inequality follows from multiplying these two and using the fact that −x 2 (ax1 + . . . + axn )(a−x 1 + . . . + an ) ≥ n . √ √ √ 2. For a, b, c > 0 a − ab + b2 + a2 − ac + c2 ≥ b2 + bc + c2 . Solution: we write it as s 2  2  2  2 r b b b b c  2  c 2  c  2  c 2 −a + + + + a− + + + ≥ 2 2 2 2 2 2 2 2 s        b c 2 b c 2 b c 2 b c 2 ≥ − + + + + + + , 2 2 2 2 2 2 2 2 and it now follows from Minkowski. 49

3. (a2 + ab + b2 )(b2 + bc + c2 )(c2 + ca + a2 ) ≥ (ab + bc + ca)3 Solution: Write it as (a2 + ab + b2 )(ac + a2 + c2 )(c2 + b2 + bc) ≥ (ab + bc + ca)3 . By CBS 2  3 1 3 1 (a + ab + b )(ac + a + c ) ≥ a 2 c 2 + a 2 b 2 + bc 2

 and

3 1 a2 c2

+

2

3 1 a2 b2

2

2

2 + bc (c2 + b2 + bc) ≥ (ab + bc + ca)3 by Holder for

3 p = , q = 3. 2 Note: by the very same method we can prove by induction the generalized version of Holder for more that two sequences of variables.

1.9

Advanced techniques with Cauchy-BuniakowskiSchwarz and Holder Inequalities

In the previous parts, we have presented the Cauchy-Schwartz inequality and the Holder’s inequality with their basic techniques in use. These inequalities are so important, however, that they deserve an additional chapter for finer applications. Firstly, let us recall the Cauchy-Schwartz inequality and the Holder’s inequality. Theorem 1. (Cauchy-Schwartz inequality) For any real numbers (a1 , a2 , . . ., an ) and (b1 , b2 , . . ., bn ), we have (a1 b1 + a2 b2 + . . . + an bn )2 ≤ (a21 + a22 + . . . + a2n )(b21 + b22 + . . . + b2n ). The equality holds if and only if ai : aj = bi : bj ∀i, j ∈ {1, 2, . . . , n} . Corollary 1. (CBS lemma) For any real numbers (a1 , a2 , . . ., an ) and (b1 , b2 , . . ., bn ) with bi > 0 ∀i = 1, 2, . . . , n, we have a21 a22 a2 (a1 + a2 + . . . + an )2 + + ... + n ≥ . b1 b2 bn b1 + b2 + . . . + bn The equality holds if and only if ai : aj = bi : bj ∀i, j ∈ {1, 2, . . . , n} . Theorem 2. (Generalized Holder’s inequality) Given positive reals xij (i = 1, m, j = 1, n). Then, for all ω1 , . . ., ωn ≥ 0 satisfy ω1 + . . . + ωn = 1, we have n Y i=1

 ωj ! m m n X X Y ω j  xij  ≥ xij . j=1

j=1

50

i=1

A special case of Holder’s inequality which we used to apply is m X

! ani

i=1

m X bni

!n−1 ≥

i=1

m X

!n ai bn−1 i

,

i=1

for all ai , bi ≥ 0. The equality holds in this inequality when

am a1 = ... = . b1 bm

In most advanced inequalities it is not immediately clear how to apply to CBS or Holder. One needs to do some more work to ”prepare” it for their use. In the following, we will describe several common techniques that help one find the way to apply Cauchy-Schwartz and Holder to problems. These techniques are not restricted only to Cauchy-Schwartz and Holder, they work for other inequalities as well, but in this paragraph we will only focus on these two particular inequalities. Balancing coefficients Grouping the coefficients to use classical inequalities is not always easy, especially in non-symmetric inequalities. In these cases, the coefficients of similar terms are normally not equal to each other and therefore there is confusion about not only the use basic inequalities properly but also taking care of the case of equality so that this case is valid throughout the process of solution. One way to deal with this is to use additional variables (”weights”) in order to apply some inequality with unknown parameters, and the solve a system of equations to find assign the corresponding values to the weights. Knowing an equality case or what should be in the right-hand side of the inequality usually gives information that is helpful in finding them. This method is called the balancing coefficients technique. It is mainly applied in conjunction with two inequalities the AM-GM inequality and Cauchy Schwartz/Holder inequality. In some sense, Holder as deduced from Young’s inequality is a consequence of p q this method for AM-GM: in the inequality ap + bq ≥ ab, p and q are ”weights” with respect to which weighted AM-GM is applied (i.e. ap appears ” p1 times” and bq appears ” 1q times”). We will focus on applying the technique for Cauchy-Schwarts/Holder inequalities in the examples. Example 1. Given nonnegative real numbers x, y, z satisfying 2x + 3y + z = 1. Determine the minimum value of P = x3 + y 3 + z 3 . Solution: According to the Holder’s inequality, we have that for all a, b, c ≥ 0, (x3 + y 3 + z 3 )(a3 + b3 + c3 )2 ≥ (xa2 + yb2 + zc2 )3 . It follows that P = x3 + y 3 + z 3 ≥

(xa2 + yb2 + zc2 )3 (a3 + b3 + c3 )2

.

Now, let us choose a, b, c, so that we can apply the given hypothesis 2x + 3y + z = 1 to the above inequality, and therefore, naturally, we will choose a, b, c, 51

√ √ a2 b2 c2 such that = = = 1, or a = 2, b = 3, c = 1. The equality holds 2 3 1 x y z x y z 2x + 3y + z 1 when = = , or = = = = , or a b c a b c 2a + 3b + c 2a + 3b + c x=

a , 2a + 3b + c

y=

b , 2a + 3b + c

z=

c . 2a + 3b + c

From now, we have P ≥

(xa2 + yb2 + zc2 )3 (a3 + b3 + c3 )2

(2x + 3y + z)3 1 =h √
i2 = √ √
2 . √
3 3 2 2+3 3+1 2 + 3 + 13

Since the equality can be attained, the minimum value of P is

1 √ √
2 . 2 2+3 3+1

Example 2. Given nonnegative real numbers x, y such that x3 + y 3 = 1. Find the maximum value of P = x + 2y. Solution: This is in a sense a converse to the previous example. By Holder’s inequality, we have that for all a, b ≥ 0 (x3 + y 3 )(a3 + b3 )2 ≥ (xa2 + yb2 )3 , hence q 3 xa + yb ≤ (x3 + y 3 )(a3 + b3 )2 . 2

2

Naturally we will choose a, b so that xa2 + yb2 ≡ P, which gives us a = 1, b = √ x y x3 y3 x3 + y 3 1 2. The equality holds when = , or 3 = 3 = 3 = 3 , or 3 a b a b a +b a + b3 x3 =

a3

a3 , + b3

y3 =

a3

b3 . + b3

Then s P ≤

3

(x3

+

y3)

 √ 3 2 r √ 3 3 3 1 + 2 = 2 2+1 .

q √
3 3 Again equality is attained and we obtain that the maximum of P is 2 2+1 . Example 3. Let a1 , a2 , . . . , an be positive real numbers. Prove that n X 1 ≤2 . ak

n X

k k X k=1

k=1

aj

j=1

Solution: By the Cauchy Schwartz inequality,  (a1 +a2 + . . . + ak )

b2 b21 b2 + 2 + ... + k a1 a2 ak 52



≥ (b1 + b2 + . . . + bk )2 ,

where bi are arbitrary positive real numbers for all i = 1, 2, . . . , n. Therefore  2  b2k b1 b22 k k + + ... + , ≤ k (b1 + b2 + . . . + bk )2 a1 a2 ak X aj j=1

from which we deduce that n X

k k X k=1

n X ci , ≤ ai i=1

aj

j=1

where ck =

kb2k (k + 1)b2k nb2k + +. . .+ , (b1 + b2 + . . . + bk )2 (b1 + b2 + . . . + bk+1 )2 (b1 + b2 + . . . + bk+1 )2

for all k = 1, 2, . . . , n. Choosing now bk = k, we get ck =

k3 k X

i

!2 +

k 2 (k + 1) !2 + . . . + k+1 X i

i=1

< = =

i=1

i=1

 1 1 1 + + . . . + k(k + 1)2 (k + 1)(k + 2)2 n(n + 1)2   1 1 1 1 2 4k + ... + − − ... − k(k + 1) n(n + 1) (k + 1)2 (n + 1)2       1 1 1 1 1 1 1 1 4k 2 + + . . . + + − − . . . − 2 k 2 (k + 1)2 2 n2 (n + 1)2 (k + 1)2 (n + 1)2   1 1 1 1 1 1 4k 2 + + + ... + 2 − − ... − 2 2 2 2 2k 2(n + 1) (k + 1) n (k + 1) (n + 1)2   1 1 4k 2 − < 2, 2k 2 2(n + 1)2

= 4k 2 =

k2 n !2 n X i



for all k = 1, 2, . . . , n. In this case, it follows that n X

k k X k=1

n X 1 ≤2 , ai

aj

i=1

j=1

which ends the proof. Example 4. Given positive real numbers a1 , a2 , . . . , an . Prove that     a1 + a2 2 a1 + a2 + . . . + an 2 4(a21 + a22 + . . . + a2n ) ≥ a21 + + ... + . 2 n Solution: By the Cauchy Schwartz inequality, for all k = 1, 2, . . . , n, bi > 0, we have  2  a2k a1 a22 + + ... + (b1 + b2 + . . . + bk ) ≥ (a1 + a2 + . . . + ak )2 , b1 b2 bk 53

It follows that    2  a2k a1 + a2 + . . . + ak 2 b1 + b2 + . . . + bk a1 a22 ≤ · + + ... + , k k2 b1 b2 bk or

 n  X a1 + a2 + . . . + ak 2 k=1

k

≤

n X

ck a2k ,

k=1

where b1 + b2 + . . . + bk+1 b1 + b2 + . . . + bn b1 + b2 + . . . + bk + + ... + . 2 2 k bk (k + 1) bk n2 bk √ √ Now, choosing bk = k − k − 1, we obtain ck =

ck = = ≤

=

=

b1 + b2 + . . . + bk b1 + b2 + . . . + bk+1 b1 + b2 + . . . + bn + + ... + k 2 bk (k + 1)2 bk n2 bk   1 1 1 1 √ + + . . . + 3/2 √ 3/2 3/2 (k + 1) n k− k−1 k      1 1 1 1 1 2  q  + . . . + 2 q  √ −q −q √ k− k−1 k − 12 k + 21 n − 21 n + 12   1 2 1 < q q 1 √ −q √ √  √ k− k−1 k − 21 n + 12 k − 21 k− k−1  √ √ √ k+ k−1 2 2 √ ≤ 4. 2k − 1

The inequality is proved. This last example, by far the hardest, √ √requires some explanation - namely, how to arrive to the choice bk = k − k − 1? There is no universal method to choose these coefficients, and hard problems require a certain degree of creativity from the solver. In this case, there are several reasons that point to the correct solution. One way would be to look at the equality case. We don’t know it (and in fact it is never achieved unless all variables are zero), but we can produce an ”asymptotic” case, i.e. an examples where the two expression are close to each other. The choice to take all ai equal fails. Then, because the inequality is not symmetric, one can try to choose ai as a function of i, and the easiest k+1 one is ai = ik . This is good, because the sum of ik is approximately ik+1 for k > −1 (this is done by tricks as the integral approximation or the Bernoulli k+1 polynomials), so that a1 + . . . + an would be approximately nk+1 . P 4 One therefore gets 4 i2k ∼ 2k+1 n2k+1 (for 2k > −1) and we compare it to P ik 2 ( k+1 ) ∼ (k+1)21(2k+1) n2k+1 . Therefore, in order for them to be equal we need (k + 1)2 = 14 so k = − 12 . Note that we cannot do any better because we need 2k > −1 - in fact we are already on the border and violate this condition for k = − 21 . This case is still very close to the equality anyway, as checked directly. 54

So now if we want the method to work, we better choose bi that ”approximately” respect the inequality case, √ i.e. that are ”approximately proportional” 1 √ to ai = i . Such a choice is bi = i. This one isn’t too nice however - the reason √ being that ck involves the sum 1 + √12 + . . . + √1k which is approximately 2 k but however is not precise. So perhaps a nicer choice √ would be the one that makes the partial sum b1 +. . .+bk k (forgetting the unnecessary constant of 2) i.e. bk = actually equal to √ √ k − k − 1. These numbers are still approximately proportional to √1k as √ √ √ k − k − 1 = √k−(k−1) = √k+1√k−1 ∼ 2√1 k , and this is how the weights are k+ k−1 found. Another idea is to apply the ”power guess” to the coefficients bk : one could try to set bk = k α and then solve for α using the same estimates to get α = − 12 .

Make it nonnegative Suppose one has to prove an inequality of the form an a1 a2 + + ... + ≥ k, bi > 0 ∀n = 1, 2, . . . , n b1 b2 bn It is similar to Cauchy Schwartz and Holder. However these inequalities may not be applicable if the quantities ai are not positive. In some cases, it is possible to resolve this inconvenience. The suggestion is to add up abii to mi so that ai + mi bi ≥ 0 and as small as possible (it is great if we choose mi so that the inequality ai + mi bi ≥ 0 has equality case), then the inequality becomes a01 a02 a0 + + . . . + n ≥ k + m1 + m2 + . . . + mn . b1 b2 bn where a0i ≥ 0, bi > 0 ∀i = 1, 2, . . . , n. In this case, the Cauchy-Schwartz or Holder inequality can be applied. In the case that ai ≥ 0 ∀n = 1, 2, . . . , n, one can subtract a positive quantity mi from abii such that ai − mi bi ≥ 0 and as small as possible, then we can apply the Cauchy Schwartz-Holder inequality more effectively than if we apply it directly (in some cases). Example 5. Let a, b, c, d be real numbers such that a2 + b2 + c2 + d2 = 1. Prove that 1 1 1 1 16 + + + ≤ . 1 − ab 1 − bc 1 − cd 1 − da 3 Solution: Rewrite the inequality in the form 

1 2− 1 − ab





1 + 2− 1 − bc



 + 2−

1 1 − cd





1 + 2− 1 − da



8 ≥ , 3

or 1 − 2ab 1 − 2bc 1 − 2cd 1 − 2da 8 + + + ≥ . 1 − ab 1 − bc 1 − cd 1 − da 3 Now, since 1 − 2ab = a2 + b2 + c2 + d2 − 2ab = (a − b)2 + c2 + d2 ≥ 0, applying 55

the Cauchy Schwartz inequality, we have

≥ =

1 − 2ab 1 − 2bc 1 − 2cd 1 − 2da + + + ≥ 1 − ab 1 − bc 1 − cd 1 − da [(1 − 2ab) + (1 − 2bc) + (1 − 2cd) + (1 − 2da)]2 (1 − 2ab)(1 − ab) + (1 − 2bc)(1 − bc) + (1 − 2cd)(1 − cd) + (1 − 2da)(1 − da) 4[2 − (a + c)(b + d)]2 . 4 − 3(a + c)(b + d) + 2(a2 + c2 )(b2 + d2 )

It suffices to prove that 3[2 − (a + c)(b + d)]2 ≥ 2[4 − 3(a + c)(b + d) + 2(a2 + c2 )(b2 + d2 )], or 3(a + c)2 (b + d)2 − 6(a + c)(b + d) + 4 − 4(a2 + c2 )(b2 + d2 ) ≥ 0, 3[1 − (a + c)(b + d)]2 + 1 − 4(a2 + c2 )(b2 + d2 ) ≥ 0. By the AM-GM inequality, we have 4(a2 + c2 )(b2 + d2 ) ≤ (a2 + b2 + c2 + d2 )2 = 1. The inequality is proved. Equality holds if and only if a = b = c = d = ± 21 . Example 6. For any real numbers a, b, c, prove that b2 − ca c2 − ab a2 − bc + + ≥ 0. a2 + 2b2 + 3c2 b2 + 2c2 + 3a2 c2 + 2a2 + 3b2 Solution: Rewrite the inequality as X cyc

or

X cyc

2

X cyc

4(a2 − bc) ≥ 0, + 2b2 + 3c2

a2

 4(a2 − bc) + 1 ≥ 3, a2 + 2b2 + 3c2

X (b − c)2 5a2 + c2 + ≥ 3. a2 + 2b2 + 3c2 a2 + 2b2 + 3c2 cyc

X (b − c)2 5a2 + c2 ≥ 0, it suffices to show that ≥ 3, a2 + 2b2 + 3c2 a2 + 2b2 + 3c2 cyc cyc but it follows from the Cauchy Schwartz inequality because Since 2

X cyc

X

5a2 + c2 a2 + 2b2 + 3c2

!

! X (5a2 + c2 )(a2 + 2b2 + 3c2 ) cyc

#2

" ≥

X

(5a2 + c2 )

cyc

!2 = 36

X cyc

56

a2

.

and !2 12

X

a2

−

X X X a2 b2 ≥ 0. a4 − 4 (5a2 + c2 )(a2 + 2b2 + 3c2 ) = 4 cyc

cyc

cyc

cyc

Equality holds if and only if a = b = c. Example 7. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that the following inequality holds 1 1 1 3(a + b + c)2 + + ≥ . a2 + bc b2 + ca c2 + ab 2(a2 + b2 + c2 )(ab + bc + ca) Solution: The inequality is equivalent to each of the following inequalities X a2 + b2 + c2 a2 + bc

cyc

X  a2 + b2 + c2 a2 + bc

cyc

3(a + b + c)2 − 3, 2(ab + bc + ca)

 −1

X b2 + c2 − bc a2 + bc

cyc

3(a + b + c)2 , 2(ab + bc + ca)

≥

≥

3(a2 + b2 + c2 ) . 2(ab + bc + ca)

≥

Now, by Cauchy Scwhartz inequality, we have that !2 X X 2 a2 − ab X b2 + c2 − bc cyc cyc ≥ X a2 + bc (a2 + bc)(b2 + c2 − bc) cyc cyc

!2 2

X

2

a −

cyc

=

! X

+

ab

ab

cyc

!2 X

X

cyc

ab

.

! X

cyc

cyc

a2

− 4abc

X

a

cyc

It suffices for us to prove that !2 2

X

a2 −

cyc

!2 X cyc

ab

X

ab

cyc

! +

X cyc

ab

≥

! X

a2

− 4abc

cyc

X

3(a2 + b2 + c2 ) . 2(ab + bc + ca)

a

cyc

To prove this inequality, we will use the pqr technique (refer to the appropriate paragraph for a general description of the method). Because the inequality is homogeneous, we can assume that a+b+c = 1. and set q = ab+bc+ca, r = abc. Then this inequality becomes (2 − 5q)2 3(1 − 2q) ≥ , 2 q + q(1 − 2q) − 4r 2q 57

or

(2 − 5q)2 3(1 − 2q) ≥ . 2 q − q − 4r 2q

If 1 ≥ 4q, then we have (2 − 5q)2 3(1 − 2q) (2 − 5q)2 3(1 − 2q) (5 − 11q)(1 − 4q) − − ≥ = ≥ 0. q − q 2 − 4r 2q q − q2 2q 2q(1 − q) If 4q ≥ 1, then from Schur’s inequality for fourth degree, we see that r ≥ (4q−1)(1−q) . Hence, 6 3(2 − 5q) (2 − 5q)2 (2 − 5q)2 = ≥ . (4q−1)(1−q) 2 q − q 2 − 4r 1−q q−q −4· 6 It remains to prove that 3(1 − 2q) 3(2 − 5q) ≥ , 1−q 2q which can be easily simplified to 3(4q − 1)(1 − 3q) ≥ 0. 2q(1 − q) a1 a2 an + + ... + ≤ k, bi > 0 b1 b2 bn ai ∀n = 1, 2, . . . , n. We can prove this inequality by adding bi to mi such that ai + mi bi is a complete square of a expression, then we can apply the CauchySchwarts to each of these terms to obtain an upper bound. Example 8. Let a, b, c be real numbers. Prove that the following inequality holds a2 − bc b2 − ca c2 − ab + + ≥ 0. 4a2 + 4b2 + c2 4b2 + 4c2 + a2 4c2 + 4a2 + b2 Solution: Rewrite the inequality in the form Sometimes one has to prove the inequality

4(ca − b2 ) 4(ab − c2 ) 4(bc − a2 ) + + ≤ 0, 4a2 + 4b2 + c2 4b2 + 4c2 + a2 4c2 + 4a2 + b2 or 

     4(bc − a2 ) 4(ca − b2 ) 4(ab − c2 ) +1 + +1 + + 1 ≤ 3, 4a2 + 4b2 + c2 4b2 + 4c2 + a2 4c2 + 4a2 + b2

or

(2b + c)2 (2c + a)2 (2a + b)2 + + ≤ 3. 4a2 + 4b2 + c2 4b2 + 4c2 + a2 4c2 + 4a2 + b2 Now, using Cauchy Schwartz inequality, we have (2b + c)2 4a2 + 4b2 + c2

= ≤ =

(b + b + c)2 (a2 + 2b2 ) + (a2 + 2b2 ) + (c2 + 2a2 ) b2 b2 c2 + + a2 + 2b2 a2 + 2b2 c2 + 2a2 2b2 c2 + . a2 + 2b2 c2 + 2a2 58

Similarly, we have (2c + a)2 4b2 + 4c2 + a2 (2a + b)2 2 4c + 4a2 + b2

≤ ≤

2c2 a2 + , b2 + 2c2 a2 + 2b2 2a2 b2 + 2 . 2 2 c + 2a b + 2c2

Adding up these inequalities, we get the result. Make it symmetric It is usually the case that a symmetric inequality is easier than cyclic inequality. This technique is based on this intuitive idea. We will use the Cauchy Schwartz-Holder inequality to make the inequality symmetric - and then it will be easier to prove. Example 9. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that r r r a b c 3 + + ≤√ . a+b b+c c+a 2 Solution: By the Cauchy Schwartz inequality, we have !2 " #" # Xr a X X a ≤ (a + c) a+b (a + b)(a + c) cyc cyc cyc ! ! X X 4 a ab =

cyc

cyc

(a + b)(b + c)(c + a)

.

Moreover, by the AM-GM inequality, ! ! X X 8 (a + b)(b + c)(c + a) = a ab − abc ≥ 9 cyc cyc Hence Xr a a+b cyc

!2

! X

a

cyc

! X

ab .

cyc

9 ≤ . 2

Equality holds if and only if a = b = c. Example 10. Let a, b, c be positive real numbers. Prove that r r r a b c + + ≤ 1. 4a + 4b + c 4b + 4c + a 4c + 4a + b Solution: By the Cauchy Schwartz inequality, we have !2 " #" # Xr X X a a ≤ (4a + b + 4c) 4a + 4b + c (4a + 4b + c)(4a + b + 4c) cyc cyc cyc ! ! X X X 2 9 a a +8 ab =

cyc

cyc

cyc

(4a + 4b + c)(4b + 4c + a)(4c + 4a + b) 59

.

It suffices to prove that ! ! X X X 2 ab ≤ (4a + 4b + c)(4b + 4c + a)(4c + 4a + b), 9 a a +8 cyc

cyc

cyc

or 7

X

a3 + 3

X

ab(a + b) ≥ 39abc.

cyc

cyc

The last inequality is trivial by the AM-GM inequality. Equality holds if and only if a = b = c. Substitutions This technique is simply the application of the ubiquitous substitution method to the Cauchy-Schwartz and Holder inequalities. Example 11. Let a, b, c be positive real numbers. Prove that a3 b3 c3 + + ≥ 1. a3 + abc + b3 b3 + abc + c3 c3 + abc + a3 b a c Solution: Set x = , y = , z = , then x, y, z > 0, xyz = 1 and the inequality a c b becomes X 1 ≥ 1, x 3 +1 cyc x + y or X 1 ≥ 1, 3 + x2 z + 1 x cyc which can be rewritten as X cyc

x2

yz ≥ 1. + yz + zx

By the Cauchy Schwartz inequality, we have !2 X

X cyc

yz yz cyc ≥X = 1. x2 + yz + zx yz(x2 + yz + zx) cyc

Example 12. Let a, b, c be positive real numbers. Prove that r r r a2 b2 c2 + + ≥ 1. a2 + 7ab + b2 b2 + 7bc + c2 c2 + 7ca + a2 b c a Solution: Set x = , y = , z = , then x, y, z > 0, xyz = 1 and the inequality a b c becomes X 1 √ ≥ 1. 2 x + 7x + 1 cyc 60

Again, since x, y, z > 0, xyz = 1 then there exist m, n, p > 0 such that x = n2 p2 p2 m2 m2 n2 , y = , z = , the inequality is transformed into m4 n4 p4 X cyc

m4 p ≥1 m8 + 7m4 n2 p2 + n4 p4

By Holder’s inequality, we have !2 "

X

m4

cyc

p m8 + 7m4 n2 p2 + n4 p4

!3

# X

8

4 2 2

4 4

m(m + 7m n p + n p ) ≥

X

cyc

3

m

cyc

It suffices to prove that !3 X

m

3

≥

cyc

X

m(m8 + 7m4 n2 p2 + n4 p4 ),

cyc

or X

(5m6 n3 + 2m3 n3 p3 − 7m5 n2 p2 ) +

sym

X

(m6 n3 − m4 n4 p) ≥ 0.

sym

The last inequality is obviously true from the AM-GM inequality. Example 13. Let a, b, c be positive real numbers. Prove that 1 1 3 1 + + ≥ √ . 3 2 2 2 a(a + b) b(b + c) c(c + a) 2 a b c Solution: Due to the homogeneity, we may assume abc = 1, then there exist x, y, z > 0 such that a = xy , b = xz , c = yz . The inequality becomes X cyc

3 y2 ≥ . 2 x + yz 2

By the Cauchy Schwartz inequality, we have !2 X

X cyc

y y2 cyc X ≥X . x2 + yz x2 y 2 + y3z cyc

Moreover

!2 X

y

2

cyc

2

−3

X

x2 y 2 =

cyc

cyc

1X 2 (x − y 2 )2 ≥ 0, 2 cyc

and !2 X cyc

y2

−3

X cyc

y3z =

1X 2 (x − z 2 − 2xy + yz + zx)2 ≥ 0. 2 cyc 61

.

Therefore !2 X

X cyc

!2 X

2

y y2 cyc X ≥X ≥ x2 + yz x2 y 2 + y3z cyc

cyc

y

2

3 !2 = . 2

cyc !2 1 3

X

y2

+

cyc

1 3

X

y2

cyc

Equality holds if and only if a = b = c. The CYH technique This is the most advanced technique that we want to present in this paper. It may not be applicable to such a wide range of inequalities as others - however it solves some very hard problems. Here is, for example, an inequality of Jack Garfunkel, famous for creating many difficult inequalities. Example 14. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that a b c 5√ √ +√ +√ ≤ a + b + c. 4 c+a a+b b+c Solution: By the Cauchy Schwartz inequality, we have X cyc

a √ a+b

!2

"

#" X

a ≤ a(5a + b + 9c) (a + b)(5a + b + 9c) cyc cyc !2 " # X X a = 5 a . (a + b)(5a + b + 9c) cyc cyc X

#

It suffices to prove that X

a

!" X cyc

cyc

# a 5 ≤ . (a + b)(5a + b + 9c) 16

This holds since since 5 − 16

X

a

!" X

cyc

cyc

# a A+B = . (a + b)(5a + b + 9c) C

where A =

X

ab(a + b)(a + 9b)(a − 3b)2 ≥ 0,

cyc

B = 243

X cyc

a3 b2 c + 835

X

a2 b3 c + 232

cyc

X

a4 bc + 1230a2 b2 c2 ≥ 0,

cyc

C = 16(a + b)(b + c)(c + a)(5a + b + 9c)(5b + c + 9a)(5c + a + 9b) > 0. Equality holds if and only if a3 = 1b = 0c or any cyclic permutations (of course, the expression 0c in the equality simply means c = 0). 62

Of course, this solution has a strategy behind it: Similar to the symmetric technique, we want to apply the Cauchy Schwartz inequality like this

X cyc

a √ a+b

!2 ≤

" X

#" X a(ma + nb + pc)

cyc

cyc

# a . (a + b)(ma + nb + pc)

where m, n, p are nonnegative real numbers that we will choose later. Now, note that the original inequality has equality for a = 3, b = 1, c = 0 then we must choose m, n, p so that this step, the equality also holds for this point. Moreover, due to the equality of Cauchy Schwartz inequality, this step, we have equality if and only if p p p a(ma + nb + pc) b(mb + nc + pa) c(mc + na + pb) r =r . =r a c b (a + b)(ma + nb + pc) (c + a)(mc + na + pb) (b + c)(mb + nc + pa) We must choose m, n, p such that this equation has a root (a, b, c) = (3, 1, 0), that is p p p 3 · (3 · m + 1 · n + 0 · p) 1 · (1 · m + 0 · n + 3 · p) 0 · (0 · m + 3 · n + 1 · p) r =r =r , 3 1 0 (3 + 1)(3 · m + 1 · n + 0 · p) (1 + 0)(1 · m + 0 · n + 3 · p) (0 + 3)(0 · m + 3n˙ + 1 · p) or 2(3m + n) = m + 3p, or 5m + 2n = 3p. Moreover, note that if the expression

X cyc

a(ma + nb + pc) = m

X

a2 + (n +

cyc

!2 p)

X cyc

ab has the form k

X

a

, then after using the Cauchy Schwartz in-

cyc

equality, it will be simpler thus it might be easier to proceed. Hence, let us choose m, n, p such that n + p = 2m. Now, due to the homogeneity, we can choose m = 5, n = 1, p = 9. That is the reason why we have the above solution for this very hard inequality. Unfortunately, the second part of the proof does not have such a general method behind it. Still, this method of choosing m, n, p can reduce the inequality to something one may have more freedom to tackle. For example, with the the same Jack Garfunkel’s inequality, we can proceed in a slightly different way : From the same Cauchy Schwartz inequality (this time for (m, n, p) = (3, 3, 1)), 63

we have that a √ a+b cyc

X

!2

! a ≤ a+b 3a + 3b + c cyc cyc " # ! X X a(a + b + c) X a = 2 a+ a+b 3a + 3b + c cyc cyc cyc ! ! ! X a X X a = a +2 , 3a + 3b + c a+b cyc cyc cyc "

X a(3a + 3b + c)

#

X

hence it suffices for us to prove that a 3a + 3b + c cyc

X

!

X a +2 a + b cyc

! ≤

25 . 16

The detailed proof of the last inequality can be found in the following book: Old And New Inequalities 2, Vo Quoc Ba Can - Cosmin Pohoata, GIL publishing house, 2008. Example 15. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that   1 1 1 9 (ab + bc + ca) + + ≥ . 2 2 2 (b + c) (c + a) (a + b) 4 Solution: The appearance of

X cyc

1 suggests the Cauchy Schwartz in(b + c)2

equality (since it is the sum of squares). Let us use the above idea to attack this inequality. By the Cauchy Schwartz inequality, we have " X (ma + nb + nc)2 cyc

#" X cyc

1 (b + c)2

# ≥

X ma + nb + nc cyc

b+c

!2 .

The equality holds if and only if ma + nb + nc mb + nc + na mc + na + nb = = . 1 1 1 b+c c+a a+b We notice that the original inequality has an equality case for a = b = 1, c = 0, hence the above solution must be satisfied at this point, that is m+n m+n 2n = = 1 ⇔ m = 3n ⇒ m = 3, n = 1. 1 1 2 And now, we have the solution as follows: 64

By the Cauchy Schwartz inequality, we have # " # !" #" X X X X X 1 1 2 2 = 11 a + 14 ab (3a + b + c) (b + c)2 (b + c)2 cyc cyc cyc cyc cyc !2 X 3a + b + c ≥ b+c cyc !2 X a = 9 1+ . b+c cyc It suffices to prove that

4 1+

X cyc

a b+c

11

!2 ≥

P a2 + 14 ab cyc cyc P . ab P

cyc

Due to the homogeneity, we may assume a + b + c = 1, setting q =

X

ab, r =

cyc



 4q − 1 abc, then by Schur’s inequality for third degree, we obtain r ≥ max 0, . 9 The inequality becomes  4

1+q −2 q−r

2 ≥

11 − 8q . q

If 1 ≥ 4q, then  4

1+q −2 q−r

2

11 − 8q − ≥4 q



2 1+q 11 − 8q (4 − 3q)(1 − 4q) −2 − = ≥ 0. q q q2

If 4q ≥ 1, then 2

  4

2 1+q 11 − 8q −2 − q−r q

1+q 11 − 8q  − 2 − 4q − 1 q q− 9 (1 − 3q)(4q − 1)(11 − 17q) ≥ 0. q(5q + 1)2

 ≥ 4 =

Equality holds if and only if a = b = c or a = b, c = 0 and any cyclic permutations. Example 16. Let a, b, c, d be positive real numbers such that   1 1 1 1 (a + b + c + d) + + + = 20. a b c d Prove that 2

2

2

2

(a + b + c + d )



1 1 1 1 + 2+ 2+ 2 2 a b c d 65

 ≥ 36.

Solution: By the Cauchy Schwartz inequality, we have X 1 a2 cyc

!" X

#

Xb+c+d−a

(b + c + d − a)2 ≥

cyc

!2 = 144.

a

cyc

Moreover, we see that 4

X

a2 =

X

cyc

(b + c + d − a)2 .

cyc

Sometimes, the above way to choose m, n, p, . . . cannot be used, and we need to another way to choose them. A nice idea to apply the Cauchy Schwartz inequality is by taken note at some well-known inequalities or taken note at the special form of the given inequalities. Here are some examples. Example 17. Let a, b, c be positive real numbers. Prove that for all k ≥ 2, we have s X Xp X 2 2 a + kab + b ≤ 4 a2 + (3k + 2) ab. cyc

cyc

cyc

Solution: By the Cauchy Schwartz inequality, we have !2 Xp a2 + kab + b2

"

# X

≤

(a + b)

X a2 + kab + b2

cyc

cyc

X

= 2

a+b

cyc

! a

cyc

X a2 + kab + b2

!

a+b

cyc

.

We need to prove that 4 2

X a2 + kab + b2 cyc

a+b

X

a2 + (3k + 2)

cyc

≤

X

ab

cyc

X

,

a

cyc

This inequality is equivalent to each of the following 3(k − 2) X X ab X 2 (a + b) + 2(k − 2) ≤4 a+ a+b cyc cyc cyc X

ab X ab cyc 2 ≤ X , a + b a cyc cyc

2

X ab(a + b + c) cyc

a+b 66

≤3

X cyc

ab,

X cyc

X cyc

3

!

a

ab ,

2abc

X cyc

X 1 ≤ ab. a+b cyc

And the last inequality follows from the Cauchy Schwartz inequality  X X 1 X 1 1 ab. ≤ 2abc + = 2abc a+b 4a 4b cyc cyc cyc Equality holds if and only if a = b = c. Example 18. Let a, b, c be positive real numbers such that a + b + c = 1. Prove that r a b c 3 √ +√ +√ . ≤ 2 c + 2a a + 2b b + 2c Solution: At this point, we can easily show that a b c 1 + + ≤ , 2a + 4b + c 2b + 4c + a 2c + 4a + b 2 with equality holding when abc = 0. (this is left as an exercise) Thus, using the Cauchy Schwartz inequality, we have that !2 !" # X X X a(2a + 4b + c) a a √ ≤ 2a + 4b + c a + 2b a + 2b cyc cyc cyc ≤

1 X a(2a + 4b + c) 2 cyc a + 2b

=

1 2

≤

1 2

! X X ca 2 a+ a + 2b cyc cyc ! X X ca 3 3X a= . 2 a+ = a 2 cyc 2 cyc cyc

Exercises 1. Let x, y ≥ 0 such that x3 + y 3 = 1. Prove that r 5 √ √ √ 6 5 x+2 y ≤ 1+2 2 . Solution: By Holder’s inequality, we have that for all a, b ≥ 0,
5 3
√ √
6 a6 + b6 x + y 3 ≥ a5 x + b5 y . Hence, 5√

5√

q q 6 6 5 6 6 3 3 a x + b y ≤ (a + b ) (x + y ) = (a6 + b6 )5 . √ Choosing a = 1, b = 5 2, then we have s 5  √ 6 5 r √ √ √ 6 6 5 5 x+2 y ≤ 16 + 2 = 1+2 2 .

67

2. Let a, b, c ≥ 0 such that a2 + b2 + c2 = 1. Find the minimum value of P = a3 + 3b3 + 2c3 . Solution: By Holder’s inequality, we have that for all m, n, p ≥ 0,  3 √ √ 3 3 (a3 + 3b3 + 2c3 )2 (m3 + n3 + p3 ) ≥ a2 m + b2 n 9 + c2 p 4 . Choosing m = 1, n =

1 √ 3 ,p 9

1 √ 3 , 4

=

then we have (a2 + b2 + c2 )3 36 = , 1 1 49 1+ + 9 4

(a3 + 3b3 + 2c3 )2 ≥

or

6 P = a3 + 3b3 + 2c3 ≥ . 7

We have equality when √ √ a b33 c32 = = , m n p or

√ √ a2 b2 3 9 c2 3 4 = = = m2 n2 p2

or

6 a= , 7

36 a2 + b2 + c2 = , 2 2 49 n p m2 + √ +√ 3 3 9 4

6 b= √ , 739

Hence

6 c= √ . 734

6 min P = . 7

3. Given a, b, c ≥ 0 satisfying a + b + c = 3. Determine the minimum value of P = a4 + 2b4 + 4c4 . Solution: By Holder’s inequality, we have that for all m, n, p ≥ 0,  4 √ √ 4 4 (a4 + 2b4 + 3c4 )(m4 + n4 + p4 )3 ≥ am3 +bn3 2+cp3 3 . Choosing m = 1, n = P

1

√ 12

2

,b =

1

√ 12

3

, then we have (a + b + c)4  4  4 #3 1 1 √ √ 14 + 12 + 12 2 3

= a4 + 2b4 + 3c4 ≥ "

=

1   . 1 1 3 1+ √ +√ 3 3 2 3 68

We have equality when √ √ a b42 c43 a+b+c 1 = = = = , n p 1 1 m n p √ m+ √ + √ √ 1 + + 4 4 3 3 2 3 2 3 or a=

1 , 1 1 √ 1+ √ + 3 3 2 3

Hence

b=

√ 3

1  , 1 1 2 1+ √ +√ 3 3 2 3

c=

√ 3

1  . 1 1 3 1+ √ +√ 3 3 2 3

1 min P =   . 1 1 3 1+ √ +√ 3 3 2 3

4. Let x1 , x2 , . . . , xn be real numbers. Prove that 1

x21 +(x1 +x2 )2 +. . .+(x1 +x2 +. . .+xn )2 ≤

2

2

Solution: By Cauchy Schwartz inequality, we have that for all ci > 0, n k X X k=1

!2

n X

≤

xi

i=1

k=1 " n X

=

k X

"

k X x2

! ci

i=1

k X x2 i=1

ci

!#

i

k=1

!#

i

i=1

Sk

ci

=

n X Si + . . . + Sn

ci

k=1

x2i .

Choosing ci such that S2 + . . . + Sn Sn S1 + S2 + . . . +Sn = = ... = , c1 c2 cn then we obtain ci = sin

iπ (i − 1)π − sin , and 2n + 1 2n + 1

S1 + S2 + . . . +Sn S2 + . . . + Sn Sn 1 = = ... = = . π 2 c1 c2 cn 4 sin 2(2n + 1) Thus n k X X k=1

i=1

!2 xi

≤

1 4 sin2

π 2(2n + 1)

n X

x2i .

k=1

5. Let a, b, c ≥ 0 such that a + b + c = 1. Determine the minimum value of r r r 1 1 1 P = a2 + 2 + b2 + 2 + c2 + 2 . b c a 69

2

·(x1 +x2 +. . .+xn ). π 4 sin 2(2n + 1) 2

Solution: We guess that the minimum attains equality when a = b = c = 1 . Hence, by Cauchy Schwartz Inequality, we have that for all m, n ≥ 0, 3 s  1 n a2 + 2 (m2 + n2 ) ≥ ma + , b b s  n 1 2 b + 2 (m2 + n2 ) ≥ mb + , c c s  1 n c2 + 2 (m2 + n2 ) ≥ mc + . a a Therefore   p 1 1 1 2 2 P m + n ≥ m(a + b + c) + n + + a b c 9n = m + 9n, ≥ m(a + b + c) + a+b+c or

m + 9n P ≥√ . m2 + n2 Choosing m, n > 0 such that

1 a 1 b 1 c 1 a=b=c= , = , = , = . 3 m bn m cn m an Then n = 9m, and we may choose m = 1, n = 9 to obtain √ 1+9·9 P ≥√ = 82. 12 + 92 √ Thus min P = 82. 6. Let a, b, c > 0 such that a2 + b2 + c2 = 3. Prove that 1 1 1 + + ≥ 3. 2−a 2−b 2−c Solution: The inequality is equivalent to       2 2 2 −1 + −1 + − 1 ≥ 3, 2−a 2−b 2−c or a b c + + ≥ 3. 2−a 2−b 2−c By the Cauchy Schwartz inequality and AM-GM inequality, we have a b c + + 2−a 2−b 2−c

a4 b4 c4 + + 2a3 − a4 2b3 − b4 2c3 − c4 (a2 + b2 + c2 )2 ≥ 2a3 + 2b3 + 2c2 − a4 − b4 − c4 (a2 + b2 + c2 )2 ≥ (a4 + a2 ) + (b4 + b2 ) + (c4 + c2 ) − a4 − b4 − c4 = a2 + b2 + c2 = 3. =

70

7. Let a, b, c be the side lengths of a triangle. Prove that a(a − b) b(b − c) c(c − a) + + ≥ 0. a2 + 2bc b2 + 2ca c2 + 2ab Solution: The inequality is equivalent to  X  a(a − b) + 1 ≥ 3, a2 + 2bc cyc or

X 2a2 − ab + 2bc a2 + 2bc

cyc

≥ 3.

Since a, b, c are the side lengths of a triangle, we have c ≥ b − a, then 2a2 − ab + 2bc ≥ 2a2 − ab + 2b(b − a) = 2(a − b)2 + ab ≥ 0. Hence, by the Cauchy Schwartz inequality, we have #2

" X X 2a2 − ab + 2bc cyc

a2

+ 2bc

≥X

(2a2 − ab + 2bc)

cyc

(2a2 − ab + 2bc)(a2 + 2bc)

cyc

It suffices to prove that #2

" X

(2a2 − ab + 2bc)

≥3

X (2a2 − ab + 2bc)(a2 + 2bc), cyc

cyc

or equivalently, X X X X X 7 a3 b + 4 ab3 ≥ 2 a4 + 3 a2 b2 + 6 a2 bc. cyc

cyc

cyc

cyc

cyc

Again, since a, b, c are the side lengths of a triangle, there exist x, y, z > 0 such that a = y + z, b = z + x, c = x + y. The inequality becomes X X X X X 2 x4 + 2 xy(x2 + y 2 ) + 3 xy 3 ≥ 6 x2 y 2 + 3 x2 yz, cyc

cyc

cyc

cyc

But it follows from the AM-GM inequality since X X X X 2 x4 ≥2 x2 y 2 , 2 xy(x2 + y 2 ) ≥ 4 x2 y 2 , cyc

cyc

cyc

cyc

cyc

3

X cyc

xy 3 ≥ 3

X cyc

Equality holds if and only if a = b = c. 8. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that 2a2 − bc 2b2 − ca 2c2 − ab + + ≥ 3. b2 − bc + c2 c2 − ca + a2 a2 − ab + b2 71

x2 yz.

Solution: Rewrite the inequality as  X  2a2 − bc + 1 ≥ 6, b2 − bc + c2 cyc or X 2a2 + (b − c)2 b2 − bc + c2

cyc

≥ 6.

By the Cauchy Schwartz inequality, we have #2

" X X 2a2 + (b − c)2 cyc

≥X

b2 − bc + c2

[2a2 + (b − c)2 ]

cyc 2

[2a + (b − c)2 ](b2 − bc + c2 )

.

cyc

It suffices to prove that #2

" X

2

2

≥6

[2a + (b − c) ]

X

[2a2 + (b − c)2 ](b2 − bc + c2 ),

cyc

cyc

or 2

X

a4 + 2abc

cyc

X cyc

a+

X

ab(a2 + b2 ) ≥ 6

cyc

X

a2 b2 ,

cyc

that is 2

X

a2 (a − b)(a − c) + 3

cyc

X

ab(a − b)2 ≥ 0.

cyc

which is true by Schur’s inequality for fourth degree. Equality holds if and only if a = b = c or a = b, c = 0 and its cyclic permutations. 9. Let a, b, c be nonnegative real numbers, not all are zero. Prove that 3a2 − bc 3b2 − ca 3c2 − ab 3 + + ≤ . 2a2 + b2 + c2 2b2 + c2 + a2 2c2 + a2 + b2 2 Solution: Rewrite the inequality as X cyc

 2(3a2 − bc) 3− 2 ≥ 6, 2a + b2 + c2

or X 3b2 + 2bc + 3c2 cyc

2a2 + b2 + c2

≥ 6,

that is 3

X cyc

X (b − c)2 bc + 8 ≥ 6. 2 2 2 2 2a + b + c 2a + b2 + c2 cyc 72

If (a−b)2 +(b−c)2 +(c−a)2 = 0, it is trivial. If (a−b)2 +(b−c)2 +(c−a)2 > 0, then by the Cauchy Schwartz inequality, we have " #2 X 2 (b − c) X (b − c)2 cyc ≥ X 2 + b2 + c2 2a (b − c)2 (2a2 + b2 + c2 ) cyc cyc

!2 X

4

a2 −

X

=

"

# X

(b − c)

ab

cyc

cyc

! X

2

cyc

2

a

+

cyc

X

a2 (b − c)2

cyc

!2 2

X

2

a −

! X

a2 −

X

cyc

ab

cyc

cyc

=

X

,

! X

ab

cyc

a2

X

+

cyc

b2 c2 −

cyc

X

a2 bc

cyc

and !2

!2 X

X

X cyc

bc bc cyc = ≥X 2a2 + b2 + c2 bc(2a2 + b2 + c2 )

bc

cyc

! X

cyc

.

! X

bc

cyc

a2

+

X

cyc

a2 bc

cyc

It suffices to prove that !2

!2 X

6

2

a −

cyc

cyc

2

a −

X cyc

ab

8

ab

cyc

! X

X

cyc

2

a

+

X cyc

2 2

b c −

X

2

a bc

bc

cyc

+

! X

X

! X

cyc

bc

cyc

≥ 6.

! X

2

a

+

cyc

Due X to the homogeneity, we may assume a + b + c = 1. Putting q = bc, r = abc, then the inequality becomes cyc

3(1 − 3q)2 4q 2 + ≥ 3, (1 − 3q)(1 − 2q) + q 2 − 3r q(1 − 2q) + r or

(3r + q − 4q 2 )2 ≥ 0, (1 − 5q + 7q 2 − 3r)(q − 2q 2 + r)

which is true. 10. Let a, b, c be the side lengths of a triangle. Prove that a(b + c) b(c + a) c(a + b) + + ≤ 2. a2 + 2bc b2 + 2ca c2 + 2ab 73

X cyc

2

a bc

Solution: The inequality is equivalent to  X a(b + c) 1− 2 ≥ 1, a + 2bc cyc or

X a2 − ab − ac + 2bc

≥1

a2 + 2bc

cyc

a2

We will show that − ab − ac + 2bc ≥ 0, or a(a − b) + (2b − a)c ≥ 0. Indeed, if 2b ≥ a, we have a(a − b) + (2b − a)c ≥ a(a − b) + (2b − a)(b − a) = 2(a − b)2 ≥ 0. If a ≥ 2b, we have a(a − b) + (2b − a)c ≥ a(a − b) + (2b − a)(a + b) = 2b2 ≥ 0. Now, by the Cauchy Schwartz inequality, we have " #2 X
2 a − ab − ac + 2bc X a2 − ab − ac + 2bc cyc ≥ X
2 + 2bc a a2 − ab − ac + 2bc (a2 + 2bc) cyc cyc

!2 X =

a2

cyc

X

.
a − ab − ac + 2bc (a2 + 2bc) 2

cyc

It suffices to show that !2 X X
≥ a2 a2 − ab − ac + 2bc (a2 + 2bc), cyc

cyc

which can be easily simplified to X ab(a − b)2 ≥ 0. cyc

which is obviously true. Equality holds if and only if a = b = c or a = b, c = 0 and its cyclic permutations. 11. Let a, b, c be the side lengths of a triangle. Prove that a b c ab + bc + ca 5 + + + 2 ≤ . 2 2 b+c c+a a+b a +b +c 2 Solution: The inequality is equivalent to X

X cyc

a 1− b+c



ab 1 cyc ≥ +X , 2 a2 cyc

74

or

!2 X Xb+c−a cyc

b+c

a

cyc

≥ 2

X

a2

.

cyc

By the Cauchy Schwartz inequality, we have ! ! " # ! X Xb+c−a X Xb+c−a 2 2 a = (b + c)(b + c − a) b+c b+c cyc cyc cyc cyc " #2 !2 X X ≥ (b + c − a) = a . cyc

cyc

Equality holds if and only if a = b = c or a = b, c = 0 or any cyclic permutations. 12. Let a, b, c be the side lengths of a triangle. Prove that b c a + + ≥ 1. 3a − b + c 3b − c + a 3c − a + b Solution: We have prove that

4a a+b−c = + 1. Hence, it suffices to 3a − b + c 3a − b + c

a+b−c b+c−a c+a−b + + ≥ 1, 3a − b + c 3b − c + a 3c − a + b which is obviously true since by Cauchy Schwartz inequality, we have a+b−c b+c−a c+a−b + + 3a − b + c 3b − c + a 3c − a + b

≥

(a + b + c)2 X

(a + b − c)(3a − b + c)

cyc

=

(a + b + c)2 X X = 1. a2 + 2 ab cyc

cyc

13. Let x1 , x2 , . . . , xn be positive real numbers such that

n X

xi = 1. Prove

i=1

that the following inequality holds n q X x2i + x2i+1 ≤ 2 − √

1 . n X 2 x2i + 2 xi+1

i=1

i=1

Solution: The original inequality is equivalent to  n  q X 1 2 2 xi + xi+1 − xi + xi+1 ≥ √ , n X 2 x2i i=1 + 2 xi+1 i=1

75

or

n X i=1

x2i xi+1

x2i 1 . ≥ q n X xi √ x2i 2 2 + xi + xi + xi+1 2+2 xi+1 xi+1 i=1

By the Cauchy Schwartz inequality, we have that n X n X i=1

x2i xi+1

x2i xi q 2 xi + x2i+1 + xi + xi+1

≥

!2 xi

i=1

 n  X xi q 2 x2i 2 + xi + xi + xi+1 xi+1 xi+1 i=1

=

1 n  X i=1

x2i xi q 2 + xi + xi + x2i+1 xi+1 xi+1

It suffices to prove that  n  X √ xi q 2 x2i 2 xi + xi+1 − ≤ 2 − 1, xi+1 xi+1 i=1

or

n X i=1

√ xi xi+1 q ≤ 2 − 1. x2i + x2i+1 + xi

q xi + xi+1 √ Since x2i + x2i+1 ≥ , it remains to show that 2 √ n X 2 xi xi+1 √
, ≤1− 2 1 + 2 xi + xi+1 i=1

or n X i=1

that is

n

X xx √ i
i+1 ≤ 1 + 2 xi + xi+1 i=1 n X i=1

! √ √ 3−2 2 2−1 xi + xi+1 , 2 2

√


2 − 1 (xi − xi+1 )2 √
  ≥ 0. 2 1 + 2 xi + xi+1

The proof is complete. Equality holds if and only if x1 = x2 = . . . = 1 xn = . n 14. Let a, b, c be positive real numbers. Prove that a4 b4 c4 a3 + b3 + c3 + + ≥ . a2 + ab + b2 b2 + bc + c2 c2 + ca + a2 a+b+c Solution: Rewrite the inequality in the form  X a4 a3 + b3 + c3 2 + ab + bc + ca − a2 − b2 − c2 , + ab − a ≥ 2 + ab + b2 a a + b + c cyc 76

.

or X cyc

a2

3abc ab3 ≥ , 2 + ab + b a+b+c

or

b2 3abc ≥ . 2 2 a + b+c a + b + ab cyc ab By the Cauchy Schwartz inequality, we have X

X cyc

(a + b + c)2 abc(a + b + c) b2 ≥ = . 2 2 2 2 X ab + bc + ca a + b + ab a + b + ab ab ab cyc

It suffices to prove that 3abc abc(a + b + c) ≥ , ab + bc + ca a+b+c or (a + b + c)2 ≥ 3(ab + bc + ca), which is obviously true by the AM-GM inequality. 15. Let a, b, c be positive real numbers such that a + b + c = 1. Prove that √

ab bc ca 1 +√ +√ ≤√ . ab + bc bc + ca ca + ab 2

Solution: By the Cauchy Schwartz inequality, we have ab √ ab + bc cyc

!2

X

" =

X√

s a+b·

cyc

" X (a + b)

≤

cyc

= 2

a2 b (a + b)(a + c)

#" X

#2

a2 b (a + b)(a + c) cyc

#

a2 b , (a + b)(a + c) cyc

X

hence it suffices to prove that 4

a2 b ≤ a + b + c, (a + b)(a + c) cyc

X

which is equivalent to 4a2 b(b + c) + 4b2 c(c + a) + 4c2 a(a + b) ≤ (a + b)(b + c)(c + a)(a + b + c), this last one being true, because it can be written as ab(a − b)2 + bc(b − c)2 + ca(c − a)2 ≥ 0. 77

16. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that r r r a2 b2 c2 + + ≤ 1. 4a2 + ab + 4b2 4b2 + bc + 4c2 4c2 + ca + 4a2 Solution: By the Cauchy Schwartz inequality, we have !2 r X a2 ≤ 4a2 + ab + 4b2 cyc # " #" 2 X X a . ≤ (4a2 + ac + 4c2 ) (4a2 + ab + 4b2 )(4a2 + ac + 4c2 ) cyc cyc It suffices to prove that " #" X X 2 2 (4a + ac + 4c ) cyc

cyc

# a2 ≤ 1. (4a2 + ab + 4b2 )(4a2 + ac + 4c2 )

By expanding, we can see the inequality is equivalent to X X X 8 a3 b3 + 8 a4 bc + 3abc ab(a + b) ≥ 66a2 b2 c2 . cyc

cyc

cyc

which follows from the AM-GM inequality. 17. Let a, b, c be positive real numbers. Prove that 1 1 1 3 √ + √ + √ ≥√ . a a+b b b+c c c+a 2abc Solution: Due to the homogeneity, we may z x exist x, y, z > 0 such that a = , b = , c = y x √ X y y p ≥ x(x2 + yz) cyc

assume abc = 1, then there y . The inequality becomes z 3 √ 2

By the Cauchy Schwartz inequality, !2 X X cyc

√ y y

y

cyc

p ≥ Xp , x(x2 + yz) xy(x2 + yz) cyc

and v ! ! u Xp X X u X xy x2 + xy xy(x2 + yz) ≤ t cyc

cyc

=

≤

cyc

cyc

v ! ! u X X X 1 u √ t 2 xy x2 + xy 2 cyc cyc cyc ! !2 X X X 1 2 √ x2 + 3 xy ≤ √ x . 2 2 cyc 3 2 cyc cyc 78

Hence X cyc

√ y y 3 p ≥√ . 2 x(x2 + yz)

Equality holds if and only if a = b = c. 18. Let a, b, c be positive real numbers such that abc = 1. Prove that 1 1 1 + + ≤ 3. a2 − a + 1 b2 − b + 1 c2 − c + 1 Solution: First of all, we will show that for any x, y, z > 0 which satisfying xyz = 1 1 1 1 + + ≥ 1. x2 + x + 1 y 2 + y + 1 z 2 + z + 1 Indeed, since x, y, z > 0 and xyz = 1 then there exist m, n, p > 0 such pm mn np that x = 2 , y = 2 , z = 2 . The inequality is transformed into m n p X cyc

m4 ≥ 1. m4 + m2 np + n2 p2

From the Cauchy Schwartz inequality and the known

X

n2 p 2 ≥

cyc

X

m2 np,

cyc

we have !2 X m4

X cyc

m4

+

m2 np

+

n2 p2

≥X

!2 X

m2

cyc 4

2

2 2

(m + m np + n p )

≥X

cyc

cyc

m2

cyc 4

m +2

X cyc

Back to the original problem, from the statement above, we have X cyc

or X cyc

1 ≥ 1, 1 1 + +1 a4 a2 a4 ≥ 1, a4 + a2 + 1

X 2(a2 + 1) ≤ 4, a4 + a2 + 1 cyc X (a2 + a + 1) + (a2 − a + 1) cyc

X cyc

(a2 + a + 1)(a2 − a + 1)

≤ 4,

X 1 1 + ≤ 4. 2 2 a +a+1 a −a+1 cyc 79

n2 p 2

= 1.

Using the above statement again, we have

X cyc

X cyc

a2

a2

1 ≥ 1, and so +a+1

1 ≤ 3. −a+1

Equality holds if and only if a = b = c = 1. 19. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that (b + c)2 (c + a)2 (a + b)2 + 2 + 2 ≥ 6. a2 + bc b + ca c + ab Solution: By the Cauchy Schwartz inequality, we obtain #" # " " #2 X X (b + c)2 X (b + c)2 (a2 + bc) ≥ (b + c)2 2 + bc a cyc cyc cyc !2 = 4

X

a2 +

X

cyc

ab

.

cyc

Hence, it suffices to show that !2 2

X

2

a +

cyc

X

≥3

ab

cyc

X

(b + c)2 (a2 + bc),

cyc

or equivalently, X X X X 2 a4 + 2abc a+ ab(a2 + b2 ) − 6 a2 b2 ≥ 0. cyc

cyc

cyc

cyc

This inequality follows by summing Schur’s inequality of fourth degree X X X 2 a4 + 2abc a≥2 ab(a2 + b2 ) cyc

cyc

cyc

to the inequality 3

X

ab(a2 + b2 ) ≥ 6

cyc

X

a2 b2 .

cyc

which follows from the AM-GM inequality. Equality holds if a = b = c or a = b, c = 0 and its cyclic permutations. 20. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that 1 1 1 2 √ +√ +√ ≥√ . 2 2 2 ab + bc + ca 4a + bc 4b + ca 4c + ab Solution: By Holder’s inequality, we have !2 " # " #3 X X X 1 √ (b + c)3 (4a2 + bc) ≥ (b + c) 2 + bc 4a cyc cyc cyc !3 X = 8 a . cyc

80

It suffices to prove that !3 X

2

! X

a

cyc

≥

ab

X (b + c)3 (4a2 + bc),

cyc

cyc

or X

X

ab(a3 + b3 ) −

cyc

a2 b2 (a + b) + 14abc

cyc

X

a2 ≥ 0,

cyc

or X

ab(a − b)2 (a + b) + 14abc

X

a2 ≥ 0,

cyc

cyc

which is trivial. Equality holds if and only if a = b, c = 0 and its cyclic permutations. 21. Let a, b, c be nonnegative real numbers, not all are zero. Prove that (b + c)2 (c + a)2 3 (a + b)2 + + ≥ . a2 + 2b2 + 3c2 b2 + 2c2 + 3a2 c2 + 2a2 + 3b2 2 Solution: By the Cauchy Schwartz inequality, we have #2

" X (a + b)(2a + b) X cyc

(a + b)2 a2 + 2b2 + 3c2

cyc

≥

X (2a + b)2 (a2 + 2b2 + 3c2 ) cyc

!2 9

X cyc

=

a2 +

X

ab

cyc

X . (2a + b)2 (a2 + 2b2 + 3c2 ) cyc

It suffices to prove that !2 X

6

cyc

2

a +

X

≥

ab

cyc

X (2a + b)2 (a2 + 2b2 + 3c2 ), cyc

which can be easily simplified to 4

X cyc

a3 b + 2

X

ab3 − 3

cyc

X

a2 b2 + 6abc

cyc

X

a ≥ 0,

cyc

or 2

X cyc

ab(a − b)2 + 2

X

a3 b +

cyc

X cyc

a2 b2 + 6abc

X

a ≥ 0,

cyc

a b c which is obviously true. Equality holds if and only if = = and its 1 0 0 cyclic permutations. 81

22. Let a, b, c be nonnegative real numbers satisfying a + b + c = 1, and moreover from which at least two are nonzero. Prove that p p p 3 a 4b2 + c2 + b 4c2 + a2 + c 4a2 + b2 ≤ . 4 Solution: With the help of the Cauchy Schwartz inequality, we get !2 X p a 4b2 + c2

"

# #" X a(4b2 + c2 ) ≤ a(2b + c) 2b + c cyc cyc # !" X X a(4b2 + c2 ) , = 3 ab 2b + c cyc cyc X

cyc

hence it suffices to prove that 3 a(4b2 + c2 ) b(4c2 + a2 ) c(4a2 + b2 ) ≥ + + , 16(ab + bc + ca) 2b + c 2c + a 2a + b or equivalently, 3 + 4abc 16(ab + bc + ca)



1 1 1 + + 2b + c 2c + a 2a + b

 ≥ 3(ab + bc + ca)

Using again the Cauchy Schwartz inequality we see that 1 1 3 1 + + ≥ = 3, 2b + c 2c + a 2a + b a+b+c and so, we are left to show that 1 + 4abc ≥ ab + bc + ca. 16(ab + bc + ca) 1 Setting q = ab + bc + ca, notice that 0 ≤ q ≤ . From the Schur’s 3 inequality, applied for the fourth degree, we have a4 + b4 + c4 + abc(a + b + c) ≥ ab(a2 + b2 ) + bc(b2 + c2 ) + ca(c2 + a2 ), and thus, in terms of q, we get abc ≥

(4q − 1)(1 − q) . 6

Therefore 1 + 4abc − ab − bc − ca = 16(ab + bc + ca) ≥ =

82

1 − q + 4abc 16x 2 1 − q + (4q − 1)(1 − q) 16q 3 (3 − 8q)(1 − 4q)2 ≥ 0. 48q

23. Let n > 2 and x1 , x2 , . . . , xn > 0 such that n X

n X 1 xk

! xk

k=1

! = n2 + 1.

k=1

Prove that n X x2k

n X 1 x2k

!

k=1

Solution: Put s =

! > n2 + 4 +

k=1

2 . n(n − 1)

n X xk . Then by Cauchy-Schwartz we have that k=1

"

n  X k=1

n 2 s − xk 2

n X 1 x2k

#

!2 n X s − n2 xk ≥ xk k=1 ! ! n #2 " n X1 X n2 − = xk xk 2 k=1 k=1   2 (n2 + 2)2 n2 = . = n2 + 1 − 2 4

!

k=1

Now, notice that n  X

s−

k=1

n

n

n

k=1

k=1

k=1

X n2 X 2 n 2 n2 X 2 xk = xk . xk = ns2 − ns xk + 2 4 4

Hence n X x2k

n2 4

!

k=1

n X 1 x2k

! ≥

k=1

(n2 + 2)2 , 4

It follows that n X k=1

! x2k

n X 1 x2k k=1

! ≥

(n2 + 2)2 . n2

It suffices to prove that (n2 + 2)2 2 > n2 + 4 + , 2 n n(n − 1) which is true because (n2 + 2)2 2 2(n − 2) − n2 − 4 − = 2 > 0. n2 n(n − 1) n (n − 1) Therefore, our inequality is proved. 83

1.10

The principle of extremality and monotonicity

To prove an inequality of the form f (x1 , x2 , . . . , xn ) ≥ (≤)0 it suffices to prove it for the multiset of x1 , x2 , . . . , xn which minimizes (maximizes) f on the given domain. This is actually a very powerful idea, because there are often methods to determine when a set of variables minimize f . However, the bulk of such examples comes from analysis, which is too advanced for the purposes of this chapter (but will be dealt with later). There are still simpler instances of this method, for which we can find the points of minimal/maximal values using elementary methods. For example, to prove the inequality a2 + b2 + c2 ≥ ab + bc + ca we can take the function f (a) = a2 + b2 + c2 − ab − ac − bc as a quadratic function in a and we must prove f (a) ≥ 0. It suffices to prove b+c it for a being the vertex of the parabola a = which minimizes f . But 2  f

b+c 2



3 = (b − c)2 ≥ 0. 4

If the function is monotone the extremal values are always attained on the ends on an interval. Therefore if for each of xi the domain is an interval, it suffices to check the inequality for xi being the extremities of the intervals. We can actually consider just one of the extremities: If f is increasing and the inequality is f ≥ 0 or f is decreasing and the inequality is f ≤ 0 we can take just the smallest extremity, otherwise just the largest extremity. For example, consider the inequality x2 − xy + y 2 ≤ 4 for x, y ∈ [0, 2]. The function f (x, y) = x2 − xy + y 2 y is increasing in both x and y, because its a quadratic trinomial and x > 2 x which is the vertex of g(x) = x2 − xy + y 2 , y > which is the vertex of 2 h(y) = y 2 − yx + x2 . Therefore the inequality must be proven only for the extremities of [1, 2]. However, as the function is increasing, we can consider only the largest extremities, which are x = y = 2, for which we get the value 4. 84

Exercises 1. If a, b, c ∈ [0, 1] then 1 ≤ a + b + c + 3(1 − a)(1 − b)(1 − c) ≤ 3. Solution: If we fix b, c then E = a + b + c + 3(1 − a)(1 − b)(1 − c) is a linear function in a so attains its extremal value at the extremes, so we need to consider only the case when a ∈ {0, 1}. Analogously it suffices to consider that b, c ∈ {0, 1}. This cases are easy to handle to obtain the desired result. 2. If x1 , x2 , x3 , . . . , xn ∈ [0, 1] then x1 + x2 + x3 + . . . + xn − x1 x2 − x2 x3 − . . . − xn x1 ≤

jnk 2

.

Solution: Again this function is linear in all xi , so it suffices to look at the cases xi ∈ {0, 1}. Now assume that there are k ones among x1 , x2 , . . . , xn and n−k zeroes. Then x1 +x2 +. . .+xn = k and x1 x2 +x2 x3 +. . .+xn x1 is the number of pairs of consecutive ones in the sequence. How small can the number of such pairs be? As there are n−k zeroes and each zero enters in two pairs of consecutive numbers, there are at most 2(n − k) pairs of consecutive numbers at least one of which is zero, so there are at least n − 2(n − k) = 2k − n pairs of consecutive ones. So our sum is at most x1 + x2 + . . . +xn = k and at most k − (2k − n) = n − k, so at most max{k, n − k} ≤ n2 .   3. If a, b, c ∈ [0, 1]; p, q, r ∈ 0, 12 ; a + b + c = p + q + r = 1 then 1 abc ≤ (pa + qb + rc). 8 Solution: we have r = 1 − p − q so the expression 1 1 abc − (pa + qb + rc) = abc − [pa + qb + (1 − p − q)c] 8 8 is linear in q for fixed p, and so takes its minimal value at the extremities. 1 1 1 The extremities are − p (which yields r = ) and . So it suffices to 2 2 2 1 look only at this case, which gives us that one of q, r is . 2 1 1 Analogously we can suppose one of p, q is and one of q, r is . It’s 2 2 1 easy to see that in this case two of p, q, r are and one is 0. WLOG 2 1 assume p = q = . Then we have to prove 16abc ≤ a + b. If we let 2   a+b 2 x = a + b ∈ [0, 1] then 16abc ≤ 16 c = 4x2 (1 − x). So we need 2 to show that 4x2 (1 − x) ≤ x or 4x(1 − x) ≤ 1 or 0 ≤ (1 − 2x)2 , which is true. 85

1.11

Breaking the inequality

Most complicated inequalities can be broken into simpler ones, by summing which we get the original inequality. We have already seen an example we have broken the inequality a2 + b2 + c2 ≥ ab + bc + ca into

a2 + b2 b2 + c2 c2 + a2 ≥ ab, ≥ bc, and ≥ ca. 2 2 2 There are different tips for breaking inequalities. If we have a homogeneous inequality of degree one, like the one b2 c2 a+b+c a2 + + ≥ , (a, b, c > 0), b+c a+c b+a 2 we can try to compare each term from the left sides with a part of the right side: a2 ≥ αa + βb + γc. b+c

We can see that α + β + γ = 12 as for a = b = c = 1 we get equality. Moreover, a2 since increases with respect to a and decreases with respect to b, c we b+c must have α ≥ 0, β, γ ≤ 0. Moreover we must have β = γ because of the symmetry of b, c. So, we must have an inequality of the form   1 a2 ≥ + 2x a − x(b + c) b+c 2 or 2

a ≥



   1 1 2 2 + 2x a(b + c) − x(b + c) or a − + 2x a(b + c) + x(b + c)2 ≥ 0. 2 2

Since equality occurs when a = b = c, this quadratic is probably a perfect 2 square that vanishes when 2a = b + c, thus it must be (a − b+c 2 ) , and indeed 1 for x = it is, thus we get the inequality 4 b+c a2 ≥a− b+c 4 with analogous ones which break up the original inequality. The most used inequalities are xy ≥ 0 if x, y have the same sign, and especially the inequality x2 ≥ 0. since every inequality can be transformed to E ≥ 0, a common way is breaking E into sum of products of squares with positive numbers. For example, the above inequality a2 + b2 + c2 ≥ ab + bc + ca is equivalent to a2 + b2 + c2 − ab − bc − ca ≥ 0 86

which breaks into 1 1 1 (a − b)2 + (b − c)2 + (c − a)2 ≥ 0, 2 2 2 clearly true. Whereas most inequalities break by sum, some of them can be broken by product. Look at this problem from Saint-Petersburg 2005:     p 3 3 3 2 2 2 x + y + z + ≥ (x + y)(y + z)(z + x). 4 4 4 Although the inequality is in three variables, it can be easily broken down into three inequalities in just two variables: s   √ 3 3 2 2 x + y + ≥ x+y 4 4 and analogous. This inequalities are actually true: keeping in mind that x = y = z = the equality, we can deduce      1 1 1 1 x+y+1 2 2 2 x + + +y + ≥ 4 2 4 2 2

1 2

give

but (x + y + 1)2 ≥ 4(x + y) by AM-GM, and from here the conclusion. Some inequalities in many variables, like Cauchy–Schwartz, can be broken down into simpler ones according to pairs of indices. For example, the inequality (a21 + . . . + a2n )(b21 + . . . + b2n ) − (a1 b1 + . . . + an bn )2 ≥ 0, if opened the brackets, can be grouped by pairs i, j of indices: X X
a2i b2j + a2j b2i − 2ai bi aj bj = (ai bj − aj bi )2 ≥ 0. 1≤i 0 then   a b c 9 (a+b+c) + + ≥ . (2a + b + c)(b + c) (2b + c + a)(c + a) (2c + b + a)(b + a) 8 Solution: We may notice that a (2a + b + c)(b + c)

= ≥

1 1 − 2(b + c) 2(a + b) + 2(a + c) 1 1 1 − − . 2(b + c) 8(a + b) 8(a + c)

By summing with the analogous terms we have

≥

a b c + + ≥ (2a + b + c)(b + c) (2b + c + a)(c + a) (2c + b + a)(b + a)   1 1 1 1 1 9 9 + + ≥ = . 4 a+b b+c c+a 4 2(a + b + c) 8(a + b + c) 88

a b c 5. Let a, b, c ≥ 0 with a2 +b2 +c2 = 1. Show that + + ≥ 2 2 1−a 1−b 1 − c2 √ 3 3 . 2 √ x 3 3 2 Solution: we notice the inequality ≥ x , which is equivalent 1 − x2 2 √ 2 1−x2 to 3 3x(1 − x2 ) ≤ 4 or 27x2 · 1−x ≤ 1, which follows by AM-GM 2 · 2 2 1−x2 for x2 , 1−x , . This inequality solves the problem. 2 2 6. If a, b, c ≥ 0 and a2 + b2 + c2 = 1 then

a 1−a4

+

b 1−b4

+

c 1−c4

≥

√ 545 4 .

Solution: Again we break the inequality according to the relation √ x 545 2 ≥ x . 1 − x4 4 √ 545 This relation is equivalent to x(1 − x4 ) ≤ 1, or raised to the fourth 4 4 1−x4 1−x4 1−x4 ≤ 55 , which follows by AM-GM. power to x4 · 1−x 4 · 4 · 4 · 4 7. Let x, y, z > 0. Then (x2 +y 2 −z 2 )(y 2 +z 2 −x2 )(x2 +z 2 −y 2 ) ≤ (x+y−z)2 (y+z−x)2 (x+z−y)2 . Solution: It suffices to consider the case when all x2 +y 2 −z 2 , x2 +z 2 −y 2 , y 2 + z 2 − x2 are positive, because at most one of them can be negative (the pairwise sums are positive, so we can’t have two negative numbers among them), and when one of them is negative, LHS is negative and RHS is positive. Now we shall break the inequality into (x2 + y 2 − z 2 )(x2 + z 2 − y 2 ) ≤ (x + y − z)2 (x + z − y)2 and the analogous ones.  2 This is equivalent to x4 − (y 2 − z 2 )2 ≤ x2 − (y − z)2 or x4 − (y + z)2 (y − z)2 ≤ x4 − 2(y − z)2 x2 + (y − z)4 , or that (y − z)4 + (y + z)2 (y − z)2 − 2(y − z)2 x2 ≥ 0 or   (y − z)2 (y + z)2 + (y − z)2 − 2x2 ≥ 0 or 2(y − z)2 (y 2 + z 2 − x2 ) ≥ 0, which is true. a2 b2 c2 a b c + + ≥ + + . b2 c2 a2 b c a Solution: As in example 3, we have a breaking that can be computed by 2 a2 1 b2 1 c2 a constructing a system of equations: + 2+ ≥ . 2 2 3b 6c 6a b

8. For a, b, c > 0 we have

89

9. For x, y, z > 0, 1 1 1 1 1 1 + + ≥ + + . 3x + y 3y + z 3z + x 2x + y + z 2y + z + x 2z + x + y Solution: We seek an inequality of form m n p 1 + + ≥ 3x + y 3y + z 3z + x 2x + y + z where m + n + p = 1. We have m n p + + 3x + y 3y + z 3z + x

m2 n2 p2 + + 3mx + my 3ny + nz 3pz + xp 1 . (3m + p)x + (3n + m)y + (3p + m)z

= ≥

By solving the system we get m = 47 , n = 71 , p = 27 . 10. Let 0 ≤ x ≤ y, a1 , a2 , . . . , an , b1 , b2 , . . . , bn > 0, n ≥ 2. Then n X

! a1+x b1−x i i

i=1

n X

n X

! ai1−x b1+x i

≤

i=1

! a1+y bi1−y i

i=1

n X

! a1−y b1+y i i

.

i=1

Solution: Like CBS, we break the inequality into pairs of indices. Indeed the LHS is n X

X

a2i b2i +

i=1

(a1+x b1−x aj1−x b1+x + ai1−x b1+x a1+x bj1−x ) i i j i j

1≤i 0 such that q ≥ n(n − 2), then p q q p √ px21 + q + px22 + q + . . . + px2n + q ≤ p + q(x1 + x2 + . . . + xn ). Solution: We try to break the inequality into √ q m+n−a 2 mx1 + n ≤ ax1 + (x2 + x3 + . . . + xn ). n−1 As the LHS depends √only on x1 , we apply AM-GM to the RHS √ to get p+q−a p+q−a the inequality ax1 + (x2 + x3 + . . . + xn ) ≥ ax + 1 n−1 x n−1 (we let x = x1 for simplicity). Now square to get a2 x2 +2a

√


1− p+q−a x

1 n−1 +


− 2 √ p + q − 2 p + qa + a2 x n−1 ≥ px2 +q,

1
− 2 √ √ 1− or (a2 −p)x2 +2a ( p + q − a) x n−1 + p + q − 2 p + qa + a2 x n−1 ≥ q. We apply now weighted AM-GM to get


1− 1
− 2 √ √ (a2 − p)x2 + 2a p + q − a x n−1 + p + q − 2 p + qa + a2 x n−1 ≥

  √ √ ≥ (a2 − p) + 2a p + q − a + p + q − 2 p + qa + a2 ·

=

    √ √ 2 1 2(a2 −p)+ 1− n−1 2a( p+q−a)+ − n−1 (p+q−2 p+qa+a2 ) √ √ (a2 −p)+2a( p+q−a)+(p+q−2 p+qa+a2 ) ·x  q  2n √ n n−1 p+qa−2 p n−1 + n−1 q . qx

So now for the desired inequality to take place, we must take n 1 p+ q √ np + q . a = n − 1√ n − 1 = p + q n p+q n(p + q) √ √ We can see that p < a < p + q so for this value the desired breaking holds and the conclusion follows. 1 1 1 + + ≤ 1. 1+a+b 1+b+c 1+c+a 1 ct Solution: Let’s try to find a breaking of the form ≤ t , 1+a+b a + bt + ct or at + bt + ct ≤ ct + ct (a + b), so at + bt ≤ ct (a + b) so a2t bt + b2t at ≤ a + b. 1 This is achieved for t = . 3

12. If a, b, c > 0 and abc = 1 then

13. If a, b, c > 0 then √

a2

a b c +√ +√ ≥1 2 2 + 8bc b + 8ac c + 8ab 91

a as ≥ s a + bs + cs a2 + 8bc 2s 2 2 s s s 2 or a (a + 8bc) ≤ a (a + b + c ) . We must have Solution: Again let’s break the inequality into √

a2s + 8a2s−2 bc ≤ a2s + b2s + c2s + 2bs cs + 2as (bs + cs ) so 8a2s−2 bc ≤ b2s + c2s + 2bs cs + 2as (bs + cs ). s s

As b2s + c2s ≥ 2bs cs and bs + cs ≥ 2b 2 c 2 , it suffices to have s s

8a2s−2 bc ≤ 4bs cs + 4as b 2 c 2 . √

bc this amount to 2a2s−2 d2 ≤ d2s + as ds . However by s 3s 4 AM-GM d2s + as ds ≥ 2a 2 d 2 , and for s = we get the desired relation. 3 If we let d =

1.12

Separating the squares

Sometimes it’s hard to break a inequality into sum of squares, and we must use a2 − b2 , some special tricks. One very good trick is the easy principle a − b = a+b which helps us deal with inequalities involving radicals. Look at p p p a2 + 2b2 + b2 + 2a2 ≥ 2 a2 + ab + b2 . The first idea is squaring, selecting the radical, and squaring again, which produces a terrible mess of computations. For another way, we first observe that if p p p a = b, a2 + 2b2 = b2 + 2a2 = a2 + ab + b2 , so we want to extract a − b as a factor. To do this, use our trick: p

a2 + 2b2 −

p a2 + b2 + ab = =

a2 + 2b2 − a2 − b2 − ab √ a2 + 2b2 + a2 + b2 + ab b(b − a) √ √ . 2 2 a + 2b + a2 + b2 + ab √

We can factor p p 2a2 + b2 − a2 + b2 + ab = √

2a2

a(a − b) √ , + + a2 + b2 + ab b2

and so the inequality is equivalent to   b a √ √ (b − a) √ −√ ≥ 0. a2 + 2b2 + a2 + b2 + ab 2a2 + b2 + a2 + b2 + ab We have selected just b − a, which is not a square, and so cannot help us: in fact it can take both positive and negative values. But, if we think that the original expression is symmetric in a, b, we can expect to factor it also symmetrically in a, b. As b − a is not symmetric but (b − a)2 is, we can be 92

pretty sure that we can extract b−a once again (this is just intuitive reasoning, but it leads us to the correct result). First, let’s clear the denominators:  √  √ √ √ a2 + 2b2 + a2 + b2 + ab − a 2a2 + b2 + a2 + b2 + ab b  √  (b − a) √ ≥ 0. √ √ 2 2 2 2 2 2 2 2 2a + b + a + b + ab a + 2b + a + b + ab The denominator is clearly positive, so let’s look at the numerator which can be written as p p p (b − a) a2 + b2 + ab + 2a2 b2 + b4 − 2a2 b2 + a4 . We have no problem in extracting b − a from p p 2a2 b2 + b4 − 2a2 b2 + a4 using our trick: p p (b − a)(b4 + b3 a + b2 a2 + ba3 + a4 ) √ √ 2a2 b2 + b4 − 2a2 b2 + a4 = . 2a2 b2 + b4 + 2a2 b2 + a4 That’s why the numerator is p  b4 + b3 a + b2 a2 + ba3 + a4 √ a2 + b2 + ab + √ (b − a)2 , 2a2 b2 + b4 + 2a2 b2 + a4 thus positive, which finishes the proof. [Note: actually this inequality can be 1 easily handled by Minkowski’s Inequality for p = ]. 2 When we have such inequalities, symmetric in a few variables, and that become equalities when all the variables are equal, its useful to select out the squares of the differences ((a − b)2 , (b − c)2 ), perhaps using the trick above. But these differences help us even sometimes when the inequalities do not transform into equalities when all the variables are equal: Consider the inequality If a, b, c are positive numbers which form a triangle (that if a < b+c, b < c+a, c < a + b), then 2ab + 2bc + 2ca > a2 + b2 + c2 . We don’t see an obvious method of writing it as a sum of squares, but since we see a2 , b2 , c2 , 2ab, 2bc and 2ca We try to get the squares (a + b)2 , (b + c)2 , (c + a)2 or (a − b)2 , (b − c)2 , (c − a)2 . We transform the inequality into 0 > a2 + b2 + c2 − 2ab − 2bc − 2ca or into a2 + b2 + c2 ≥ (a − b)2 + (b − c)2 + (c − a)2 which breaks into the inequalities a2 ≥ (b − c)2 ,

b2 ≥ (c − a)2 , 93

c2 ≥ (a − b)2 .

They are true: since b < a + c we get a > b − c; since c < a + b we get a > c − b, hence |a| ≥ |b − c| and a2 ≥ (b − c)2 , the other inequalities being analogous. When trying to extract the squares we must reason logically and usually this leads to a solution. For example, let’s try to break the inequality a3 + b3 + c3 + 3abc − a2 b − b2 a − b2 c − c2 b − a2 c − c2 a ≥ 0. If we write it in the form P1 (a, b, c)(a − b)2 + P2 (a, b, c)(b − c)2 + P3 (a, b, c)(c − a)2 then P1 , P2 , P3 must be linear, thus we must have P1 (a, b, c) = ax + by + cz, and from the symmetry in a, b ce suppose x = y. Further, from the total symmetry we may assume P2 (a, b, c) = x(b + c) + za, P3 (a, b, c) = x(a + c) + zb. To compute, x, z, set a = b. Then the relation reduces to c3 −2c2 a+ca2 = c(a−c)2 = (P2 (a, a, c)+P3 (a, a, c))(a−c)2 = (2(x+z)a+2xc)(c−a)2 . 1 1 This yields us x = , z = − and so the inequality is equivalent to 2 2 (b + c − a)(b − c)2 + (a + c − b)(a − c)2 + (a + b − c)(a − b)2 . Actually one of these terms may be negative, but the problem is simpler to solve in this form. Indeed, as it’s symmetric, we may assume that a ≥ b ≥ c. If b + c ≥ a, we are done. Otherwise, we must prove that (a − b − c)(b − c)2 ≤ (a + c − b)(a − c)2 + (a + b − c)(a − b)2 . However a − b − c < a + c − b and b − c ≤ a − c and from here the result follows. The method of extracting squares of differences is conventionally called SOS (Sum of Squares) method, and is usually used to solve hard inequalities. Sometimes it’s hard to see the square. In this case it’s good to find an in√ 2ab termediary expression. For example, if we want to express ab − with a+b 2 respect to (a − b) , we might remember that √ √ a+b √ ( a − b)2 (a − b)2 − ab = =  √ 2 √ 2 2 2 a+ b and

hence

a+b 2ab (a − b)2 − = 2 a+b a+b √

2ab (a − b)2 ab − = a+b 2



1 1 √ − a + b a + b + 2 ab

94

 ≥ 0.

Exercises √ 1 ≥2+ 2 ab √ a2 + b2 Solution: The inequality can be rewiteen as 2 − a − b ≤ − 2 or abr ! r 2 p a b a b (a − b)2 2 2 √ ≤ − = as 2(a + b )−a−b ≤ + −2 or as b a b a a+b+ 2 √ (a − b)2 which transforms to ab ≤ a + b + 2 which is obviously true. ab √ 1 2. If a, b, c > 0 and a2 + b2 + c2 = 1 then a + b + c + ≥4 3 abc √ √ 1 − 3 3. It’s Solution: We write the conclusion as 3 − (a + b + c) ≤ abc clear that both sides are positive (from AM-GM). This is equivalent to

1. If a2 + b2 = 1 and a, b > 0 then a + b +

1 − 27a2 b2 c2 3 − (a + b + c)2 √ √
. ≤ 3+a+b+c abc 1 + 3 3abc Now we remember that a2 + b2 + c2 = 1 and we use the identities 3(a2 + b2 + c2 ) − (a + b + c)2 = (a − b)2 + (b − c)2 + (c − a)2 and (x + y + z)3 − 27xyz =

X x + y + 7z (x − y)2 (for x = a2 , y = b2 , z = 2 cyc

c2 ) to transform our inequality into X 2 2 2 (a − b)2 (a + b)2 a +b2+7c cyc

√


abc 1 + 3 3abc

X (a − b)2 ≥√

cyc

3+a+b+c

.

√ a2 + b2 + 7c2 1 + 6c2 = ≥ 6c from AM2 2 X √ 1 √ GM, and abc ≤ 3 3 , implying that LHS is at least 2 6(a − b)2 ≥ However (a + b)2 ≥ 4ab,

cyc

X (a − b)2 √ , which is, in turn, greater than RHS. 3 cyc 3. Let a, b, c be nonnegative numbers. Show that n√ √ √ √ √ √ o a+b+c √ 3 − abc ≤ max ( a − b)2 , ( a − c)2 , ( b − c)2 . 3 Solution: Set a = x6 , b = y 6 , c = z 6 . WLOG suppose x ≥ y ≥ z. We make use of the identity   1 u3 + v 3 + w3 − 3uvw = (u + v + w) (u − v)2 + (v − w)2 + (w − u)2 , 2 rewriting the conclusion as   1 2 (x + y 2 + z 2 ) (x2 − y 2 )2 + (y 2 − z 2 )2 + (x2 − z 2 )2 ≤ (x3 − z 3 )2 , 6 95

or (x2 + y 2 + z 2 )(x + y)2 (x − y)2 + (x2 + y 2 + z 2 )(y + z)2 (y − z)2 + +(x2 + y 2 + z 2 )(x + z)2 (x − z)2 ≤ 6(x2 + z 2 + zx)(z − x)2 . However we have x2 + y 2 + z 2 ≤ 2(x2 + z 2 + xz), (x − y)2 + (y − z)2 ≤ (x − z)2 , (y − z)2 ≤ (x − z)2 , Therefore we have 6(x2 + xz + z 2 )2 (x − z)2 − [(x2 + y 2 + z 2 )(x + y)2 (x − y)2 + +(x2 + y 2 + z 2 )(y + z)2 (y − z)2 + (x2 + y 2 + z 2 )(x + z)2 (x − z)2 ] ≥   ≥ 2(x2 + zx + x2 ) (2x2 + xz + 2z 2 )(x − z)2 − (y + z)2 (y − z)2 − (x + y)2 (x − y)2 . Therefore it suffices to show that (2x2 + 2z 2 + xz)(x − z)2 ≥ (x + y)2 (x − y)2 + (y + z)2 (y − z)2 . Now set u = x − y, v = y − z, u + v = x − z. The inequality to show transforms then into   2(z + u + v)2 + 2z 2 + z(z + u + v) (u+v)2 ≥ (2z+2u+v)2 u2 +(2z+v)2 v 2 , or     z 2 5(u + v)2 − 4u2 − 4v 2 + 5(u + v)3 − 4(2u + v)u2 − 4v 3 z + +2(u + v)4 − (2u + v)2 v 2 − v 4 ≥ 0. It is now pretty clear that 5(u+v)2 −4u2 −4v 2 , 5(u+v)3 −4(2u+v)u2 −4v 3 , 2(u + v)4 − (2u + v)2 v 2 − v 4 are positive, so the proof is finished. 4. Let a, b, c > 0. Show that  4(ab + bc + ca)

1 1 1 + + 2 2 (a + b) (b + c) (c + a)2

 ≥ 9.

Solution: We have 4(ab + bc + ca) = (a + b)2 + (b + c)2 + (c + a)2 − (a − b)2 − (b − c)2 − (c − a)2 so 

 1 1 1 4(ab + bc + ca) + + −9= (a + b)2 (b + c)2 (c + a)2     1 1 1 2 2 2 = − (a − b) + (b − c) + (c − a) + + + (a + b)2 (b + c)2 (c + a)2     1 1 1 2 2 2 + (a + b) + (b + c) + (c + a) + + −9 (a + b)2 (b + c)2 (c + a)2     1 1 1 = − (a − b)2 + (b − c)2 + (c − a)2 + + + 2 2 (a + b) (b + c) (c + a)2 (a − b)2 (a + b + 2c)2 (b − c)2 (b + c + 2a)2 (a − c)2 (a + 2b + c)2 + + + (a + c)2 (b + c)2 (a + b)2 (a + c)2 (a + b)2 (b + c)2 96

 (we used here the identity (x + y + z)

1 1 1 + + x y z

 =

(x − y)2 + xy

(x − z)2 (y − z)2 + .) xz yz 1 1 2 (a + b + 2c)2 = + + , we finally 2 2 2 2 (a + c) (b + c) (a + c) (b + c) (a + c)(b + c) write   1 1 1 −9= 4(ab + bc + ca) + + (a + b)2 (b + c)2 (c + a)2     2 1 2 1 2 2 = (a − b) − + (a − c) − + (a + c)(b + c) (a + b)2 (a + b)(b + c) (a + c)2   2 1 +(b − c)2 . − (a + b)(a + c) (b + c)2 As

2 1 − ≥ 0 and (a + b)(b + c) (a + c)2 2 1 2 1 − ≥ 0. If − ≥ 0 then we 2 (a + c)(b + c) (a + b) (a + b)(a + c) (b + c)2 1 2 − < 0. Then we rewrite are done. So assume that (a + b)(a + c) (b + c)2 the inequality as     2 1 2 1 2 2 (a − b) − + (a − c) − ≥ (a + c)(b + c) (a + b)2 (a + b)(b + c) (a + c)2   1 2 2 ≥ (b − c) − . (b + c)2 (a + b)(a + c) Assume that a ≥ b ≥ c. Then we have

We shall prove that     2 1 1 2 2 2 (a−c) − ≥ (b−c) − . (a + b)(b + c) (a + c)2 (b + c)2 (a + b)(a + c) After clearing denominators and cancelling common terms we reduce this to (a − c)2 (b − c)2 2 (2a2 +3ac+2c2 −b2 −ab−bc) ≥ (a +ab+ac−2b2 −2c2 −3bc). a+c b+c However

(a−c)2 a+c

≥

(b−c)2 b+c

as this is equivalent to   (a − b) (a + b)(b + c) − 4c2 ≥ 0

and 2a2 + 3ac + 2c2 − b2 − ab − bc ≥ a2 + ab + ac − 2b2 − 2c2 − 3bc as this is equivalent to a2 + b2 + 2c2 − 2ab + 2ac + 2bc ≥ 0, clearly true. 5. If a, b, c > 0 then show that (a3 + b3 + c3 )2 ≥ (a4 + b4 + c4 )(ab + bc + ca). Solution: We need to write the difference (a3 + b3 + c3 )2 − (a4 + b4 + c4 )(ab + bc + ca) in SOS style?. 97

We might remember that (a3 + b3 + c3 )2 − (a4 + b4 + c4 )(a2 + b2 + c2 ) can be written in this form according to Lagrange Identity, and clearly (a4 + b4 + c4 )(a2 + b2 + c2 − ab − bc − ca), can be written in that form, too. So (a3 + b3 + c3 ) − (a4 + b4 + c4 )(ab + bc + ca) = = (a3 + b3 + c3 )2 − (a4 + b4 + c4 )(a2 + b2 + c2 ) + +(a4 + b4 + c4 )(a2 + b2 + c2 − ab − bc − ca) = −(a − b)2 a2 b2 − (b − c)2 b2 c2 − (c − a)2 c2 a2 +   (a − b)2 + (b − c)2 + (c − a)2 4 4 4 +(a + b + c ) 2 1 = [(a − b)2 (a4 + b4 − 2a2 b2 + c4 ) + (b − c)2 (b4 + c4 − 2b2 c2 + a4 ) + 2 +(c − a)2 (c4 + a4 − 2a2 c2 + b4 )] ≥ 0. 6. If a, b, c > 0 then 9(a4 + b4 + c4 )2 ≥ (a5 + b5 + c5 )(a + b + c)3 . Solution: We use the same method as in the previous problem. Keeping in mind that (a4 +b4 +c4 )2 −(a5 +b5 +c5 )(a3 +b3 +c3 ) = −a3 b3 (a−b)2 −b3 c3 (b−c)2 −c3 a3 (a−c)2 and 9(a3 + b3 + c3 ) − (a + b + c)3 = = 2(a3 + b3 + c3 − 3abc) + 3(a3 + b3 − a2 b − ab2 ) + 3(b3 + c3 − b2 c − bc2 ) + +3(a3 + c3 − a2 c − ac2 ) = (a − b)2 (a + b + c + 3(a + b)) + (b − c)2 (a + b + c + 3(b + c)) + +(c − a)2 (a + b + c + 3(a + c)), we write 9(a4 + b4 + c4 )2 − (a5 + b5 + c5 )(a + b + c)3 as (a − b)2 [(a5 + b5 + c5 )(4a + 4b + c) − 9a3 b3 ] + +(b − c)2 [(a5 + b5 + c5 )(4b + 4c + a) − 9b3 c3 ] + +(c − a)2 [(a5 + b5 + c5 )(4a + 4c + b) − 9a3 c3 ]. As (a5 +b5 +c5 )(4a+4b+c) ≥ 4(a5 +b5 )(a+b) = 4(a6 +b6 +a5 b+ab5 ) ≥ 16a3 b3 , we conclude that all terms accompanying the squares are nonnegative so we are done.   9 1 1 1 1 1 1 7. If a, b, c > 0 then + + + ≥4 + + . a b c a+b+c a+b b+c c+a Solution: We have   1 1 1 9 1 1 1 + + + −4 + + = a b c a+b+c a+b b+c c+a   (a − c)2 (b − c)2 2 2 2 1 (a − b)2 9 = + + + − − − . 2 ab(a + b) ac(a + c) bc(b + c) a+b+c a+b b+c a+c 98

Now knowing the identity   (x − y)2 (x − z)2 (y − z)2 11 1 −9= (x + y + z) + + + , xy z xy xz yz we obtain

=

9 2 2 2 − − − = a+b+c a+b b+c a+c   1 (b − c)2 (a − b)2 (a − c)2 , + + a + b + c (a + b)(a + c) (c + a)(c + b) (b + a)(b + c)

so finally   1 1 1 9 1 1 1 = + + + −4 + + a b c a+b+c a+b b+c c+a    1 1 4 = (a − b)2 − + 2 ab(a + b) (a + c)(b + c)(a + b + c)   1 2 2 +(a − c) − + ac(a + c) (a + b)(b + c)(a + b + c)   1 2 2 +(c − b) − . cb(c + b) (a + c)(b + a)(a + b + c) 2 1 ≤ ? ab(a + b) (a + c)(b + c)(a + b + c) This is equivalent to ab(a + b) ≥ c(a2 + b2 + 3ab) + 2c2 (a + b) + c3 so ab c < a+b and c is the smallest of a, b, c. In what cases do we have the inequality

Now assume that a ≥ b ≥ c and write the inequality as   2 1 2 (a − b) − ≤ (a + c)(b + c)(a + b + c) ab(a + b)   1 2 ≤ (a − c)2 − + ac(a + c) (a + b)(b + c)(a + b + c)   1 2 2 +(c − b) − . cb(c + b) (a + c)(b + a)(a + b + c) However we may note that (a − c) ≥ (a − b) and 1 2 2 1 − ≥ − . ac(a + c) (a + b)(b + c)(a + b + c) (a + c)(b + c)(a + b + c) ab(a + b) This is equivalent to   1 1 1 1 + ≥2 + . ab(a + b) ac(a + c) (a + c)(b + c)(a + b + c) (a + b)(b + c)(a + b + c) However we may prove that 1 1 2 8 + ≥ p ≥ ab(a + b) ac(a + c) a(b + c)(2a + b + c) a bc(a + b)(a + c) 99

and 1 1 + < (a + c)(b + c)(a + b + c) (a + b)(b + c)(a + b + c) 1 1 2 < + = . a(b + c)(a + b + c) a(b + c)(a + b + c) a(b + c)(a + b + c) 8 4 So it suffices to prove that ≥ a(b + c)(2a + b + c) a(b + c)(a + b + c) which is equivalent to 2(a + b + c) ≥ 2a + b + c, true! 8. Let a, b, c > 0 and a + b + c = 1. Prove that √ √ √ √ √  2 √ a + abc + b + abc + c + abc ≥ a+ b+ c . 3 Solution: Let’s substitute x2 y2 z2 a= 2 , b = , c = . x + y2 + z2 x2 + y 2 + z 2 x2 + y 2 + z 2 We have reduced the inequality to proving X p 3 x (x2 + y 2 )(x2 + z 2 ) ≥ 2(x + y + z)(x2 + y 2 + z 2 ). cyc

Lemma: 2(x3 + y 3 + z 3 ) −

X

x2 y =

sym

Proof : Immediately from

x3

+

X

(x + y)(x − y)2 .

sym

y3

− xy(x + y) = (x + y)(x − y)2 .

Now we see that X (x2 + y 2 ) + (x2 + z 2 ) 1X 3 x − RHS = (x + y)(x − y)2 2 2 cyc cyc using Lemma. However 2 p X (x2 + y 2 ) + (x2 + z 2 ) 3 X p 2 − LHS = 3 x x + y 2 − x2 + z 2 . 2 2 cyc cyc So we are left to prove that 2 p X X p 2 2 2 2 2 (x + y)(x − y) ≥ 3 x x +y − x +z . cyc

cyc

We prove that p 2 p 3z(x − y)2 (x + y)2 z 2 + y 2 − z 2 + x 2 = p 2 . √ z 2 + y 2 + z 2 + x2 p 2 √ Thus we have to prove z 2 + y 2 + z 2 + x2 ≥ 3z(x + y) or p p 2z 2 + x2 + y 2 + 2 z 2 + x2 z 2 + y 2 ≥ 3z(x + y). p √ However z 2 + x2 z 2 + y 2 ≥ z 2 + xy by CBS. Thus we reduce it to

(x+y)(x−y)2 ≥ 3z

4z 2 + x2 + y 2 + 2xy ≥ 3z(x + y) or 4z 2 + (x + y)2 ≥ 3z(x + y). However 4z 2 + (x + y)2 ≥ 4z(x + y) by AM-GM and so we are done.

100

1.13

The Dual Principle

This is very simple idea, and yet it prove to have quite a few applications - and not only in inequalities, but also in computational geometry. In particular, it is useful in solving algebraic inequalities, as we will see in the examples.

The positive reals a, b, c are sides of a triangle if and only if there are positive reals x, y, z with a = y + z, b = x + z, c = x + y. Indeed, if a = y + z, b = x + z, c = x + y, then a = y + z < 2x + y + z = b + c b+c−a and analogously b < c + a, c < a + b. Conversely, setting x = , 2 a+c−b a+b−c y= ,z= we satisfy the requirements. 2 2 Let’s look at one already met inequality through the prism of dual principle:

2(ab + bc + ca) ≥ a2 + b2 + c2

is equivalent, by dual principle, to

2 [(x + y)(y + z) + (x + y)(x + z) + (y + z)(x + z)] ≥ (x+y)2 +(z+x)2 +(y+z)2 ≥ 0

or to

2(x2 + y 2 + z 2 ) + 3(xy + yz + zx) ≥ 2(x2 + y 2 + z 2 ) + 2(xy + yz + zx)

which is obvious. 101

Exercises 1. If a, b, c are sides of a triangle then a(b − c)2 + b(c − a)2 + c(a − b)2 + 4abc > a3 + b3 + c3 . Solution: After substituting a = y + z, b = x + z, c = x + y by cancelling common terms the inequality becomes xyz > 0. b c 2 a + + > . 2a + b + c 2b + c + a 2c + a + b 3 Solution: After applying the CBS Lemma we get

2. If a, b, c form a triangle then

a b c (a + b + c)2 + + ≥ 2a + b + c 2b + c + a 2c + a + b 2(a2 + b2 + c2 ) + 2(ab + bc + ca)   so we left to prove that 3(a+b+c)2 > 2 2(a2 + b2 + c2 ) + 2(ab + bc + ca) or 2(ab + bc + ca) > a2 + b2 + c2 which we have already proven. 3. If a, b, c are the sides of a triangle and ax+by+cz = 0 then xy+yz+zx ≤ 0. Solution: We can suppose that x, y > 0, because the inequality is symmetric and doesn’t change if we replace x, y, z by −x, −y, −z. We have xy + yz + zx = xy + z(x + y) = xy −

(ax + by)(x + y) . c

So we have to prove that cxy ≤ (ax + by)(x + y). But cxy ≤ (a + b)xy ≤ (ax + by)(x + y). We see that in this case we managed better without using the dual principle. 4. Show that in a triangle, the length of a median from a vertex is not less than the length of the bisector from the same vertex. r 2b2 + 2c2 − a2 Solution: We use the formula ma = and 4 s bc(b + c)(b + c − a)(b + c + a) la = . So we have to prove (b + c)2 2b2 + 2c2 − a2 bc(b + c − a)(b + c + a) ≤ . 4 (b + c)2 Now by using the dual principle we have 2b2 + 2c2 − a2 4

2(x + z)2 + 2(y + z)2 − (x + y)2 4 x2 + y 2 − 2xy + 4z 2 + 4zy + 4zx = 4 ≥ z 2 + zy + zx = z(x + y + z). =

102

However la2 = =

(b + c)2 (b + c − a)(b + c + a) bc(b + c − a)(b + c + a) 4 ≤ (b + c)2 (b + c)2 (b + c − a)(b + c + a) = z(x + y + z). 4

rc ra ra rb rb rc + ≥ 3. + ma mb mb mc mc ma Solution: We pass by dual principle to x, y, z. Then we must prove

5. In a triangle, show that

X x(x + y + z) mb mc

cyc

We prove more

X x(x+y+z) cyc

m2b +m2c

≥ 3.

3 ≥ . Which transforms to 2

x(x + y + z) 3 ≥ 2 + (y − z)2 (y + z)(x + y + z) + (x − z) 2 cyc

X or

x2

X cyc

y + z + [(x −

z)2

+ (x −

y)2 ]

3 ≥ . x 2 x+y+z

Now applying (Cauchy) this reduces to (x + y + z)2 ≥ 3(xy + yz + zx) +

which is just

X x+y (x − y)2 x + y + z cyc

z (x − y)2 ≥ 0 true. x + y + z cyc

X

6. Prove that: In the triangle ABC: ma · cos

B C 3 A + mb · cos + mc · cos ≥ (a + b + c). 2 2 2 4

Solution: We pass to dual principle. It’s easy to see that s 2 (y − z) A x(x + y + z) m2a = x(x + y + z) + , cos = . 4 2 (x + y)(x + z)   A x x Now let’s prove the following: ma cos ≥ p + - where 2 x+y x+z p = x + y + z = a+b+c 2 . The inequality would follow then by summing the desired terms. First, we consider the expression (1)

p 2 p m2a − x(x + y + z) (y − z)2 i p = h ma − x(x + y + z) = p ma + x(x + y + z) 4 ma + x(x + y + z) 103

Denote this expression by E. Then we must have (2)

  i px(x + y + z) hp x x x(x + y + z) + E p −p = + x+y x+z (x + y)(x + z) p r 2 r E x(x + y + z) 1 x x = p − p − . x+y x+z (x + y)(x + z) 2

We are to prove this expression is greater than zero. However r r x x − x+y x+z

= =

p p x(x + z) − x(x + y) p (x + y)(x + z) x(x + z) − x(x + y) hp ip . p x(x + z) + x(x + y) (x + y)(x + z)

Squaring we get (3)

x2 (y − z)2 . hp i2 p x(x + z) + x(x + y) (x + y)(x + z)

Substituting 1) and 3) into 2) we produce p x(x + y + z)(y − z)2 x2 (x + y + z)(y − z)2 h i ≥ . hp i2 p p 4 ma + x(x + y + z) 2 x(x + z) + x(x + y) (x + y)(x + z) Now cancelling common terms and grouping leads to h ip p p
2 √ √ 2 ma + x(x + y + z) x(x + y + z) ≤ (x + y)(x + z) x + y + x + z . p p However (x + y)(x + z) ≥ x(x + y + z) thus is suffices to prove that i h p p
2 √ √ x + y + x + z = 2x+y+z+2 (x + y)(x + z). 2 ma + x(x + y + z) ≤ p This breaks into the inequality 2ma < 2x + y + z and x(x + y + z) ≤ p (x + y)(x + z) which are true (the first one is the well known ma ≤ b+c ). 2

1.14

Substitutions

b+c−a The Dual Principle actually consists of substituting x = , y = 2 a+b−c a+c−b , z = . There are many other substitutions, which solve 2 2 many inequalities impossible to solve in other way. Basically every inequality is solved by substituting some variables into a known inequality, possibly more than one time. 104

When we have some numbers with product 1, say xyz = 1, it’s convenient to b c a set x = , y = then z = . b c a For example the inequality x y x + + ≥ x + y + z, xyz = 1 y z z reduces to

ab2 + bc2 + ca2 ac ab bc + + ≥ , b2 c2 a2 abc

or a3 c3 + a3 b3 + b3 c3 ≥ a2 b3 c + ab2 c3 + a3 bc2 , or u3 + v 3 + w3 ≥ u2 v + v 2 w + w2 u, where u = ab, v = ac, w = bc, and this is a known inequality - we’ve done it before. For the inequality x1 x2 xn + + ... + > 1, 1 + x1 x2 1 + x2 x3 1 + xn x1 ai an where x1 x2 . . . xn = 1, we can substitute xi = , xn = , but this turns ai+1 a1 xi ai+1 ai ai+2 . The substitution xi = into a messy , however, 1 + xi xi+1 ai+1 (ai + ai+2 ) ai ai+1 ai+1 > and it’s done. turns it into ai + ai+2 a1 + a2 + . . . + an Another popular substitution is to work with the inverses of the numbers instead of them. Look at this example: 1 1 1 3 If abc = 1 then 3 + + ≥ . a (b + c) b3 (a + c) c3 (a + c) 2 1 1 1 This time it’s convenient to set x = = bc, y = = ac, z = = ab, the a b c 2 2 2 x y z 3 inequality becoming then + + ≥ . However we know that y+z x+z x+y 2 x2 y2 z2 x+y+z 3 + + ≥ ≥ . y+z x+z x+y 2 2 There are also trigonometric substitutions, based on the following relations: b2 + c2 − a2 r 2bc r A (p − b)(p − c) yz sin = = 2 bc (x + y)(x + z) s r A p(p − a) x(x + y + z) cos = = 2 bc (x + y)(x + z)

cos A =

and other analogous ones. (The notations are usual: a, b, c are the sides of the a+b+c b+c−a triangle, A, B, C its angles, p = its semiperimeter, x = , 2 2 a+c−b a+b−c y= ,z= ). 2 2 105

3 π Via this substitutions, the inequality sin x+sin y +sin z ≤ , for x+y +z = , 2 2 for example, transforms to a purely algebraic one: if we consider the triangle with angles 2x, 2y, 2z then r r r z z y y x x sin x + sin y + sin z = + + x+yx+z x+yy+z x+zy+z   1 y 3 z x z x y ≤ = . + + + + + 2 x+y x+z x+y y+z x+z y+z 2 The list of substitutions can be extended further, for example if xyz = x + 1 1 1 + + = 1 so y + z + 2 then 1+x 1+y 1+z 1 a = , 1+x a+b+c

1 b = , 1+y a+b+c

1 c = 1+z a+b+c

b+c a+c a+b thus x = , y= , z= . a b c Thus, for example, we can prove that if x, y, z > 0 and xyz = x + y + z + 2 b+c then xy + yz + zx ≥ 2(x + y + z). Indeed, after the substitution x = , a a+c a+b y= ,z= , we have to prove that b c (a + b)(a + c) (b + a)(b + c) (c + a)(c + b) b a a c b c + + ≥ 2( + + + + + ), bc ac ab a b c a c b or a2 b2 c2 b a a c b c + + +3≥ + + + + + . bc ac ab a b c a c b After clearing denominators it becomes a3 + b3 + c3 + 3abc ≥ ab(a + b) + bc(b + c) + ca(c + a), or Schur’s Inequality. Exercises:   (1 − a2 )(1 − b2 ) 4ab ≤ 1. 1. If a, b ∈ R\{−1, 1} then 1+ 2 2 2 2 (1 + a )(1 + b ) (1 − a )(1 − b ) Solution: If a = tan x, b = tan y then   (1 − a2 )(1 − b2 ) 4ab (1 + a2 )(1 + b2 ) 1 + (1 − a2 )(1 − b2 ) =   (1 − tan2 x)(1 − tan2 y) 4 tan x tan y = | cos(x − y)| ≤ 1. = 1 + (1 + tan2 x)(1 + tan2 y) (1 − tan2 x)(1 − tan2 y) a2 b2 c2 3 + + ≥ . (a + b)(a + c) (b + a)(b + c) (c + a)(c + b) 4 Solution: If we let x = b + c, y = a + c, z = a + b the inequality turns to

2. If a, b, c > 0 then

(y + z − x)2 (x + z − y)2 (x + y − z)2 + + ≥3 yz xz xy 106

or after cancelling common terms to y2 z2 x y x z y z x2 + + +3≥ + + + + + , yz xz xy y x z x z y which multiplying by xyz comes to Schur. a2 1 1 3. Let a0 = , and ak+1 = ak + k . Show that 1 − < an < 1. 2 n n nb2k 1 1 = Solution: Let bk = . Then bk+1 = = bk − 1 1 ak nbk + 1 + 2 b nbk  k bk bk 1 1 . As b0 = 2, we . Now if bk > 1 then ∈ , nbk + 1 nbk + 1 n+1  n  k k can prove by induction on 1 ≤ k ≤ n that bk ∈ 2 − , 2 − . n n+1   n+2 Particularly for k = n, bk ∈ 1, , which implies the problem. n+1 1 1 1 + + ≥ 4(x+y +z). x y z 1 1 1 1 1 1 1 1 1 Solution: As , , satisfy + + +2 · · = 1, we have numbers x y z x y z x y z 1 b+c 1 a+c 1 a+b a, b, c with = , = , = . The inequality to prove x a y b z c now becomes   a c b b+c a+b a+c + + ≥4 + + a c b b+c a+b a+c

4. If x, y, z > 0 with xy +yz +zx+2xyz = 1, then

a a a ≤ + and the analogous relations. b+c b c √ √ √ y+z z+x x+y 4 (x + y + z) 5. Show that + + ≥ p for x y z (y + z) (z + x) (x + y) every three positive reals x, y, z. which follows from 4

Solution: For simplifying the solution set √ √ x → y + z, y → z + x,

z→

√

x + y.

Then x2 , y 2 , z 2 are sides of a triangle and hence so do x, y, z that form an acute-angled triangle. The inequality transforms then to x x2 + y 2 + z 2 ≥ . y 2 + z 2 − x2 xyz cyc

X

We can solve this inequality in two ways: a) Applying cosine theorem to the acute-angled triangle formed by x, y, z X x x2 + y 2 + z 2 we transform the inequality to ≥2 and by mulyz cos A xyz cyc X x2 tiplying it by xyz we have to prove ≥ 2(x2 + y 2 + z 2 ). cos A cyc 107

 1 1 1 are clearly ordered the However and , , cos A cos B cos C same way hence by Chebyshev (see the appropriate chapter on ordering) we have 1 1 1 + + X x2 X 2 cos A cos B cos C ≥ x cos A 3 cyc cyc (x2 , y 2 , z 2 )



and the desired inequality follows now from 1 1 1 9 + + ≥ cos A cos B cos C cos A + cos B + cos C 3 (by CBS) and from the well-known inequality cos A + cos B + cos c ≤ . 2 b) The inequality is  X X x(x2 + yz − y 2 − z 2 ) x x − ≥ 0. ≥ 0 or y 2 + z 2 − x2 yz yz(y 2 + z 2 − x2 ) cyc cyc Now using the fact that x, y, z form a triangle it’s quite clear that x(x2 + 1 are ordered the same way with x, y, z yz − y 2 − z 2 ), 2 yz(y + z 2 − x2 ) X and hence by Chebyshev inequality it suffices to prove that x(x2 + cyc

yz − xy − xz) ≥ 0 which is just the renowned Schur’s Inequality. 6. If x1 , x2 , . . . , xn > 0 with

1 1 1 + + ... + = n. Show that x1 x2 xn 

x 1 . . . xn − 1 ≥

n−1 n

n−1 (x1 + . . . + xn − n).

n X 1 Solution: Let yi = . Then yi = n. Now multiplying the inequality xi i=1 by y1 . . . yn we get a new inequality where we are allowed even to have yi = 0.

Transform our inequality to f (y1 , y2 , . . . , yn ) ≥ 0 where f (y1 , y2 , . . . , yn ) = = nn−1 (1 − y1 y2 . . . yn )+ +(n−1)n−1 (ny1 y2 . . . yn −y2 y3 . . . yn −y1 y3 . . . yn −. . .−y1 y2 . . . yn−1 ). Consider y1 , y2 , . . . , yn with minimal value of f and of all such n-uples, one with minimal y12 + y22 + . . . + yn2 . Now let’s try to decrease f . Pick up arbitrary non-zero yi , yj and fix the others. Thus yi + yj is constant. Then f will have form ayi yj + b linear in yi yj and takes its minimum at the extremes, hence when yi = yj or one of them is 0. Thus we may replace yi , yj by their mean (type 1) or make one of them 0 (type 2). 108

We keep doing transformations. We see that we can perform only finitely many operations of type 2 because they increase the number of zeros in the sequence. Thus at some time we shall have l zeros and some other non-zero number every two of them we can replace by their arithmetic mean non-increasing f . Now we can suppose the non-zero numbers are also equal(otherwise replacing them by their mean decreases the sum of squares) hence they n are . For this very particular case we can perform a trivial manual n−l checking. 7. Prove that for any seven numbers we may find two numbers x, y for 1 x−y ≤√ . which 0 ≤ 1 + xy 3 Solution: Let the numbers be a1 , a2 , . . . , a7 and let ai = tan bi , where π 0 ≤ bi ≤ π. We may partition [0, π] into six intervals of length . Then 6 some two numbers bi , bj will fit into the same interval then if bi < bj π 1 π so 0 ≤ tan(bj − bi ) ≤ tan = √ , thus we will have 0 ≤ bj − bi ≤ 6 6 3 aj − ai 0≤ and we may take x = aj , y = ai . 1 + aj ai

1.15

Homogenization and dehomogenization

If we have to solve a non-homogeneous inequality,we may want to make it homogeneous to deal with more familiar homogeneous inequalities (this is because we already have a luggage of known inequalities to which we can reduce it). Look, for example, at the inequality       1 1 1 8 1 1 1 2 2 2 a + b + c + ≥ (a+b+c) + + , for a, b, c > 0, abc = 1. b c a 9 a b c The inequality is not homogeneous, but we can use the condition abc = 1 to 1 1 1 by bc, ac, ab. It becomes then make it homogeneous, by replacing , , a b c 8 (a2 + ac)(b2 + ab)(c2 + bc) ≥ (a + b + c)(ab + bc + ca). 9 We can the remove abc = 1 from the left-hand side and then we get the already known homogeneous inequality 8 (a + b)(b + c)(c + a) ≥ (a + b + c)(ab + bc + ca). 9 In other cases is better to dehomogenize an inequality to reduce the number of variables or add a good condition on them. For example, if the inequality is in just two variables, we may set one of them be 1 (for example the inequality x3 + x y 3 ≥ x2 y +xy 2 reduces to x3 +1 ≥ x2 +x be dividing by y 3 and letting x → ). y 109

The inequality then may be reduced to a polynomial in one variable and solved by algebraic methods applicable to polynomials (i.e. finding the roots). Other example of dehomogenization is the AM-GM inequality x1 + x2 + . . . + xn ≥ x1 xn √ n n x1 x2 . . . xn . By replacing x1 , x2 , . . . , xn by √ ,..., √ we n x ...x n x ...x 1 n 1 n y2 yn y1 . can assume x1 x2 . . . xn = 1 and then write x1 = , x2 = , . . . , xn = y2 y3 y1 The inequality then becomes y1 y2 yn + + ... + ≥ n, y2 y3 y1 and it can be solved by unimonotonic sequences (refer to the chapter on ordering). Exercises 1. If ai , bi , ci > 0 then (a1 b1 c1 + . . . + an bn cn )3 ≤ (a31 + . . . + a3n )(b31 + . . . + b3n )(c31 + . . . + c3n ). Solution: The quantity (a1 b1 c1 + . . . + an bn cn )3 (a31 + . . . + a3n )(b31 + . . . + b3n )(c31 + . . . + c3n ) doesn’t change if we replace each ai by tai . Therefore we may assume that a31 + . . . + a3n = 1 and analogously b31 + . . . + b3n = 1, c31 + . . . + c3n = 1. a3 + b3i + c3i Then ai bi c≤ i . 3 By summing this inequalities we get a1 b1 c1 + . . . + an bn cn ≤ 1 and the conclusion follows. Remark: This also easily follows from Holder’s inequality. The solution we gave above is essentially the same as the inequality ai bi ci ≤ 13 (pa3i + a31 +...+a3n b31 +...+b3n qb3i +rc3i ) where p = √ ,q = √ 3 3 3 3 3 3 3 3 3 3 3 3 √ 3

c31 +...+c3n

(a1 +...+an )(b1 +...+bn )(c1 +...+cn )

(a31 +...+a3n )(b31 +...+b3n )(c31 +...+c3n )

(a1 +...+an )(b1 +...+bn )(c31 +...+c3n )

- except this method makes it easier to

see. Of course, this is also how we prove Holder’s inequality. 2. If x, y, z > 0 then x3 y + y 3 z + z 3 x ≥ xyz(x + y + z). Solution: Assume that x is the smallest of x, y, z. Then as the inequality is homogeneous we can assume x = 1, y = 1 + u, z = 1 + v. The inequality now after cancelling common terms becomes (u − v)2 + (u3 + v 3 − uv 2 ) + u2 + v 2 + 2u2 v + u3 v ≥ 0 and is clearly true. 2 2 2 3. For every positive X a a, b and c such that a + b + c = 1 prove the inequality: > 3 (the inequality is strict). a3 + bc cyc

110

,r =

Solution: By letting x = a2 , y = b2 , z = c2 and homogenizing we have X x(x + y + z) p to prove the following ≥ 3 where t = xyz(x + y + z). x2 + t cyc Which is in turn equivalent to X (x + y + z) x(y 2 + t)(z 2 + t) − 3(x2 + t)(y 2 + t)(z 2 + t) ≥ 0. cyc

Now let u = x + y + z, v = xy + yz + zx, w = xyz. We transform our inequality to √ g(x, y, z) = u(vw + (uv − 3w) uw + u3 w) − 3w2 − √ √ −3(v 2 − 2uw) uw − 3(u2 − 2v)uw − 3uw uw ≥ 0. Now let’s assume that this inequality fails for some x, y, z thus g(x, y, z) < 0. Now keep u, w fixed and let’s try to vary v in order to minimize g. It can run on a set intervals of reals that ensures that the polynomial p(x) = x3 − ux2 + vx − w = 0 has solutions or equivalently that the line w y = −v intersects the graph of function x2 − ux − in three points. x Now g(x, y, z) transforms to a quadratic expression in v with leading coefficient negative thus it takes minimum at some interval. However the extremum for v realizes when the line x = −v touches the graph of the function somewhere which implies that the equation has a double root. This means that two of x, y, z are equal. So we need to check the relation only for x = y, z = 1. This is quite simple to check: √ after clearing 2 denominators this expression transforms to (x +4x−2) 2x + 1 ≥ 4x2 + x − 2. This inequality √ can be easily proved as follows: If 4x2 +x−2 ≥ 0 then we 33 − 1 shall have x ≥ > 0.6 and necessarily x2 + 4x − 2 ≥ 0 and thus 8 by squaring we deduce that we have to prove 2x4 +x3 +24x2 −5x−4 > 0 which is clearly true for x > 0.6. If 4x2 +√x − 2 < 0 we must look only at the case x2 + 4x − 2 < 0 which is x < 6 − 2. By squaring we must prove the inequality 2x4 + x3 + 24x2 − 5x − 4 < 0 √ which again easily follows from x < 6 − 2.

1.16

Unimonotonic sequences

Suppose that x1 > x2 and y1 > y2 . Then we have (x1 y1 + x2 y2 ) − (x1 y2 + x2 y1 ) = (x1 − x2 )(y1 − y2 ) ≥ 0. Now take numbers x1 , x2 , . . . , xn , y1 , y2 , . . . , yn and consider the quantity Q(y1 , y2 , . . . , yn ) = x1 y1 + . . . + xn yn . 111

If xi > xj but yi < yj then by exchanging yi with yj we increase Q by (xi − xj )(yj − yi ) > 0. If yi > yj then by exchanging yi with yj we decrease Q by (xi − xj )(yi − yj ). Now suppose x1 > x2 > . . . > xn . Let z1 , z2 , . . . , zn be the permutation that maximizes Q and let z10 , z20 , . . . , zn0 be the permutation that minimizes Q. If zi > zj for some i < j then we could increase Q which contradicts the maximality of Q(z1 , z2 , . . . , zn ). That’s why z1 ≥ z2 ≥ . . . ≥ zn . Analogously we deduce z10 ≤ z20 ≤ . . . ≤ zn0 (and so z10 = zn , z20 = zn−1 , . . . , zn0 = z1 ). So, we proved the following theorem: Theorem (Rearrangement Inequality). If x1 ≥ x2 ≥ . . . ≥ xn and y1 , y2 , . . . , yn are real numbers then x1 z1 + . . . + xn zn ≥ x1 y1 + x2 y2 + . . . + xn yn ≥ x1 zn + . . . + xn z1 , where z1 ≥ z2 ≥ . . . ≥ zn is a decreasing permutation of y1 , y2 , . . . , yn . The sequences x1 , x2 , . . . , xn and y1 , y2 , . . . , yn are called unimonotonic if they are ordered in the same way (i.e (xi − xj )(yi − yj ) ≥ 0 for any i, j). If they are inversely ordered (i.e. (xi − xj )(yi − yj ) ≤ 0) we call them antimonotonic. By the above theorem, the quantity x1 y1 + . . . + xn yn is maximized by unimonotonic sequences and minimized by antimonotonic sequences. 1 For example, if yi = where xi > 0 then x1 , x2 , . . . , xn and y1 , y2 , . . . , yn are xi antimonotonic sequences so Q(y1 , y2 , . . . , yn ) ≤ Q(y2 , . . . , yn , y1 ). Thus

x1 xn + ... + ≥n x2 x1

which proves the AM-GM Inequality as it is equivalent to it (see the last example from the previous chapter - or the first exercise from this one). If y1 ≥ y2 ≥ . . . ≥ yn and x1 ≥ x2 ≥ . . . ≥ xn then x1 y1 + x2 y2 + . . . xn yn ≥ x1 y1+i + x2 y2+i + . . . + xn yn+i (we consider yn+k = yk ). Summing this i = 0, 1, . . . , n − 1 we obtain Theorem (Chebyshev’s Inequality) n(x1 y1 + . . . + xn yn ) ≥ (x1 + . . . + xn )(y1 + y2 + . . . + yn ) If y1 < y2 < . . . < yn , it changes sign. A consequence of Chebyshev’s inequality is the following inequality (which can also be proved by induction): If x1 , x2 , . . . , xn and y1 , y2 , . . . , yn are antimonotonic then x1 xn x1 + . . . + xn + ... + ≥n . y1 yn y1 + . . . + yn 112

Proof : By Chebyshev’s inequality x1 xn 1 + ... + ≥ (x1 + x2 + . . . + xn ) y1 yn n Since



1 1 + ... + y1 yn

 .

1 1 n2 + ... + ≥ , y1 yn y1 + . . . + yn

the conclusion follows. Exercises √ 1. Show that if x1 , x2 , . . . , xn > 0 then x1 + x2 + . . . + xn ≥ n n x1 x2 . . . xn . Solution: As the inequality is homogeneous, we can assume x1 x2 . . . xn = a1 a2 an 1 then we can write x1 = , x2 = , . . . , xn = . a2 a3 a1 a1 an And the inequality becomes + ... + ≥ n, so it follows from the a2 a1   1 1 1 , ,..., antimonocity of the sequences (a1 , a2 , . . . , an ) and . a1 a2 an b2 − a2 c2 − b2 a2 − c2 + + ≥ 0. c+a a+b b+c Solution: We can rewrite the inequality as

2. Show that if a, b, c > 0 then

b2 c2 a2 a2 b2 c2 + + ≥ + + c+a a+b b+c c+a a+b b+c   1 1 1 2 2 2 which follows from the unimonotonicity of (b , a , c ) and , , . a+c b+c a+b 3.

a3 b3 c3 + + ≥ a + b + c. b2 − bc + c2 a2 − ac + c2 a2 − ab + b2 Solution: We can suppose that a ≥ b ≥ c. Now note that we have (b2 − bc + c2 ) − (a2 − ac + c2 ) = (b − a)(b + a − c). Hence we have two cases to consider: a) a, b, c form a triangle. In this case (a3 , b3 , c3 ) and 

1 1 1 , 2 , 2 2 2 2 b − bc + c a − ac + c a − ab + b2



are ordered the same way, so  a3 b3 c3 2 2 + + ≥ b − bc + c2 a2 − ac + c2 a2 − ab + b2 b3 + c3 a3 + c3 a3 + b3 ≥ 2 + + = a + b + c. b − bc + c2 a2 − ac + c2 a2 − ab + b2 

113

1 1 , y = 2 , z = 2 − bc + c a − ac + c2 1 . Then x ≥ z ≥ y. Hence a3 x+b3 y +c3 z ≥ a3 y +b3 x+c3 z ≥ 2 a − ab + b2 a3 y+b3 z+c3 x and also a3 x+b3 y+c3 z ≥ a3 z+b3 y+c3 x ≥ a3 z+b3 x+c3 y.

b) a > b + c. Let’s denote x =

b2

By summing these two relation we get again   a3 b3 c3 2 2 + + ≥ b − bc + c2 a2 − ac + c2 a2 − ab + b2 b3 + c3 a3 + c3 a3 + b3 ≥ 2 + + = a + b + c. b − bc + c2 a2 − ac + c2 a2 − ab + b2 4. If α > 0 and x, y, z > 0 then xα+2 y α+2 z α+2 1 + + ≥ (xα + y α + z α ). (x + y)(x + z) (y + x)(y + z) (z + y)(z + x) 4 Solution: We can see that the sequences   x2 y2 z2 , , (xα , y α , z α ) and (x + y)(x + z) (y + x)(y + z) (z + x)(z + y) are ordered the same way, hence

≥

xα+2 y α+2 z α+2 + + ≥ (x + y)(x + z) (y + x)(y + z) (z + y)(z + x)   1 x2 y2 z2 + + (xα + y α + z α ). 3 (x + y)(x + z) (y + x)(y + z) (z + x)(z + y)

Moreover, we have

x2 y2 z2 3 + + ≥ (x + y)(x + z) (y + x)(y + z) (z + x)(z + y) 4

from CBS Lemma. 5. Let x1 ≤ x2 ≤ . . . ≤ xn and y1 ≤ y2 ≤ . . . ≤ yn be real numbers. For any permutations (z1 , z2 , . . . , zn ) of (y1 , y2 , . . . , yn ), prove that (x1 −y1 )2 +(x2 −y2 )2 +. . .+(xn −yn )2 ≤ (x1 −z1 )2 +(x2 −y2 )2 +. . .+(xn −nn )2 . Solution: Notice that y12 +y22 +. . .+yn2 = z12 +z22 +. . .+zn2 . After expansion and simplification, the desired inequality is equivalent to x1 y1 + x2 y2 + . . . + xn yn ≥ x1 z1 + x2 z2 + . . . + xn zn , which is just the Rearrangement inequality. 6. (Nesbitt’s Inequality) Let a, b, c be positive real numbers. Prove that a b c 3 + + ≥ . b+c c+a a+b 2 Solution: Without loss of generality, we can suppose that a ≥ b ≥ c, then 1 1 1 ≥ ≥ . b+c c+a a+b 114

Therefore, applying the Rearrangement inequality, we obtain a b c + + b+c c+a a+b

= a·

1 1 1 +b· +c· b+c c+a a+b 1 1 1 ≥ a· +b· +c· a+b b+c c+a a b c = + + , a+b b+c c+a

b c a + + b+c c+a a+b

= a·

and 1 1 1 +b· +c· b+c c+a a+b 1 1 1 ≥ a· +b· +c· c+a a+b b+c a b c = + + , c+a a+b a+b

Adding up these two inequalities, we obtain   a b c a+b b+c c+a 2 + + ≥ + + = 3. b+c c+a a+b a+b b+c c+a This completes the proof. 7. Let a1 , a2 , . . . , an be distinct positive integers. Prove that a1 a2 an 1 1 + 2 + ... + 2 ≥ 1 + + ... + . 2 1 2 n 2 n Solution: Let (a01 , a02 , . . . , a0n ) be a cyclic permutation of (a1 , a2 , . . . , an ) such that a01 < a02 < . . . < a0n . Then we have a0i ≥ i for all i = 1, 2, . . . , n. Now, we have that a01 < a02 < . . . < a0n and

1 1 1 > 2 > ... > 2. 2 1 2 n

Therefore, according to the Rearrangement inequality, we obtain a1 a2 an + 2 + ... + 2 2 1 2 n

1 1 + a2 · 2 + . . . + an · 2 1 2 1 1 0 0 ≥ a1 · 2 + a2 · 2 + . . . + a0n · 1 2 1 1 1 ≥ 1 · 2 + 2 · 2 + ... + n · 2 1 2 n 1 1 = 1 + + ... + . 2 n = a1 ·

1 n2 1 n2

This completes our proof. Equality holds if and only if ai = i for all i = 1, 2, . . . , n. 8. Let a, b, c be the side lengths of a triangle. Prove that a2 (b + c − a) + b2 (c + a − b) + c2 (a + b − c) ≤ 3abc. 115

Solution: We will firstly show that if a ≥ b, then a(b + c − a) ≤ b(c + a − b). Indeed, we have b(c + a − b) − a(b + c − a) = (a − b)(a + b − c) ≥ 0. Now, we see that the inequality is symmetric, hence we can assume that a ≥ b ≥ c. Then, from the above inequality, we have a ≥ b ≥ c and c(a + b − c) ≥ b(c + a − b) ≥ a(b + c − a). Thus, according to the Rearrangement inequality, we obtain a · a(b + c − a) + b · b(c + a − b) + c · c(a + b − c) ≤ ≤ b · a(b + c − a) + c · b(c + a − b) + a · c(a + b − c), and a · a(b + c − a) + b · b(c + a − b) + c · c(a + b − c) ≤ ≤ c · a(b + c − a) + a · b(c + a − b) + b · c(a + b − c). Adding both these inequalities, we have   2 a2 (b + c − a) + b2 (c + a − b) + c2 (a + b − c) ≤ 6abc, which is a2 (b + c − a) + b2 (c + a − b) + c2 (a + b − c) ≤ 3abc. 9. Let a, b, c be the side lengths of a triangle. Prove that a2 b(a − b) + b2 c(b − c) + c2 a(c − a) ≥ 0. Solution: Without loss of generality, we can suppose that a = max {a, b, c} and consider 2 cases Case 1. If a ≥ b ≥ c, then from the above problem, we have 1 1 1 ≤ ≤ and a(b + c − a) ≤ b(c + a − b) ≤ c(a + b − c). a b c And the Rearrangement inequality yields that 1 1 1 · a(b + c − a) + · b(c + a − b) + · c(a + b − c) a b c 1 1 1 ≥ · a(b + c − a) + · b(c + a − b) + · c(a + b − c) c a b a2 b(a − b) + b2 c(b − c) + c2 a(c − a) = a+b+c− . abc

a+b+c =

This implies that a2 b(a − b) + b2 c(b − c) + c2 a(c − a) ≥ 0. 116

Case 2. If a ≥ c ≥ b, then from the above problem, we have 1 1 1 ≤ ≤ and a(b + c − a) ≤ c(a + b − c) ≤ b(c + a − b). a c b Thus, by Rearrangement inequality, we see that 1 1 1 · a(b + c − a) + · c(a + b − c) + · b(c + a − b) a c b 1 1 1 · a(b + c − a) + · c(a + b − c) + · b(c + a − b) ≥ c b a a2 b(a − b) + b2 c(b − c) + c2 a(c − a) = a+b+c− . abc

a+b+c =

Then a2 b(a − b) + b2 c(b − c) + c2 a(c − a ≥ 0, and our proof is complete. 10. Let a, b, c be positive real numbers. Prove that a+c b+a c+b a b c + + ≥ + + . b c a b+c c+a a+b Solution: We rewrite our inequality as       a b b c c b c a a − + − + − ≥ + + , b b+c c c+a a a+b a+b b+c c+a ca ab bc b c a + + ≥ + + . b(b + c) c(c + a) a(a + b) a+b b+c c+a Applying Cauchy Schwarz inequality,  2 b c a + + ≤ a+b b+c c+a    ab bc ca bc ca ab ≤ + + + + . c(a + b) a(b + c) b(c + a) a(a + b) b(b + c) c(c + a) It remains to show that ab bc ca ca ab bc + + ≤ + + . c(a + b) a(b + c) b(c + a) b(b + c) c(c + a) a(a + b) Let (x, y, z) be a cyclic permutation of (a, b, c) such that x ≥ y ≥ z, then xy zx yz 1 1 1 ≥ ≥ and ≤ ≤ . z y x x+y z+x y+z Therefore, according to the Rearrangement inequality, we have 1 ab 1 ca 1 bc · + · + · a a+b c c+a b b+c

≥ = =

xy 1 zx 1 yz 1 · + · + · z x+y y z+x x y+z xy yz zx + + z(x + y) x(y + z) y(z + x) ab bc ca + + . c(a + b) a(b + c) b(c + a)

Our proof is complete. Equality holds if and only if a = b = c. 117

11. Let a, b, c be positive real numbers. Prove that a+b b+c c+a (a + b + c)2 + + ≤ . b+c c+a a+b ab + bc + ca Solution: Rewrite our inequality as (a + b) [a(b + c) + bc] (b + c) [b(c + a) + ca] (c + a) [c(a + b) + ab] + + ≤ (a+b+c)2 , b+c c+a a+b bc(a + b) ca(b + c) ab(c + a) + + ≤ ab + bc + ca. b+c c+a a+b Let (x, y, z) be a cyclic permutation of (a, b, c) such that x ≥ y ≥ z, then x + y ≥ z + x ≥ y + z and

xy zx yz ≥ ≥ . x+y z+x y+z

Therefore, according to the Rearrangement inequality, we obtain ab + bc + ca = xy + yz + zx zx yz xy + zx · + yz · = xy · x+y z+x y+z ab ca bc ≥ (c + a) · + (b + c) · + (a + b) · a+b c+a b+c bc(a + b) ca(b + c) ab(c + a) = + + . b+c c+a a+b Our proof is complete. 12. Let a, b, c, d be nonnegative real numbers such that a + b + c + d = 4. Prove that a2 bc + b2 cd + c2 da + d2 ab ≤ 4. Solution: Let (x, y, z, t) is a cyclic permutation of (a, b, c, d) such that x ≥ y ≥ z ≥ t, then xyz ≥ xyt ≥ xzt ≥ yzt. Therefore, according to the Rearrangement inequality, we obtain x · xyz + y · xyt + z · xzt + t · yzt ≥ a · abc + b · bcd + c · cda + d · dab = a2 bc + b2 cd + c2 da + d2 ab. That is a2 bc + b2 cd + c2 da + d2 ab ≤ x · xyz + y · xyt + z · xzt + t · yzt   xy + tz + xz + ty 2 = (xy + tz)(xz + ty) ≤ 2  2  2 (x + t)(y + z) (x + y + z + t)2 = ≤ = 4. 2 8 Remark. We can also prove this inequality by AM-GM inequality as follows: 118

We have a2 bc + b2 cd + c2 da + d2 ab = ac(ab + cd) + bd(bc + da). If ab + cd ≥ bc + da, then a2 bc + b2 cd + c2 da + d2 ab ≤ ac(ab + cd) + bd(ab + cd)   ac + bd + ab + cd 2 = (ac + bd)(ab + cd) ≤ 2 2  2  (a + d)(b + c) (a + b + c + d)2 = ≤ = 4. 2 8 If bc + da ≥ ab + cd, then a2 bc + b2 cd + c2 da + d2 ab ≤ ac(bc + da) + bd(bc + da)   ac + bd + bc + da 2 = (ac + bd)(bc + da) ≤ 2 2  2  (a + b + c + d)2 (a + b)(c + d) ≤ = 4. = 2 8 This completes the proof. 13. Let a, b, c, d be positive real numbers. Prove that 

a a+b+c

2

 +

b b+c+d

2

 +

c c+d+a

2

 +

d d+a+b

2

4 ≥ . 9

Solution: Let (x, y, z, t) is a cyclic permutation of (a, b, c, d) such that x ≥ y ≥ z ≥ t, then 1 1 1 1 ≥ ≥ ≥ . (x + y + z)2 (x + y + t)2 (x + z + t)2 (y + z + t)2 Therefore, according to the Rearrangement inequality, we obtain 1 1 1 1 + y2 · + z2 · + t2 · ≤ (x + y + z)2 (x + y + t)2 (x + z + t)2 (y + z + t)2 2  2  2  2  a b c d ≤ + + + . a+b+c b+c+d c+d+a d+a+b x2 ·

It remains to show that x2 y2 z2 t2 4 + + + ≥ . 2 2 2 2 (x + y + z) (x + y + t) (x + z + t) (y + z + t) 9 The inequality being homogeneous, hence we can assume that x + y + z + t = 1. The above inequality becomes     x2 y2 t2 z2 4 + + + ≥ . 2 2 2 2 (1 − t) (1 − x) (1 − z) (1 − y) 9 119

Setting u = x + t (0 < u < 1), and v = xt, we have x2 t2 2v 2 − 2(u − 1)(2u − 1)v + u2 (1 − u)2 + = = f (v), (1 − t)2 (1 − x)2 (1 − u + v)2 and   2(1 − u) (3 − 2u)v − (1 − u)3 f (v) = , (1 − u + v)3 (1 − u)3 . f 0 (v) = 0 ⇔ v = 3 − 2u 0

From now, by making variation board, we obtain   (1 − u)3 −2u2 + 4u − 1 f (v) ≥ f = ∀v > 0. 3 − 2u (2 − u)2 Similarly, if we put w = y + z = 1 − u, then we also have y2 z2 −2w2 + 4w − 1 −2(1 − u)2 + 4(1 − u) − 1 1 − 2u2 + ≥ = = . (1 − z)2 (1 − y)2 (2 − w)2 (1 + u)2 (1 + u)2 It remains to show that −2u2 + 4u − 1 1 − 2u2 4 + ≥ , 2 2 (2 − u) (1 + u) 9 which can be easily simplified to (1 − 2u)2 (11 + 10u − 10u2 ) ≥ 0, 9(1 + u)2 (2 − u)2 which is clearly nonnegative and our proof is complete. Remark. We can also prove this inequality using Holder inequality. See Old And New Inequalities 2, Vo Quoc Ba Can - Cosmin Pohoata, GIL publishing house, 2008. 14. Let a, b, c be positive real numbers such that a2 + b2 + c2 = 3. Prove that a3 b2 + b3 c2 + c3 a2 ≤ 3. Solution: Let (x, y, z) be a cyclic permutation of (a, b, c) such that x ≥ y ≥ z, then x2 y 2 ≥ x2 z 2 ≥ y 2 z 2 . Hence, the Rearrangement inequality yields that a3 b2 + b3 c2 + c3 a2 = a · a2 b2 + b · b2 c2 + c · c2 a2 ≤ x · x2 y 2 + y · x2 z 2 + z · y 2 z 2
= y x3 y + yz 3 + x2 z 2   2 2 2 2 2 x +y 2 y +z 2 2 +z · +x z ≤ y x · 2 2 3 1 = y(x2 + z 2 )(x2 + y 2 + z 2 ) = y(x2 + z 2 ) 2v 2 !3 u 2 2 2 2 u y 2 + x +z + x +z 2 2 ≤ 3t = 3. 3 120

This completes our proof. Equality holds if and only if a = b = c = 1. Remark. We can also prove this inequality using Cauchy Schwartz inequality. See Old And New Inequalities 2. 15. Let a, b, c be nonnegative real numbers such that a + b + c = 1. Prove that 4 a2 b + b2 c + c2 a ≤ . 27 ) Solution: Let (x, y, z) be a cyclic permutation of (a, b, c) such that x ≥ y ≥ z, then xy ≥ xz ≥ yz. Hence, the Rearrangement inequality yields a2 b + b2 c + c2 a = a · ab + b · bc + c · ca = x · xy + y · xz + z · yz = y(x2 + z 2 + xz) ≤ y(x + z)2   1 2y + x + z + x + z 3 4 ≤ = . 2 3 27 16. Let a, b, c be positive real numbers such that a2 + b2 + c2 = 3. Prove that a2 b + b2 c + c2 a ≤ 2 + abc. Solution: Similar to the above problem, let (x, y, z) be a cyclic permutation of (a, b, c) such that x ≥ y ≥ z, then by Rearrangement inequality, we also have that a2 b + b2 c + c2 a ≤ y(x2 + z 2 + xz). And we can deduce our inequality to y(x2 + z 2 + xz) ≤ 2 + xyz, y(x2 + z 2 ) ≤ 2, which is true because y(x2 + z 2 ) = y(3 − y 2 ) = 2 − (y + 2)(y − 1)2 ≤ 2. 17. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that a2 + b2 + c2 4abc + ≥ 2. ab + bc + ca a2 b + b2 c + c2 a + abc Solution: Similar to the above problem, let (x, y, z) be a cyclic permutation of (a, b, c) such that x ≥ y ≥ z, then by Rearrangement unequality, we also have that a2 b + b2 c + c2 a ≤ y(x2 + z 2 + xz). 121

And we can deduce our inequality to x2 + y 2 + z 2 4xyz + ≥ 2, 2 2 xy + yz + zx y(x + z + xz) + xyz 2(x2 + z 2 ) x2 + y 2 + z 2 ≥ , xy + yz + zx (x + z)2 which can be easily simplified to (xy + yz − x2 − z 2 )2 ≥ 0, Equality holds if and only if a = b = c or a = b, c = 0 and its cyclic permutations. Remark. By the same manner, we can prove that 4(a2 + b2 + c2 ) 9abc + 2 ≥ 7, ab + bc + ca a b + b2 c + c2 a abc 2(a2 + b2 + c2 ) + 2 ≥ 1. 2 (a + b + c) a b + b2 c + c2 a 18. Let a, b, c be positive real numbers. Prove that √ √ √ a a+ b+ c b c √ √ ≥ . +√ +√ c+a a+b b+c 2 √ √ √ Solution: Setting x = a, y = b, z = c and squaring both sides, we can rewrite our inequality as X x4 X x2 y 2 p 2 + 4 ≥ (x + y + z)2 . 2 + y2 2 + y 2 )(y 2 + z 2 ) x (x cyc cyc Let (m, n, p) be a cyclic permutation of (x, y, z) such that m ≥ n ≥ p, then m2 n2 p2 m2 n2 p 2 √ ≥p ≥p m 2 + n2 p2 + m2 n2 + p 2 and

1 1 1 ≤p ≤p . 2 2 2 2 +n p +m n + p2 Therefore, using Rearrangement inequality, we obtain X x2 y 2 = x2 + y 2 cyc √

m2

m2 n2 p2 m2 n2 p 2 + + m2 + n2 p2 + m2 n2 + p2 m 2 n2 1 p2 m2 1 n2 p 2 1 = √ ·√ +p ·p +p ·p 2 2 2 2 2 2 2 2 2 2 2 m +n m +n p +m p +m n +p n + p2 x2 y 2 1 z 2 x2 1 y2z2 1 ≤ p ·p +√ ·p +p ·p 2 2 2 2 2 2 2 2 2 2 2 z +x x +y y +z x +y y +z y + z2 X x2 y 2 p = . (x2 + y 2 )(y 2 + z 2 ) cyc =

122

It remains to show that 2

X x2 y 2 x4 + 4 ≥ (x + y + z)2 , 2 + y2 2 + y2 x x cyc cyc

X

which is obviously true because 2

x4 x2 + y 2 cyc

X

=

X x4 + y 4 cyc

= =

+

x2 + y 2

cyc

X x4 + y 4

+

x2 + y 2 cyc X x4 + y 4 cyc

x2 + y 2

X x4 − y 4

X

x2 + y 2 (x2 − y 2 )

cyc

,

and X x4 + y 4

+4

X x2 y 2 = x2 + y 2 cyc

x2 + y 2 X  x4 + y 4 + 4x2 y 2 x2 + y 2 + 4xy  X x2 + y 2 + 4xy − + = x2 + y 2 2 2 cyc cyc !2 !2 X X X (x − y)4 + x ≥ x . = 2(x2 + y 2 ) cyc cyc cyc cyc

Remark. In the same manner, we can prove that r r r √ √ √ a3 b3 c3 a+ b+ c √ + + ≥ . a2 + ab + b2 b2 + bc + c2 c2 + ca + a2 3 19. Let x, y, z be positive real numbers. Prove that √ s x (x + y)(y + z)(z + x) y z 3 3 √ +√ +√ ≤ · . x+y y+z 4 xy + yz + zx z+x Solution: We cam rewrite our inequality as X cyc

√ 3 3 p ·p ≤ √ . 4 xy + yz + zx (x + y)(x + z) (x + y)(y + z) x

1

Let (a, b, c) be a cyclic permutation of (x, y, z) such that a ≥ b ≥ c, then a b c p ≥p ≥p (a + b)(a + c) (b + c)(b + a) (c + a)(c + b) and 1 1 1 p ≤p ≤p . (a + b)(a + c) (b + c)(b + a) (c + a)(c + b) 123

Hence, by Rearrangement inequality, we obtain x 1 p ·p ≤ (x + y)(x + z) (x + y)(y + z) cyc

X

a 1 b 1 p ·p +p ·p + (a + b)(a + c) (c + a)(c + b) (b + c)(b + a) (b + c)(b + a) c 1 +p ·p (c + a)(c + b) (a + b)(a + c) " # b 1 1+ p . = p (b + c)(b + a) (b + c)(b + a)

≤

It remains to show that s " # √ b ab + bc + ca 3 3 1+ p ≤ . (b + c)(b + a) 4 (b + c)(b + a) Indeed, setting u = √

b (b+c)(b+a)

≤ 1, then

q

xy+yz+zx (y+z)(y+x)

=

√

1 − u2 . Ap-

plying AM-GM inequality, we have s " # p p ab + bc + ca b 1+ p = (1 + u) 1 − u2 = (1 + u)3 (1 − u) (b + c)(b + a) (b + c)(b + a) r (1 + u) · (1 + u) · (1 + u) · 3(1 − u) = 3 s √ (3/2)4 3 3 = . ≤ 3 4 Our inequality is proved. Equality holds if and only if x = y = z. 20. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that a 5√ b c √ ≤ +√ +√ a + b + c. 4 c+a a+b b+c Solution: After squaring, we can rewrite our inequality as X X a2 ab 25 X p +2 ≤ a, a+b 16 cyc (a + b)(b + c) cyc cyc X p 2 cyc

ab (a + b)(b + c)

≤

X ab 9X a+ . 16 cyc a+b cyc

Let (x, y, z) be a cyclic permutation of (a, b, c) such that x ≥ y ≥ z, then √

xy zx yz ≥√ ≥√ , x+y y+z z+x

√

1 1 1 ≥√ ≥√ . y+z x+y z+x

and

124

Therefore, according to the Rearrangement inequality, we obtain ab bc ca p +p +p = (a + b)(b + c) (b + c)(c + a) (c + a)(a + b) 1 bc 1 ca 1 ab ·√ +√ ·√ +√ ·√ = √ c+a c+a a+b b+c b+c a+b xy 1 zx 1 yz 1 ≤ √ ·√ +√ ·√ +√ ·√ x+y y+z y+z x+y z+x z+x y(x + z) xz = p + . (x + y)(y + z) x + z From now, we can deduce our inequality to X xy 2y(x + z) 2xz 9X p + ≤ x+ , 16 cyc x+y (x + y)(y + z) x + z cyc 2y(x + z) 9 xy yz zx p ≤ (x + y + z) + + − . 16 x+y y+z z+x (x + y)(y + z) Now, using AM-GM inequality, we have that x+y+z 2 2 p ≤ + . 2(x + y)(y + z) x + y + z (x + y)(y + z) and it remains to show that y(x + z)(x + y + z) 2y(x + z) 9 xy yz zx + ≤ (x+y +z)+ + − , 2(x + y)(y + z) x+y+z 16 x+y y+z z+x y(x + z)(x + y + z) 2y(x + z) 9 y(xy + yz + 2xz) zx + ≤ (x+y+z)+ − , 2(x + y)(y + z) x+y+z 16 (x + y)(y + z) z+x Since this inequality is homogeneous, we may assume that x + z = 1, put t = xz, then t − y(1 − y) = xz − y(x + z − y) = (y − x)(y − z) ≤ 0. Hence 0 ≤ t ≤ y(1 − y). The above inequality becomes y(y + 1) 2y 9 y(y + 2t) + ≤ (y + 1) + 2 − t, 2 2(y + y + t) y + 1 16 y +y+t 9 2y y(4t + y − 1) (y + 1) − + − t ≥ 0. 16 y + 1 2(t + y + y 2 ) After expanding, we can see that our inequality has the form g(t) = At2 +Bt+C ≥ 0, with A < 0. Therefore, in order to prove this inequality, we just need to prove that it holds when t = 0 and t = y(1 − y), or in the other words, we must prove that f (0) ≥ 0 and f (y(1 − y)) ≥ 0. But it is true because 9 2y y(y − 1) (1 − 3y)2 f (0) = (y + 1) − + = ≥ 0, 16 y + 1 2(y + y 2 ) 16(1 + y) 9 2y y [4y(1 − y) + y − 1] f (y(1 − y)) = (y + 1) − + 16 y + 1 2 [y(1 − y) + y + y 2 ] 5 + 2y + 13y 2 − 16y 3 = > 0 (since 1 ≥ y ≥ 0). 16(y + 1) f (t) =

Thus f (t) ≥ 0 and our proof is complete. 125

21. Let a, b, c be positive real numbers such that a + b + c = 3. Prove that r r r a b c 3 + + ≤ . 2 2 2 b +3 c +3 a +3 2 Solution: Let (x, y, z) be a cyclic permutation of (a, b, c) such that x ≥ y ≥ z, then √

x≥

√

y≥

√

z and √

1 1 1 ≥p ≥√ . 2 2 +3 x +3 y +3

z2

Therefore, by Rearrangement inequality, we obtain r r r a b c + + = 2 2 2 b +3 c +3 a +3 √ √ √ 1 1 1 + b· √ + c· √ = a· √ b2 + 3 c2 + 3 a2 + 3 √ √ 1 1 1 √ ≤ + y·p + z·√ x· √ z2 + 3 x2 + 3 y2 + 3 r r r x z y = + + . z2 + 3 x2 + 3 y2 + 3 And we can deduce our inequality to r r r 3 x z y + + ≤ . 2 2 2 z +3 x +3 y +3 2 Setting t = xz, then 0 ≤ t ≤ inequality, we have r

x + z2 + 3

r

z x2 + 3

(x+z)2 4

=

(3−y)2 . 4

s ≤

 (x + z)

By Cauchy-Schwartz

1 1 + z 2 + 3 x2 + 3



s

(3 − y)(x2 + z 2 + 6) x2 z 2 + 3(x2 + z 2 ) + 9

s

(3 − y)(y 2 − 6y + 15 − 2t) . t2 − 6t + 3y 2 − 18y + 36

=

= Consider the function f (t) =

y 2 − 6y + 15 − 2t t2 − 6t + 3y 2 − 18y + 36

We have 0

f (t) =

  2 t2 − (y 2 − 6y + 15)t + 9

, (t2 − 6t + 3y − 18y + 3)2  √ y 2 −6y+15+(3−y) y 2 −6y+21 t = t1 = 2 √ f 0 (t) = 0 ⇒  . y 2 −6y+15−(3−y) y 2 −6y+21 t = t2 = 2 126

If y ≥ 1 then we have t1 ≥ t2 ≥

(3−y)2 , 4

therefore since 0 ≤ t ≤

we have f 0 (t) ≥ 0, and f (t) is increasing for all 0 ≤ t ≤   8 (3 − y)2 = 2 . f (t) ≤ f 4 y − 6y + 21 Hence r

x + 2 z +3

r

z + 2 x +3

Since s

8(3 − y) + 2 y − 6y + 21

r

r

y ≤ 2 y +3

y 2 y +3

8(3 − y) + 2 y − 6y + 21

s ≤

= If 1 ≥ y ≥ 0, then we have t1 ≥ f (t) ≤ f (t2 ) =

s

(3−y)2 . 4

r y2

(3−y)2 , 4

Thus

y 3 ≤ . +3 2



 4(3 − y) y (2 + 1) 2 + y − 6y + 21 y 2 + 3 s (y 2 + 15)(y − 1)2 3 3 1− 2 ≤ . 2 2 (y + 3)(y − 6y + 21) 2

(3−y)2 4

≥ t2 > 0, therefore we obtain

2 p . (3 − y) y 2 − 6y + 21 + y − 3

Hence s r r r r x z y 2 y + + ≤ p + . 2 2 2 2 z +3 x +3 y +3 y +3 y 2 − 6y + 21 + y − 3 We need to prove s

r

2

p + y 2 − 6y + 21 + y − 3

3 y ≤ . y2 + 3 2

We have 4(y 2 − 6y + 21) − (9 − y)2 = 3(y − 1)2 ≥ 0, It follows that p y+3 9−y y 2 − 6y + 21 + y − 3 ≥ +y−3= . 2 2 and it suffices to prove that r 2 y 3 √ + ≤ , y2 + 3 2 y+3 which is true because r 2 y √ + 2 y +3 y+3

s ≤

=



(2 + 1) 3 2

s 1−



(y − 1)2 (y + 1) 3 ≤ . (y + 3)(y 2 + 3) 2

Equality holds if and only if a = b = c = 1.

127

2 y + 2 y+3 y +3

1.17

Working backwards

Consider the following hard problem: Non-negative reals x, y, z satisfy x2 + y 2 + z 2 + xyz = 4. Then xy + yz + zx − xyz ≤ 2. The condition is hard to use, however the conclusion is quite simple. For this reason let us reverse the problem: Assume that xy + yz + zx − xyz = 2, we shall then prove that x2 + y 2 + z 2 + xyz ≥ 4. This new question implies the original one: if we would find a, b, c with a2 + b2 + c2 + abc = 4 and ab+bc+ca−abc > 2, we could take a 0 < t < 1 with t2 (ab+bc+ca)−t3 abc = 2, then setting x = ta, y = tb, z = tc we would get x2 + y 2 + z 2 + xyz < a2 + b2 + c2 + abc = 4, contradiction. So, let’s prove the new problem. Assume that x ≤ y ≤ z. We have z = Now if x + y < xy so

2 − xy . x + y − xy

1 1 + < 1 we deduce y + z < yz, x + z < xz so x y

x(y + z − yz) + y(x + z − xz) + z(x + y − xy) ≥ 0 so 2(xy + yz + zx − xyz) − xyz ≤ 0 so xyz ≥ 4 and then x2 + y 2 + z 2 + xyz ≥ xy + yz + zx − xyz + 2xyz ≥ 10 > 4. So assume x + y − xy > 0. Set xy = a, x + y − xy = b. Then we must prove that if a + bz = 2 then (a + b)2 − 2a + z 2 + az ≥ 4. 2−a the relation to prove becomes By replacing z = b
b2 − b + 1 a2 + 2a(b3 − b2 + b − 2) + (b2 − 2)2 . Regarded as an equation in a, its discriminant is −b2 (b − 1)(b2 − 5b + 5) so it can be positive for b < 1 or b ∈

√ √ ! 5− 5 5+ 5 , . 2 2

If b < 1. Notice that we must have (x + y)2 ≥ 4xy or 4a ≤ (a + b)2 or a2 + 2a(b − 2) + b2 ≥ 0, 128

which yields a2 ≥ 2a(2 − b) − b2 . And we can deduce our inequality to
  b2 − b + 1 2a(2 − b) − b2 + 2a(b3 − b2 + b − 2) + (b2 − 2)2 ≥ 0, or (1 − b)(4 + 4b − b2 − 4ab) ≥ 0. Moreover, as z ≥ y ≥ z, we have 2z ≥ x + y = a + b, therefore 2 = a + bz ≥ (a + b)b , or 0 ≤ a ≤ 2 − b. It follows that 4 + 4b − b2 − 4ab ≥ 4 + 4b − b2 − a+ 2 4(2 − b)b = −4b + 3b2 +!4 = 2b2 + (b − 2)2 > 0. √ √ 5− 5 5+ 5 If b ∈ , our inequality is trivial because b3 − b2 + b − 2 > 0. , 2 2 This completes our proof. Exercise: 1. If x2 + y 2 + z 2 = 2 then x + y + z − xyz ≤ 2. Solution: If one of x, y, z is negative, for example z < 0, then we have x2 + y 2 z2 z − xyz = z(1 − xy) ≤ 0 as from AM-GM 1 − xy ≥ 1 − = . 2 p 2 Therefore x + y + z − xyz = x + y + z(1 − xy) ≤ x + y ≤ 2(x2 + y 2 ) ≤ 2 from Cauchy Schwartz. If x, y, z are all non-negative. As above, we prove that if x+y+z−xyz = 2 2−x−y then x2 + y 2 + z 2 ≥ 2. z = . Let a = x + y, b = xy. Then 1 − xy 2−a (2 − a)2 z = , and we need to prove that a2 − 2b + ≥ 2 or that 1−b (1 − b)2 a2 (b − 1)2 + (a − 2)2 − 2b(b − 1)2 ≥ 2(b − 1)2 , or (a − 2)2 + a2 (b − 1)2 − 2(1 − b)2 (1 + b) ≥ 0, If b ≥ 1 then a2 ≥ 4b ≥ 2(1 + b) and the expression in clearly nonnegative. If b < 1 then the discriminant, looked as an equation in a, is 2b2 (b − 1)3 , so negative, and in this case the conclusion also follows.

1.18

Mixing variables

In the previous chapter we took a quantity Q(y1 , y2 , . . . , yn ) = x1 y1 + x2 y2 + . . . + xn yn and increased or decreased it by exchanging some of the yj . In this chapter we will take a similar quantity as a function of some variables (usually it’s symmetric in these variables) and change the variables thus increasing or decreasing the function. We distinguish two sorts of change: one is replacing a variable by a constant or a value that equals other variable (i.e. set xi = xj ), and the other by replacing two (or more variables) by a new common value 129

(usually some sort of mean) - this way is particularly useful when we have some conditions on the variables. This method is called the method of mixing variables and it is used to reduce the number of variables until we can prove the inequality directly. Let’s illustrate this method. Take Schur’s inequality: a3 + b3 + c3 + 3abc − a2 b − ab2 − a2 c − ac2 − b2 c − bc2 ≥ 0. Take now f (a, b, c) = a3 + b3 + c3 + 3abc − a2 b − ab2 − a2 c − ac2 − b2 c − bc2 . Then f (a, a, c) − f (a, b, c) = (a − b)(b2 + 2ac − a2 − bc − c2 ). Since the inequality is symmetric in a, b, c, we can suppose that c < a < b. Then a − b < 0 and b2 + 2ac − a2 − bc − c2 = (b2 − a2 ) + (ac − bc) + (ac − c2 ) = (b − a)(a + b − c) + (ac − c2 ) ≥ 0 so f (a, a, c) < f (a, b, c) and so the inequality is sufficient to prove for the case when the two variables are equal. But in this case it reduces to c(a − c)2 ≥ 0, true! Take now a harder inequality: If x2 + y 2 + z 2 + xyz = 4, x, y, z > 0 then x + y + z ≤ 3. The condition is quite uncomfortable to tackle. On the contrary the conclusion sounds simple. We have seen such inequalities in the section on working backwards That’s why let’s reverse them: prove that if x + y + z = 3 then x2 + y 2 + z 2 + xyz ≥ 4. This would provide a solution for the original inequality: if we would have x y z x + y + z > 3 then dividing x, y, z by a constant a > 1 such that + + = 3 a a a  x 2  y 2  z 2 xyz we would have + + + < x2 + y 2 + z 2 + xyz = 4 a a a aaa contradiction. So, let’s prove the new inequality. Let f (x, y, z) = x2 + y 2 + z 2 + xyz.  x+y x+y Since x + y + z is fixed, consider f , , z we have 2 2   x+y x+y (x − y)2 z(x − y)2 f (x, y, z) − f , ,z = − >0 2 2 2 4 if we suppose z < 2 which we can do since x + y + z = 3 and the inequality is x+y x+y symmetric in x, y, z. So f (x, y, z) ≥ f ( , , z). The inequality is just 2 2 enough to prove when two of the variables are equal, so y = x, z = 1 − 2x. The condition now becomes (x − 1)2 (5 − 2x) ≥ 0 and is true. 130

Finally, one can not make two variables equal but even make their difference bigger. Consider the inequality a2 b + b2 c + c2 a ≤

4 if a + b + c = 1, a, b, c ≥ 0. 27

Let f (a, b, c) = a2 b + b2 c + c2 a. Then f (0, a + b, c) − f (a, b, c) = (ca2 + 2abc − c2 a − a2 b) = a(ac + 2bc − c2 − ab) = a [bc − (c − a)(c − b)] . So if c is the between b and a then f (0, a + b, c) ≥ f (a, b, c). But we can clearly assume this since the inequality is cyclic. The inequality is so sufficient to prove when one variable is 0. So if a = 0, b = x, c = 1 − x it’s equivalent to 4 x x x2 (1 − x) ≤ which follows from the AM-GM inequality for , , 1 − x. 27 2 2 Exercises 2

1. If a, b, c > 0 then 2(a2 + b2 + c2 ) + 3(abc) 3 ≥ (a + b + c)2 . Solution: Let’s try to mix a, b into their geometric mean, trying to 2

invariate the more complicated (abc) 3 . 2

The difference 2(a2 + b2 + c2 ) + 2(abc) 3 − (a + b + c)2 will then change by  √ √  −2(a − b)2 − (2 ab + c)2 + (a + b + c)2 = −(a − b)2 + 2 a + b − 2 ab c = −(a − b)2 +

2(a − b)2 c √ ≤0 a + b + 2 ab

when c is the smallest of a, b, c. So we may assume that a = b. Take a = b = 1, c = x3 ≤ 1. The inequality becomes 4 + 2x6 + 3x2 − (2 + x3 )2 ≥ 0 or x6 − 4x3 + 3x2 ≥ 0 or x2 (x − 1)2 (x2 + 2x + 3) ≥ 0, which is true. 2. If a, b, c > 0 with abc = 1 then a2 + b2 + c2 + 3 ≥ a + b + c + ab + bc + ca. Solution: Let f (a, b, c) = a2 +b2 +c2 +3−ab−bc−ca−a−b−c. To preserve the condition abc = 1, let’s mix b, c into their geometric mean. We com   √ √  √ √ 2 √ √ 2 pute f (a, b, c) − f a, bc, bc = b− c b+ c −1−a . √  √ √ 2 Now if a = min{a, b, c} ≤ 1 then b + c ≥ 4 bc ≥ 4 ≥ 1 + a, so f 1 decreases. So assume b = c = , a = x2 . x We then compute f to be x4 + 3 +

1 2 − − x2 − 2x = x2 x = 131

x6 − x4 − 2x3 + 3x2 − 2x + 1 x2 (x − 1)2 (x4 + 2x3 + 3x2 + 1) ≥ 0. x2

3. Let a, b, c, d ≥ 0 with a + b + c + d = 1. Show that abc + bcd + cda + dab ≤

1 176 + abcd. 27 27

Solution: This is a classic example of mixing variables. Since the inequality is symmetric, we remember the principle saying ”in almost all cases, the extreme values of a symmetric function are achieved when almost all variables are equal”. As we have that the sum of a, b, c, d, it is natural to try to mix two numbers into their arithmetic mean. 1 176 So, let f (a, b, c, d) = abc + bcd + cda + dab − − abcd. Then 27 27   a+b a+b f , , c, d − f (a, b, c, d) = 2 2     2 (a + b) 176 (a + b)2 = − ab (c + d) − cd − ab . 4 27 4   a+b a+b (a + b)2 − ab is non-negative, f , , c, d ≥ f (a, b, c, d) Since 4 2 2 176 (c + d)2 a+b a+b when c+d ≥ cd. As cd ≤ , f( , , c, d) ≥ f (a, b, c, d) 27 4 2 2 2 176 (c + d) 27 when c + d ≥ or ≥ c + d. This certainly happens when 27 4 44 a+b . a ≥ b ≥ c ≥ d, so we have f (a, b, c, d) ≤ f (m, m, c, d) where m = 2 1 27 Then m+d ≤ < , so we can replace (m, c) with their mean n, to get 2 44 f (a, b, c, d) ≤ f (m, m, c, d) ≤ f (m, n, n, d). Analogously, we can replace any two of the three greater numbers (except d) by their arithmetic mean.   a+b+c a+b+c a+b+c We can now conclude that f (a, b, c, d) ≤ f , , ,d 3 3 3 by passing to the limit (remember the topic ”limits in inequalities”). So 1 a+b+c 1 we must prove that f (x, x, x, 1 − 3x) ≤ 0 where ≤ x = ≤ . 4 3 3 This is easy to do, as this equals 3x2 (1 − 3x) + x3 −

1 176 3 − x (1 − 3x) = 27 27 =

1 (528x4 − 392x3 + 81x2 − 1) 27 1 (3x − 1)(4x − 1)2 (11x + 1). 27

4. Let a, b, c, d ≥ 0. Show that a4 + b4 + c4 + d4 + 2abcd ≥ a2 b2 + b2 c2 + c2 d2 + d2 a2 + b2 b2 + a2 c2 . Solution: We shall present two solutions by mixing here. The first is the mix by product, which invariates the term 2abcd. Indeed, let f (a, b, c, d) = a4 +b4 +c4 +d4 +2abcd−(a2 b2 +b2 c2 +c2 d2 +d2 a2 +b2 b2 +a2 c2 ). 132

Then f (a, b, c, d) − f

√

 √ ab, ab, c, d =

= a4 + b4 − 2a2 b2 − (c2 + d2 )(a2 + b2 − 2ab)   = (a − b)2 (a + b)2 − c2 − d2 . This is nonnegative when (a+b)2 ≥ c2 +d2 so √ if we suppose a ≥ b ≥ c ≥ d, then f (a, b, c, d) ≥ f (m, m, c, d) where m = ab. As in the problem above, we can mix any two of the larger three number into their geometric product, hence by passing to the limit it suffices to prove the inequality for a = b = c ≥ d. As the inequality is homogeneous, we can set a = b = c = 1, d = x ≤ 1. We have to prove that x4 +3+2x ≥ 3 + 3x2 , or that x3 + 2 ≥ 3x, which is obvious by AM-GM. The second solution is the unusual sort of mix, in which we replace one variable by another. Define f as above. Let’s look at f (a, b, c, d) − f (a, a, c, d). It equals b4 − a4 + 2acd(b − a) − (b2 − a2 )(a2 + c2 + d2 ) =   = (b − a) (b + a)(b2 + a2 ) + 2bcd − (a2 + c2 + d2 )(b + a) . So it’s positive when b ≥ a and (b2 + a2 )(b + a) + 2acd ≥ (a2 + c2 + d2 )(a + b), or when b ≤ a and (b2 + a2 )(b + a) + 2acd ≤ (a2 + c2 + d2 )(a + b). As (a2 + b2 )(a + b) + 2acd − (a2 + c2 + d2 )(a + b) = = b2 (a + b) + 2acd − (c2 + d2 )(a + b) = (b2 + ab − c2 − d2 )b − (c − d)2 a, we can see that this is positive when b > a are the largest among a, b, c, d and is negative when a > b are the smallest among a, b, c, d. Then, if we assume a ≥ b ≥ c ≥ d, by our conclusion we deduce f (a, b, c, d) ≥ f (b, b, c, d) ≥ f (b, b, c, c). Finally, f (b, b, c, c) = (b2 − c2 )2 ≥ 0.

1.19

Limits in inequalities

Limits is a subject that apparently has nothing to do with inequalities. It does, as we shall list above three applications for inequalities. All of these applications are based on the following crucial fact: (F): If xi is a sequence converging to a constant a, f, g are continuous functions and f (xi ) ≥ g(xi ), then f (a) ≥ g(a). 133

The proof of this fact is very simple: if we suppose that g(a) − f (a) =  > 0 then we may choose some n with   |f (xi ) − a| < , |g(xi ) − a| < 2 2 and thus (f (xi ) − g(xi )) − (f (a) − g(a)) ≤ |f (xi ) − f (a)| + |g(xi ) − g(a)| < . But this is as f (xi ) − g(xi ) is a nonnegative number, but f (a) − g(a) = −. This fact can easily be extended to functions of more variables. An important note here is that the inequalities f (xi ) ≥ g(xi ) can be strict, but it doesn’t follow from here that the inequality f (a) ≥ g(a) is strict. Look for example 1 1 at the inequalities > 0, but lim = 0. n→∞ n n bq ap + ≥ ab, • Sometimes we have to prove an inequality like Young’s one: p q 1 m+n m+n 1 + = 1, p, q > 0. We can prove them for p = ,q = being p q n n rationals, as multiplying to m + n this inequality is a direct consequence of AM-GM. Now, as every real can be approximated by rationals, we can find 1 1 1 1 rationals pi , qi with + = 1, |pi − p| < , |qi − q| < . As pi tend to p, pi qi i i qi tend to q, from (F) we deduce that the inequality is true for p, q. • Sometimes we have to prove an inequality f (x1 , x2 , . . . , xn ) ≥  x1 + x2 + . . . + xn x1 + x2 + . . . + xn x1 + x2 + . . . + xn ≥f , ,..., n n n and we can prove that   xi + xj xi + xj f (x1 , x2 , . . . , xn ) ≥ f x1 , x2 , . . . , xi−1 , , xi+1 , . . . , xj−1 , , xj+1 , . . . , xn 2 2 for any 1 ≤ i < j ≤ n. By such operations we can actually make our sequence of numbers tend to a1 + a2 + . . . + an a1 + a2 + . . . + an ,..., . n n Indeed, consider the invariant E(x1 , x2 , . . . , xn ) =

n−1 X

X

(xi − xj )2 .

i=1 j=i+1n

Using the inequality   a+b 2 (x − a) + (x − b) ≥ 2 x − , 2 2

2

xi + xj , we decrease E by at least 2 2 (xi − xj ) . Thus, taking |xi − xj | be the greatest of all possible differences we can prove that replacing xi , xj by

134

 2 E. So between two numbers, we can decrease E to at most 1 − n(n − 1) we can make E as close to zero as possible, particularly making any of the possible differences between two number as small as possible. Since our transformations preserve the sum, it’s now clear that al the numbers get as close a1 + a2 + . . . + an to as desired. Since f is continuous, we now deduce the n claim. Like this, for example, we can prove AM-GM: 

a1 + a2 + . . . + an ≥

√ n

a1 a2 . . . an ,

and even in two ways. The first way is almost identical to what we’ve told x1 x2 . . . xn above. However, we can do it another way: take f (x1 , x2 , . . . , xn ) = √ . n x x ...x 1 2 n Then √ √ f (x1 , x2 , . . . , xn ) ≥ f (x1 , . . . , xi−1 , xi xj , xi+1 , . . . , xj−1 , xi xj , xj+1 , . . . , xn ) (the product of the number is the same, but the sum decreases from AM-GM for two numbers). Next, we can proceed analogously as above (for example, work with ln xi instead of xi ). • Limits are also applicable with integral inequalities, as the integral is in fact itself a limit. Like this, many of the classic inequalities translate into integral ones. For example, CBS (Cauchy-Schwartz). Inequality rewrites as b

Z

2

b

 Z

2

f (x)dx b

Z

h(x)dx = lim

n→∞

a

Z a

g (x)dx a

a

Indeed, as



b

2

b

Z ≥

f (x)g(x)dx

.

a

f (x1 ) + f (x2 ) + . . . + f (xn ) (n − i)a + ib , where xi = , n n

 Z b  Z b 2 f 2 (x)dx g 2 (x)dx − f (x)g(x)dx a

a

is the limit of   f 2 (x1 ) + . . . + f 2 (xn ) g 2 (x1 ) + . . . + g 2 (xn ) f (x1 )g(x1 ) + . . . + f (xn )g(xn ) 2 − , n n n which is positive from CBS. Using (F), we deduce our claim.

1.20

Derivatives in Inequalities

When we have to prove a inequality of the form f (a) ≥ f (b), a > b, we can prove that f is increasing on [b; a]. But to prove that f is increasing on [b, a] it suffices to prove that f 0 (x) ≥ 0. Consider for example the already discussed inequality an1 + an2 + . . . + ann ≥ na1 a2 . . . an , an > 0 (AM-GM). 135

We can suppose that a1 ≥ a2 ≥ . . . ≥ an . Now let f (a1 , a2 , . . . , an ) = an1 + an2 + . . . + ann − na1 a2 . . . an . Regarded as a function in a1 , it is increasing, as f 0 (a1 ) = na1n−1 − (n − 1)a2 a3 . . . an ≥ 0. So f (a1 , a2 , . . . , an ) ≥ f (a2 , a2 , a3 , . . . , an ). Now we can regard it as a function in a2 . It’s derivative is again positive, as it is 2an−1 − 2a2 a3 . . . an . Now, the method of proof is clear: by induction on 2 k we prove that f (a1 , a2 , . . . , an ) ≥ f (ak , ak , . . . , ak , ak+1 , ak+2 , . . . , an ). Indeed, the basis is proven, and for the induction step we note that f (ak , ak , . . . , ak , ak+1 , ak+2 , . . . , an ) ≥ f (ak+1 , ak+1 , . . . , ak+1 , ak+1 , ak+2 , . . . , an ) because the function g(x) = g(x, x, . . . , x, ak+1 , ak+2 , . . . , an ) is increasing as its derivative is nkxn−1 − nkxk−1 ak+1 ak+2 . . . an ≥ 0, g(ak ) ≥ g(ak+1 ). Finally, because f (an , an , . . . , an ) ≥ 0, we deduce our claim. Another use of derivatives in inequalities comes from the combination of Weierstrass Theorem with Fermat’s Theorem. Recall that Weierstrass’ theorem states that every continuous function on a compact set attains its extremes, and Fermat’s Theorem says that if x ∈ (a, b), is an extremal point of a continuous derivable function f then f 0 (x) = 0. As proving an inequality f (x1 , x2 , . . . , xn ) ≥ 0 is equivalent to min{f } ≥ 0, it suffices to consider x1 , . . . , xn that realize this minimum (Weierstrass’s theorem ensures that it is achieved). Using Fermat’s theorem, we obtain information about these points, and often we can explicitly find then and verify the inequality directly. Let’s solve the already familiar inequality x2 y + y 2 z + z 2 x ≤

4 ,x + y + z = 1 27

If we suppose x ≥ y ≥ z and substitute z = 1 − x − y then we must prove f (x, y) = x2 y + y 2 (1 − x − y) + (1 − x − y)2 x ≤

4 for x, y 27

in the compact set bounded by the conditions x ≥ y ≥ 0, y ≥ 1 − x − y and x + y ≤ 1. 136

Now let (x0 , y0 ) maximize f (this is true by Weierstrass Theorem). Therefore, from Fermat’s Theorem, either some of these conditions is an equality, either f 0 (x) = f 0 (y) = 0. In the first case, we either have one of the variables (x, y, z) zero, or two variables equal. If one is zero, we have discussed this case before, if 3 2 2 x = y then z = 1−2x and the inequality becomes  x +x (1−x)+(1−2x) x ≤  4 4 1 or f (x) = 4x3 −2x2 +x ≤ , for x ∈ 0, . As f 0 (x) = 12x2 −4x+1 ≥ 0, 27 27 2 1 we must prove this for x = , which is true. Finally, let’s get to the cream of 2 the solution: the last case, when fx0 (x, y) = fy0 (x, y) = 0. First, write out f explicitly: x3 − y 3 + 3x2 y − 2x2 − 2xy + y 2 + x . Then fx0 (x, y) = (3x − 1) (x + 2y − 1) fy0 (x, y) = (3x + 3y − 2)(x − y) 1 or x+2y = 1 and fy0 (x, y) = 0 when 3x+3y−2 = 0 3 1 or x = y. If x = y, then since fx0 (x, y) = 0 we must have x = y = which 3 1 gives us x = y = z = and if 3x + 3y = 2 = 2(x + y + z), we will have that 3 x + y = 2z, which also gives us x = y = z since x ≥ y ≥ z. Therefore, in 1 this case, we can see that extreme attains when x = y = z = but with this 3 4 value, it is trivial that f (x, y) < . This ends our proof. 27 So, fx0 (x, y) = 0 when x =

Exercises 1. Show that for a, b ≥ 0 we have 3a3 + 7b3 ≥ 9ab2 . Solution: If b = 0 it’s trivial so assume b > 0. By putting b = 1, a = x, we need to prove f (x) = 3x3 + 7 − 9x ≥ 0. As f 0 (x) = 9(x2 − 1), it suffices to look at the case x = 1, and in this case we get a positive value. q p √ √ 2. Let a, b, c > 0. Then 4 a + 3 b + c ≥ 40 abc.   p p √ 10 √ 10 Solution: Put it as a + 3 b + c ≥ abc. Let f (x) = x + 3 b + c − bcx. As f 00 (x) ≥ 0 we conclude f 0 (x) ≥ f 0 (0). Now we prove f 0 (0) ≥ 0.  p √ 9 √ 3 As f 0 (x) = 10 x + 3 b + c −bc, we have to prove that 10 (b + c) ≥ bc. √ Now take g(x) = 10(x+ c)3 −cx, then g 00 (x) ≥ 0 and so g 0 (x) ≥ g 0 (0) = 29c ≥ 0 so g 0 is positive and g is increasing so g(b) ≥ g(0) ≥ 0. Hence f 0 (0) ≥ 0 so f 0 is increasing and positive and then f (a) ≥ f (0) > 0, as desired. 137

3. Show that

1 1 + ... + < ln 3. n+1 3n

1 , for x ≥ 1. Indeed, by x+1 1 Lagrange Theorem there is y ∈ [x, x + 1] with ln(x + 1) − ln(x) = ≥ y 1 . x+1 Thus Solution: We prove that ln(x + 1) − ln x ≥

1 1 + ... + n+1 3n

< ln(n + 1) − ln(n) + . . . + ln(3n) − ln(3n − 1) = ln(3n) − ln(n) = ln 3.

4. Let a, b, c > 0 with a + b + c = 6. Find the minimum value of     1 1 1 P = a+ b+ c+ . a b c Solution: The expressions P is quite complicated to link with the relation a+b+c = 6, so we must to try to simplify P , looking at just two variables.    1 1 Thus, let’s compute the minimal value of a + b+ when a+b = b a   1 1 t−a+ . When t is constant. This is a function in a: a + a t−a a is close to the extreme values 0 and t, the expression tends to infinity, so the minimum is attained somewhere in the interior of the interval, thus at a = x where the derivative vanishes. The derivative is, as easily to compute,     1 1 1 1 1 1 g(x) = t − 2x + − − 2 t−x+ + x + . t−x x x t−x (t − x)2 x By symmetry, we see that when a = b i.e. 2x = t the derivative vanishes, so let’s extract t − 2x out of the expression:   1 1 1 2x − t + x+ +x−t− g(x) = (t − 2x) + + x(t − x) x2 x t−x    1 1 1 + x+ − 2 2 x (t − x) x 2x − t (t − 2x)(x2 − tx + 1) t(x2 + 1)(t − 2x) = (t − 2x) + + − x(t − x) x3 (t − x) x3 (t − x)2  2  (t − 2x) x (x − t)2 − t2 − 1 = . x2 (t − x)2 t or when x2 (t − x)2 = t2 + 1. 2    1 1 Thus we have established that for fixed a + b, a + b+ takes a b its minimal value when a = b or a2 b2 = (a + b)2 + 1. It’s quite hard to

So, the derivative vanishes when x =

138

compute a that satisfies the second relation and compare this value for t the value in . So we can’t actually compute the minimal value for all 2 t. t2 However, AM-GM helps us: we know that x(t − x) ≤ , so x2 (t − x)2 ≤ 4 4 t4 t . So when t2 +1 > , the minimum value occurs exactly when a = b. 16 16 t4 t4 We note that all t < 4 satisfy t2 + 1 > because t2 ≥ . Since 16 16 a + b + c = 6, some two of a + b + c sum to at most 4, that’s why we can suppose they are equal. Indeed, let a ≤ b ≤ c realize the minimum of P , then as a + b ≤ 4 we must have a = b otherwise  we  couldchange a, b to keep a + b (and c) 1 1 fixed but decrease a + b+ . Set a = b = x. If c = x then a b x = 2. If not, by the same reasoning we deduce that a2 c2 = (a + c)2 + 1 so x2 (6 − 2x)2 = (6 − x)2 + 1, or 4x4 − 24x3 + 5x2 + 12x = 37. This 3 cannot actually happen for x ≤ 2, because if x < then 4x4 < 24x3 and 2 5x2 + 12x < 5 · 1.52 + 12 · 1.5 = 29.25 < 37, and if

3 ≤ x ≤ 2 then 4x4 − 24x3 ≤ 8x3 − 24x3 = −16x3 , and clearly 2 −16x3 + 5x2 + 12x < 0 < 37.

So, x = 2 and the minimum value realizes for a = b = c = 2 and equals 125 . 8

1.21

Convexity

Convexity, despite being slightly more complicated to define than monotonicity or continuity, is a very important property of a function. Here is one of the most used definitions for convexity: Definition: A function defined on a interval I is called convex if for any 0 < α, β < 1, α + β = 1, x, y ∈ I f (αx + βy) ≤ αf (x) + βf (y). If the inequality is strict or α, β ∈ {0; 1}, the function is called strictly convex. (Note that when one of α, β is negative the inequality changes sign). When −f is convex, f is called concave. Convexity is much more used with continuous and twice derivable functions, and here’s why: • For a continuous twice derivable function f , f is convex if f 00 (x) ≥ 0. If f 00 (x) > 0 for all x, the function is called strictly convex. 139

The proof of this fact goes as follows: clearly f 0 (x) is increasing hence a+t

Z

Z

0

a+p+kt

f (x)dx ≤

f 0 (x)dx,

a+p

a

if a < b, p, t > 0. This means k(f (a + t) − f (a)) ≤ f (a + p + kt) − f (a + p). Now if we let p = t we get kf (a + t) − kf (a) ≥ f (a + (k + 1)t) − f (a + t) or f (a + t) ≤

kf (a) + f (a + kt) , k+1

and this is exactly the definition of convexity. • A continuous function f is convex if  f (x) + f (y) ≥ 2f

x+y 2



. One implication is obvious. For the other, we firstly prove by induction that k for any n ∈ [0, 1], k, n ∈ N, we have 2   2n − k kx + (2n − k)y k f (x) + f (y) ≥ f . 2n 2n 2n The basis of the induction is k = 1 and follows form the condition. Now, if we have proven the statement for n − 1, let’s prove it for n. k k 2 and we apply the induction hypothesis. If k is even, then n = n−1 2 2 If k is odd, then we k − 1, k + 1 are even and so   k−1 2n − k + 1 k−1 2n − k + 1 f (x) + f (y) ≥ f x+ y 2n 2n 2n 2n and

k+1 2n − k − 1 f (x) + f (y) ≥ f 2n 2n



 k+1 2n − k − 1 x + y . 2n 2n

Summing these two identities and using the fact that       k−1 2n − k + 1 k+1 2n − k − 1 k 2n − k f x+ y +f x+ y ≥ 2f x+ y , 2n 2n 2n 2n 2n 2n we get the induction step done. Therefore, we have proven αf (x) + βf (y) ≥ f (αx + βy), α + β = 1 140

k for α of the form n . Since every real can be represented as the limit of 2 k numbers of form n , using the continuity of the function we deduce the formula 2 (see ”Limits in Inequalities”). A useful reformulation of the definition goes as follows: if z is between x and y, then z−x z−y f (x) + f (y) f (z) ≤ x−y y−x z−y z−x z−y z−x (you may check that + = 1 and that x +y = z). x−y y−x x−y y−z In this setting one can define convexity for functions that are not necessarily defined on an interval (even though in practice we don’t really deal with such functions). A very important property of the convex function is that they realize the maximal value on a interval at the extremities of the interval (and hence concave functions realize their minimal value at the extremities of the interval). This follows directly from the definition and this is sometimes used in problems like this one: Let a, b, c ∈ (0, 1). Then a b c + + + (1 − a)(1 − b)(1 − c) ≤ 1. b+c+1 c+a+1 a+b+1 We may consider this as a function in a : f (x) =

c b x + + + (1 − x)(1 − b)(1 − c). b+c+1 x+b+1 x+c+1

We can easily see by taking the second derivative that it is convex, so takes its maximal values at 0 or 1. Analogously for b and c we conclude it’s enough to investigate the cases when all a, b, c are either 0 or 1, which are easy to handle. But the most important property of convex functions is Jensen’s inequality, which is presented next.

1.22

Jensen’s Inequality

The notion of convexity gets most of its importance in lights of the famous Jensen’s Inequality. Suppose that f is continuous. We have seen from the definition of the inequality that if z is a number between x and y then f (z) is smaller from that a linear combination of f (x) and f (y), and moreover this combination applied to x and y gives z. Let’s look at this situation from a geometrical point of view: Draw the graph of f (x). Consider the points X(x, f (x)), Y (y, f (y)) and Z(z, f (z)). If z is between x and y, then the condition of convexity tells us that f (z) ≤

z−y z−x f (x) + f (y) x−y y−x 141

which actually means that Z is below the line segment XY . From here we can deduce that every point in (XY ) is above the graph of f . This means that the area above the graph of f is a convex subset (hence why this definition). It’s clear that the reverse implication holds: if the area above the graph is convex, then function if convex (the statements ”f (z) ≤ z−y z−x f (x) + f (y)” and ”Z is below XY ” are equivalent). A good geox−y y−x metric image of convex functions is therefore a kind of ”trough” representing the graph of the function, like a parabola with positive leading coefficient. Now, take n points A1 (x1 , f (x1 )), A2 (x2 , f (x2 )), . . . , An (xn , f (xn )) on the graph of f , and assign to them positive weights λ1 , . . . , λn with λ1 +λ2 +. . .+λn = 1. The point G = (λ1 x1 + . . . + λn xn , λ1 f (x1 ) + . . . + λn xn ) is then the center of gravity of the system of points. But it’s known that the center of gravity of some points (with positive weights) lies inside the convex hull of these points. Indeed, if it wasn’t true, there would be a line l such that it separates the points A1 , . . . , An and G. But then taking a system of coordinates with l being the ordinate, we would see that G has positive abscise but Ai have negative abscises. This contradicts the obvious condition that the abscise of G is a linear combination with positive coefficients λ1 , . . . , λn of abscises of A1 , A2 , . . . , An and thus must be negative. Therefore G must lie inside the convex hull of A1 , . . . , An . As the area above the graph of f is convex, G must lie above the graph of f . This means that λ1 f (x1 ) + . . . + λn f (xn ) ≥ f (λ1 x1 + . . . + λn xn ). Thus, we have proved the following inequality: Theorem(Jensen’s Inequality) If f is a convex function on a interval I and x1 , x2 , . . . , xn ∈ I, then for any λ1 , . . . , λn ≥ 0 with sum 1 we have λ1 f (x1 ) + . . . + λn f (xn ) ≥ f (λ1 x1 + . . . + λn xn ). Jensen’s inequality has a lot of applications in inequalities. It clearly gets reversed when f is concave. For example, the function ex is convex, as its second derivative is ex > 0. 1 Taking now λi = and applying Jensen we deduce n n X

exi

i=1

n

Pn

i=1 xi

≥e

n

.

Now taking ai = exi , we obtain exactly AM-GM inequality. Alternatively, one can prove the AM-GM inequality using the concave function ln x. Many functions are either convex or concave. Here is a list of most commonused concave and convex functions: − − − −

ax is convex for a > 1 and concave for 0 < a < 1. ln x is concave. So is loga x when S a > 1. When a < 1, loga x is convex. t x is convex when t ∈ (−∞, 0) (1, ∞) and concave when t ∈ (0, 1) sin x is convex on (π, 2π) and concave on (0, π). 142

  π π π 3π and concave on − , − cos x is convex on , . 2 2 2 2    π  S 3π  π  S  3π  − tan x or cot x is convex on π, . ,π , 2π and concave on 0, 2 2 2 2 − |x − a| is convex, although not differentiable at a. 

Exercises 3 1. In a triangle show that cos A + cos B + cos C ≤ . 2 Solution: Suppose that A ≤ B ≤ C, then 0 < A ≤ B < π2 . As the funch πi A+B tion cos x is concave on 0, , we have cos A + cos B ≤ 2 cos = 2 2 C C 3 2 sin .It suffices to prove that 2 sin + cos C ≤ , which can be trans2 2 2  2 C 1 2 sin − 1 ≤ 0, true! formed into − 2 2 2. If a, b, c ∈ (0, 1) then p p p p a − a2 + 2b − b2 + 3c − c2 ≤ 6(a + b + c) − (a + b + c)2 . Solution: Set a = x, b = 2z, c = 3z. The inequality becomes s p

p p x − x2 + 2 y − y 2 + 3 z − z 2 ≤ 6

x + 2y + z − 6



x + 2y + 3z 6

It follows from Jensen’s inequality applied to the concave function

√

2 .

x − x2 .

3. If a, b, x, y, z > 0 and x + y + z = 1 then     b b b a+ a+ ≥ (a + 3b)3 . a+ x y z Solution: inequality follows from Jensen applied to the convex func The  b tion ln a + . x 4. Show that for a convex function f , we have       2x + y 2y + z 2z + x f (x) + f (y) + f (z) + f +f +f ≥ 3 3 3        x + 2y y + 2z z + 2x ≥2 f +f +f . 3 3 3 

   2z + x z + 2x ≥ 2f by Jensen. By 3 3 summing with the analogously deduced relations we get the result.

Solution: We have f (x) + f

143

n π π  X π 5. Let a0 , a1 , . . . , an ∈ , with tan ai − . Show that 4 2 4 i=0

tan a0 tan a1 . . . tan an ≥ nn+1 . π n−1 , then 0 < bi < 1 and b0 + . . . + bn ≥ . 4 n+1 n Y 1 + bi ≥ nn+1 . This is straightforward by We have to show that 1 − bi i=0   1+x Jensen, because the increasing function f (x) = ln is convex (it 1−x 2 has increasing derivative ). 1 − x2 We can also show this by using just AM-GM: if we set xi = 1 − bi then as n X bi ≥ n − 1, Solution: Set bi = tan ai −

i=0

we get 1 + bk ≥

n X

v u Y n u n xi . Mulbi . Hence by AM-GM 1 + bi ≥ t i=0,i6=k

i=0,i6=k

tiplying this by all i and dividing by x0 x1 . . . xn , we get the conclusion.

1.23

The shrinking principle and Karamata’s Inequality

Take a convex function f and two numbers x, y. For any α, β ∈ [0, 1] with α + β = 1 we have αf (x) + βf (y) ≥ f (αx + βy). Also βf (x) + αf (y) ≥ f (βx + αy). Summing we get f (x) + f (y) ≥ f (αx + βy) + f (βx + αy). Noticing the fact that (αx + βy) + (βx + αy) = x + y, we can reformulate this result as follows: If z, t are between x, y and z + t = x + y then f (x) + f (y) ≥ f (z) + f (t), for any convex f . We see that z, t are actually obtained by taking x, y and ”pulling” them closer one to another, like ”shrinking” the interval (x; y). That’s why we can call this result ”the shrinking principle”. 144

If we take now a sum f (x1 ) + . . . + f (xn ) and ”shrink” some pair of them xi , xj to a pair x0i , x0j between xi and xj we decrease the sum f (x1 ) + . . . + f (xn ). We can then perform more shrinkings. Thus, if y1 , y2 , . . . , yn are obtained from x1 , x2 , . . . , xn by some shrinkings, we have f (x1 ) + f (x2 ) + . . . + f (xn ) ≥ f (y1 ) + f (y2 ) + . . . + f (yn ). It is now natural to ask how to characterize all the sequences y1 , . . . , yn that are obtained from x1 , x2 , . . . , xn by several applications of shrinking. A first observation is that the sum is constant. Another observation is, as y1 , . . . , yn are ”closer” than x1 , . . . , xn , the greatest of xi is bigger than the greatest of yi , and the least of xi is smaller than the least of yi . We now come to the following definition: Definition. A sequence x1 ≥ x2 ≥ . . . ≥ xn majorizes another sequence y1 ≥ y2 ≥ . . . ≥ yn if and only if the following conditions hold: (i) x1 + x2 + . . . + xn = y1 + y2 + . . . + yn (ii) x1 + x2 + . . . + xm ≥ y1 + y2 + . . . + ym for any 1 ≤ m ≤ n. We claim that a sequence y1 ≥ y2 ≥ . . . ≥ yn can be obtained from x1 ≥ x2 ≥ . . . ≥ xn by shrinking if and only if it is majorized by x1 , · · · , xn . Let’s prove the first implication: that any sequence obtained by is majorized by the original sequence. Since the relation of majorization is transitive, it’s enough to prove that shrinking once yields a sequence majorized by the original one. Indeed, suppose we replace xi > xj by xi > yi > yj > xj , xi + xj = yi + yj . Then the condition (i) clearly holds, so let’s prove (ii): x1 + x2 + . . . + xm ≥ S, where S is the sum of m largest numbers from the new sequence. If m < i then actually x1 + . . . + xm = S. If j > m ≥ i but the number yj is not between the m th largest numbers of the new sequence, a strict inequality holds. Finally if m < j again equality holds. Now let’s prove the second implication: if (yi ) is majorized by (xi ), it can be obtained from (xi ) by some consecutive applications of shrinking. Pick up the least k s.t. xk 6= yk and the greatest l s.t. xl 6= yl . Clearly xk > yk , xl < yl . If xk +xl ≥ yk +yl , then we can shrink xk , xl to yk , xk +xl −yk otherwise we shrink xk , xl to xk + xl − yl , yl . It’s easy to prove that the new sequence is majorized by (xi ) and majorizes (yi ), but it has more common terms with (yi ). Hence, performing some such operations we can get al terms of the sequence equal to the terms of (yi ), as we desired. An immediate corollary of this reasoning is the Karamata’s inequality: Theorem (Karamata’s Inequality) If (xi ) majorizes (yi ) and f is a convex function then n n X X f (xi ) ≥ f (yi ). i=1

i=1

Another inequality, proved just like Karamata with the help of the shrinking principle, is Muirhead’s inequality: 145

If (pi ) majorizes (qi ) and f is a convex function, then ! ! n n X X X X f pi xσ(i) ≥ f qi xσ(i) . σ∈Sn

i=1

σ∈Sn

i=1

The proof is left to the reader as an exercise. The suggestion isPthat the shrinkages P that turn (pi ) into (qi ) induce shrinkages turn that turn ( pi xσ(i) ) into ( qi xσ(i) ). A particular case of this inequality, obtained for the function ex , is very commonly used in solving symmetric homogeneous olympiad-style inequalities, and has the following form: If x1 , . . . , xn are positive numbers and (pi ) majorizes (qi ), then X Y p X Y q i i xσ(i) ≥ xσ(i) . σ∈Sn

σ∈Sn

For example, taking the sequence (4, 1, 0) that majorizes (3, 1, 1), we get that x4 y + xy 4 + x4 z + z 4 x + y 4 z + z 4 y ≥ 2(x3 yz + y 3 xz + z 3 xy). One of Karamata’s applications is Popoviciu’s Inequality:          x+y y+z z+x x+y+z ≥2 f +f +f . f (x)+f (y)+f (z)+3f 3 2 2 2 Exercises 1. Assume that f is a convex increasing positive-valued function, and x1 ≥ x2 ≥ . . . ≥ xn ,

y1 ≥ y2 . . . ≥ yn

satisfy x1 ≥ y1 , x1 +x2 ≥ y1 +y2 , . . . , x1 +x2 +. . .+xn ≥ y1 +y2 +. . .+yn . Then show that f (x1 ) + f (x2 ) + . . . + f (xn ) ≥ f (y1 ) + f (y2 ) + . . . + f (yn ). Solution: We use induction on n. Let ci = x1 + x2 + . . . + xi − y1 − y2 − . . . − yi . If one of ci is zero, then the inequality follows from Karamata applied to (x1 , x2 , . . . , xi ) and (y1 , y2 , . . . , yi ) and induction hypothesis applied to (xi+1 , . . . , xn ) and (yi+1 , . . . , yn ). If all of the ci are greater than zero, let ck be the smallest. We can see that   ck  ck  f (y1 )+. . .+f (yn ) ≤ f y1 + +. . .+f yk + +f (yk+1 )+. . .+f (yn ), k k   ck ck and y1 + , . . . , yk + , yk+1 , . . . , yn still satisfies the condition with n n respect to (x1 , x2 , . . ., xn ). Now the inequality follows from Karamata for ck ck  (x1 , x2 , . . . , xk ) and y1 + , . . . , yk + and induction hypothesis for n n (xk+1 , . . . , xn ) and (yk+1 , . . . , yn ). 146

2

2

2

2. Show that 2a + 2b + 2c ≥ 2ab + 2ac + 2bc . Solution: It follows from the previous exercise applied to the increasing convex function 2x . Another solution is possible by applying Jensen to x the function ee . 3. Show that for a, b, c > 0 we have 9 1 1 1 + + + ≥4 a b c a+b+c



1 1 1 + + a+b b+c c+a

 .

Solution: This is just Popoviciu’s Inequality applied to the convex func1 tion . Compare this solution to the one we given in the SOS section. x 4. Prove that for any positive real numbers a1 , . . . , an we have the inequality: n n X X a1 a21 ≥ . 2 2 a + . . . + an a + . . . + an i=1 2 i=1 2 Solution: Lemma Consider positive reals a1 ≥ . . . ≥ an and b1 ≥ . . . ≥ bn with ai bi a1 + . . . + an = b1 + . . . + bn and ≥ for i < j. Then a1 , . . . , an aj bj majorizes b1 , . . . , bn . Proof : The proof is by induction. We see that

ai bi ≤ for i = 1, . . . , n. a1 b1

S S ≤ hence a1 ≥ b1 (S = a1 +. . .+an = b1 +. . .+bn ) a1 b1 Hence a1 ≥ b1 . Further, we look at the new sequences a2 , a3 , . . . , an and b2 , . . . , bn . By induction hypothesis, we get a2 , a3 , . . . , an majorizes b2 , . . . , bn . Combined with a1 ≥ b1 the lemma is proven.

Summing we get

Now let xi =

a21

a2i ai , yi = . We see that a1 + · · · + an + · · · a2n

xi a2i = 2 2 2 2 1 − xi a1 + . . . + ai−1 + ai+1 + . . . + an and

ai yi = . a1 + . . . + ai−1 + ai+1 + . . . + an 1 − yi

If we suppose a1 ≥ a2 ≥ . . . ≥ an then x1 ≥ . . . ≥ xn and y1 ≥ y2 ≥ xi a2 ai yi = 2i ≥ = for i < j. Hence by Lemma . . . ≥ yn . Further xj aj yj aj x x1 , . . . , xn majorizes y1 , . . . , yn and by noticing that f (x) = is 1−x convex, we apply Karamata QED.

147

1.24

Schur’s Inequality

A lot of contest inequalities, after homogenization, clearing denominators and other algebraic manipulations, reduce to symmetric homogeneous inequalities in some variables. One useful tool in handling these inequalities is, as we have seen above, Muirhead’s inequality. Another tool, also very helpful, is Schur’s inequality: a3 + b3 + c3 + 3abc − a2 b − b2 c − c2 a − ab2 − bc2 − ca2 ≥ 0. It can be rewritten as a(a − b)(a − c) + b(b − a)(b − c) + c(c − a)(c − b) ≥ 0. Now, a generalization follows: If f is a convex or monotonic positive valued function, then for any a, b, c from the value domain the following inequality holds: f (a)(a − b)(a − c) + f (b)(b − c)(b − a) + f (c)(c − a)(c − b) ≥ 0. Proof. Let’s suppose that a ≥ b ≥ c. Then the inequality to prove is f (a)(a − b)(a − c) + f (c)(a − c)(b − c) ≥ f (b)(a − b)(b − c). If f is increasing, then f (a) ≥ f (b), a − c ≥ b − c so f (a)(a − b)(a − c) ≥ f (b)(a − b)(b − c). As f (c)(a − c)(b − c) ≥ 0, the inequality follows. If f is decreasing, analogously we break the inequality into f (c)(a − c)(b − c) ≥ f (b)(a − b)(b − c) and f (a)(a − b)(a − c) ≥ 0. Finally, when f is convex then Jensen’s inequality for b between a and c gives b−c a−b f (a) + f (c) ≥ f (b), a−c a−c which multiplied by (a − b)(b − c) gives (b − c)2 (a − b) (a − b)2 (b − c) + f (c) ≥ f (b)(b − a)(b − c). a−c a−c To conclude the proof, it suffices to notice that f (a)

(b − c)2 (a − b) ≤ (a − c)(a − b) a−c and

(a − b)2 (b − c) ≤ (a − c)(b − c). a−c This inequality is particularly used for function f of the form at , for example for f (x) = x2 we get a4 +b4 +c4 +abc(a+b+c) ≥ a3 (b+c)+b3 (a+c)+c3 (a+b). In this form, the inequality has already been encountered and used several times in the book, mostly for t = 1 but also for some other values.

148

1.25

The generalized Means

We have met the Arithmetic Mean, the Quadratic Mean and the Harmonic Mean. All these means have the form 

n X

t

 1t

ai    i=1   M (t) =   n    Moreover, the inequalities between them for the case of two variables are M (−1) ≤ M (1) ≤ M (2). So, we might propose that M (x) is an increasing function in x for any fixed system of numbers a1 , a2 , . . . , an . The aim of this paragraph is to prove this inequality. The function is defined for all x 6= 0, so to prove our assertion we need to prove that M 0 (x) > 0 for x 6= 0 and to prove that M (x) ≥ M (y) for x > 0 > y. The last inequality follows from AM-GM, as AM-GM for ati gives M (t) ≥ G for t > 0 and M (t) ≤ G for t < 0, where G is the geometric mean of ai . n X ati Now, let’s compute M 0 (x). Note F =

i=1

n

, for simplicity. Then

  lnF 0 F 0 x − F ln F M (x) = e x = M (x) . x2 F 0

So, we must prove that F 0 x − F ln F ≥ 0. It rewrites as 



n X



X  n  i=1 1 x   a ln a x − ln i i  n n   i=1

and is clearly equivalent to 





axi  n X

   x  a i  ,  i=1 



X   n  naxi  1 x    ≥ 0. ai ln  n  X n  i=1   axi i=1

n2 Multiplying this by Pn

x i=1 ai

we have to prove

ti =

n X i=1

naxi ax1 + ax2 + . . . + axn 149

ti ln ti ≥ 0, where

thus

n X

ti = n.

i=1

This follows from the convexity of the function x ln x, whose second derivative 1 is > 0. Moreover, this function is strictly convex hence if not all ai are equal x then M (x) is strictly increasing, as we wish. We have remarked that f is increasing and continuous in any point except x = 0, where it is not defined. However we know that for x > 0 f (x) > G and for x < 0 f (x) < G. So, let’s prove that lim M (x) = G and then we x→0 !1 n x X can put M (0) = G. Dividing by G, we have to prove that lim axi = x→0

i=1

1, if a1 a2 . . . an = 1. Now we use Taylor expansion for the function axk as ∞ X xi 1 + ln ak x + lni ak . i! i=2 If we denote by A the maximum number among | ln ai |, then we see that |axk

∞ X (Ax)i − (1 − ln ak x)| < nx A < x2 A2 < x (i + 2)! 2

2

i=0

for any  if x is sufficiently small. ! n n X X Therefore axi − 1 + ln ak x < nx. i=1

i=1

Dividing by n and using the fact that

n X

ln ai = 0 we get 1 ≤

i=1 1 x) x

n X

axi ≤ 1 + x.

i=1

e

Finally using the fact that (1 + < we deduce that 1 < M (x) < e for x sufficiently small and this means that M (x) tends to 0 as desired. Finally, it’s easy to prove that M (+∞) = lim M (x) = max{ai } and M (−∞) = lim M (x) = min{ai }. x→+∞

x→−∞

Concluding, M (x) is a non-decreasing (actually strictly increasing unless all variables are equal in which case it is constant) function that spans all values between min{ai } and max{ai } i.e. √ min{ai } = M (−∞) < M (0) = n a1 a2 . . . an < M (+∞) = max{ai }.

1.26

Inequalities between the symmetric sums

All polynomial expressions that are symmetric in n variables can be expressed in terms of their symmetric sums - this is a content of a well-known theorem in algebra. Therefore, in solving symmetric inequalities one would like to compare the symmetric sums among each other. More precisely, take x1 , x2 , . . . , xn be some real numbers and consider the polynomial P (X) = (X − x1 )(X − x2 ) . . . (X − xn ). 150

Its coefficient at xn−k is clearly X

(−1)k σk = (−1)k

xi1 xi2 . . . xik .

1≤i1 k + 2 vanish also as after the ”reversal” (the passing from P (k) to Q) there are not among the three leading coefficients and vanish. Thus, the coefficients that are left are (−1)k σk , (−1)k+1 σk+1 , (−1)k+2 σk+2 . Now one can compute that the remaining quadratic is actually k!(n − k)! (k + 2)!(n − k − 2)! (−1)k σk x2 +(k+1)!(n−k−1)!(−1)k+1 σk+1 x+ σk+2 . 2 2 This is written simpler by passing to mi :   1 k 2 1 k+2 k+1 k (−1) n! m x − mk+1 x + mk+2 . 2 k 2 2(k+1)

It’s discriminant is mk+1 − mkk mk+2 k+2 ≥ 0. This can be also rewritten as     mk k mk+1 k+2 ≤ . mk+1 mk+2 Raising this to k + 1th power and multiplying them for k = 0, 1, . . . , i + 1 and   mi i(i+1) cancelling common terms we deduce 1 ≤ so mi ≥ mi+1 . mi+1 The inequality mi ≥ mi+1 is a pretty strong inequality. For example, the inequality 3(x2 + y 2 + z 2 ) ≥ (x + y + z)2 is equivalent to 3(σ12 − 2σ2 ) ≥ σ12 i.e. σ12 ≥ 3σ2 which follows directly from the inequality m1 ≥ m2 for n = 3. We invite the reader to play with these relations and find more inequalities. Note: other symmetric inequalities, like Schur’s inequality, can also be written in terms of symmetric sums. In the following chapter, we will concentrate on the case n = 3, which gives most applications.

1.27

The pqr technique

This inequality is devoted to a special kind of substitution. It is a particular case of the symmetric sums from the previous chapter, for three variables. 151

Recall from the previous chapter that if x1 , x2 , . . . , xn are variables, there are special expressions called the symmetric sums: the coefficients of the polynomial (t − x1 )(t − x2 ) . . . (t − xn ) Explicitly, they are σ1 =

n X i=1

xi , σ2 =

X

xi xj , . . . , σn = x1 x2 . . . xn

1≤i 0 such that a + b + c = 1. Show that 81abc ≥ 15.

4 + ab + bc + ca

4 + 81r ≥ 15 .Now, since a, b, c q are positive real numbers, the Schur’s inequality in degree three (i.e. k = 1) gives us 9r ≥ p(4q − p2 ) = 4q − 1. Using this inequality, it suffices to show that 4q + 9(4q − 1) ≥ 15, or 1q + 9q ≥ 6 which is obviously true by AM-GM inequality.

Solution: Our inequality is equivalent to

Example 2. If a, b, c are positive reals such that abc = 1, then r 2 2 2 5 a + b + c 3

a+b+c ≥ 3

Solution: Our inequality is equivalent to p5 ≥ 81(p2 −2q). Since q 2 ≥ 3pr = 3p, 2 2 2 we have 3 ≤ qp . Therefore 81(p2 − 2q) ≤ 27q (pp −2q) . It suffices to prove that p6 ≥ 27q 2 (p2 − 2q), or equivalently p6 + 54q 3 ≥ 27p2 q 2 . But this inequality is true since by AM-GM inequality, we have p p6 + 54q 3 = p6 + 27q 3 + 27q 3 ≥ 3 3 p6 · 27q 3 · 27q 3 = 27p2 q 2 . Example 3. Let a, b, c be real numbers such that a2 + b2 + c2 = 2. Prove that a + b + c ≤ 2 + abc. 153

2

Solution: From the given condition, we get q = p 2−2 and we may write our inequality as r + 2 − p ≥ 0. If p ≤ 1, then the inequality is trivial since √ √ 3 3 by AM-GM inequality, we have 2 = a2 + b2 + c2 ≥ 3 a2 b2 c2 ≥ 2 a2 b2 c2 , then |r| ≤ 1. If p > 1, then we get inequality for fourth degree,  by Schur’s 2−2  2 + 2)(p2 − 4)  p (p 6pr ≥ (p2 − q)(4q − p2 ) = p2 − 2(p2 − 2) − p2 = , 2 2 2 2 −4) (p2 + 2)(p2 − 4) or r ≥ . It suffices to show that (p +2)(p + 2 − p ≥ 0, or 12p 12p (4p + p2 − 2)(p − 2)2 ≥ 0 which is obviously true since p ≥ 1. Example 4. Let a, b, c > 0 such that abc = 1. Show that (a + b)(b + c)(c + a) ≥ 4(a + b + c − 1). Solution: We may write our inequality as pq − r ≥ 4(p − 1), or p(q − 4) + 3 ≥ 0. If q ≥ 4, then the inequality is trivial. If q ≤ 4, since q 2 ≥ 3pr = 3p. We have 1 q2 ·(q −4) +3 = (q −3)(q 2 −q −3). On the other hand, we have p(q −4) +3 ≥ 3 3 √ 3 q = ab+bc+ca ≥ 3 a2 b2 c2 = 3. Therefore q 2 −q −3 ≥ 3q −q −3 = 2q −3 > 0. Our proof is complete. As we said above, sometimes the above estimations are not always enough, for example with the following inequality 1 1 1 + + + 48(ab + bc + ca) ≥ 25. a b c for any a, b, c > 0 such that a + b + c = 1. We cannot apply the above estimations
to prove this inequality since it has a ”strange” equality case, that is 21 , 14 , 14 , but all above estimations are just used for the case a = b = c or a = b, c = 0 or b = c = 0. That is the main reason. Here is a stronger estimate: 2 2 Denote a + b + c = p, ab + bc + ca = p −t 3 , abc = r, where t ≥ 0, then we have (p + t)2 (p − 2t) (p − t)2 (p + 2t) ≤r≤ . 27 27 This estimate attains equality when (a − b)(b − c)(c − a) = 0. Exercises 1 1. Let a, b, c > 0 such that a + b + c = 1. Show that a3 + b3 + c3 + 4abc ≥ . 4 Solution: After transforming the inequality into pqr form, we need to prove 1 − 3q + 7r ≥ 14 . By Schur’s inequality for third degree, we have 1 1 1 3 4q ≤ 1 + 9r, hence 1 − 3q + 7r ≥ 1 − (1 + 9r) + 7r = + r ≥ . 4 4 4 4 2. Let a, b, c be real numbers such that a2 + b2 + c2 = 9. Prove that 2(a + b + c) − abc ≤ 10. Solution: With this problem, we will get confused since the original inequality attains equality for (a, b, c) = (2, 2, −1). With this equality 154

point, we cannot apply the Schur’s inequality to prove it since the Schur’s Inequality attains equality for (t, t, 0). But there is a key for us to apply it, that is to change the equality case from (2, 2, −1) to (t, t, 0). To do this, we will use the substitution a = x + 1, b = y + 1, c = z + 1, then the original condition can be rewritten as (x−1)2 +(y −1)2 +(z −1)2 = 9, or p2 − 2p − 2q = 6.And we need to prove 2 [(x − 1) + (y − 1) + (z − 1)] − p2 − 2p − 6 (x − 1)(y − 1)(z − 1) ≤ 10, or p + q − r ≤ 15, or p + − r ≤ 15, 2 2 2 2 2 or 2r ≥ p − 36. From (x − 1) + (y − 1) + (z − 1) = 9, we have s  √ (x − 1)2 + (y − 1)2 + (z − 1)2 3 |(x − 1)(y − 1)(z − 1)| ≤ = 3 3. 3 Hence 2 [(x − 1) + (y − 1) + (z − 1)] − (x − 1)(y − 1)(z − 1) ≤ ≤ 2 [(x − 1) + (y − 1) + (z − 1)] + |(x − 1)(y − 1)(z − 1)| √ = 2p + |(x − 1)(y − 1)(z − 1)| − 6 ≤ 2p + 3 3 − 6. √ √ Thus, if 2p ≤ 16 − 3 3, then our inequality is proved. If 2p ≥ 16 − 3 3, then by Schur’s inequality for fourth degree, we have    p2 − 2p − 6 p2 − 2p − 6 2 2 2 2 6pr ≥ (p − q)(4q − p ) = p − 4· −p 2 2 1 = (p + 2)(p − 6)(p2 + 2p + 6), 2 or (p + 2)(p − 6)(p2 + 2p + 6) 2r ≥ . 6p (p + 2)(p − 6)(p2 + 2p + 6) ≥ p2 − 36, or (p − 6p √ 3 6)2 (p2 + 4p − 2) ≥ 0 which is obviously true since p ≥ 16−3 > 1. 2

It suffices to show that

3. Let a, b, c ≥ 0 such that a+b+c = 2. Prove that a2 b2 +b2 c2 +c2 a2 +abc ≤ 1. Solution: We have p = 2 and our inequality is equivalent to q 2 − 3r ≤ 1. By Schur’s inequality for third degree, we have 9r ≥ p(4q−p2 ) = 8(q−1),   8 + 9r 8 + 9r 2 3 2 or q ≤ . Hence q −3r ≤ −3r = 1+ r(27r−16) ≤ 1 as 8 8 64 3  a+b+c 8 16 by AM-GM inequality, we have r ≤ = < . 3 27 27 4. Let a, b, c be positive real numbers such that abc = 1. Prove that 1 1 1 1 1 1 + + ≤ + + . 1+a+b 1+b+c 1+c+a a+2 b+2 c+2 Solution: After some computations (with notice that r = 1), we can find that 1 1 1 p2 + 4p + 3 + q + + = 2 , 1+a+b 1+b+c 1+c+a p + 2p + pq + q 155

and

1 1 1 4p + 12 + q + + = . a+2 b+2 c+2 4p + 9 + 2p

p2 + 4p + 3 + q 4p + 12 + q ≥ 2 , 4p + 9 + 2p p + 2p + pq + q or (3q − 5)p2 + (q 2 + 6q − 24)p − q 2 − 3q − 27 ≥ 0. By AM-GM inequality, we have p, q ≥ 3, therefore q 2 + 6q − 24 > 0. Hence Hence, our inequality is equivalent to

(3q − 5)p2 + (q 2 + 6q − 24)p − q 2 − 3q − 27 ≥ ≥ (3q − 5) · 32 + (q 2 + 6q − 24) · 3 − q 2 − 3q − 27 = 2(q − 3)(q + 24) ≥ 0. 5. If a, b, c > 0 then

a b c abc 5 + + + ≥ . 3 3 3 b + c c + a a + b 2(a + b + c ) 3

Solution: We normalize for p = 1, then we have   a b c 1 1 1 + + = (a + b + c) + + −3 b+c c+a a+b b+c c+a a+b q+1 − 3, = q−r and a3

abc r = . 3 3 +b +c 1 − 3q + 3r

q+1 r 14 + ≥ . By q−r 2(1 − 3q + 3r) 3 q+1 4q − 1 . Hence ≥ Schur’s inequality for third degree, we have r ≥ 9 q−r 4q−1 q+1 9(q + 1) r 9 = = , and ≥  4q − 1 4q − 1 5q + 1 2(1 − 3q + 3r) q− 2 1 − 3q + 3 · 9 9 4q − 1 9(q + 1) 4q − 1 14 (17 − 50q)(1 − 3q) . We need to prove + ≥ , or ≥ 6(2 − 5q) 5q + 1 6(2 − 5q) 3 2(5q + 1)(2 − 5q) 1 17 > ≥ q. 0 which is true since 50 3 Hence, our inequality is equivalent to

6. Let a, b, c ≥ 0 such that ab + bc + ca + abc = 4. Prove that a + b + c ≥ ab + bc + ca. Solution: From the given condition, we have q + r = 4. And we need to prove p ≥ q. If p ≥ 4, then it is trivial since p ≥ 4 ≥ q. If 4 ≥ p, by AMGM inequality, we can easily check that p ≥ 3. And the Schur’s inequality p(4q − p2 ) p(4q − p2 ) for third degree gives us r ≥ . Then q + ≤ 4, or 9 9 p3 + 36 p3 + 36 (4 − p)(p2 − 9) q≤ . Hence p − q ≥ p − = ≥ 0. 4p + 9 4p + 9 4p + 9 7. Let a, b, c ≥ 0. Prove that

a2 + b2 + c2 8abc + ≥ 2. ab + bc + ca (a + b)(b + c)(c + a) 156

Solution: Also, we normalize that p = 1, then our inequality becomes 1 − 2q 8r 8r 4q − 1 + ≥ 2, or ≥ .By Schur’s inequality for third q q−r q−r q   (4q − 1)(1 − q) 4q−1 . degree and fourth degree, we have r ≥ max , 9 6 8r 8r 72r 12(4q − 1)(1 − q) Then ≥ = ≥ . It suffices to 4q−1 q−r 1 + 5q 1 + 5q q− 9 12(4q − 1)(1 − q) 4q − 1 (1 − 3q)(1 − 4q)2 show that ≥ , or ≥ 0 which 1 + 5q q q(1 + 5q) is true. 8. Let a, b, c ≥ 0 such that ab+bc+ca+6abc = 9. Prove that a+b+c+3abc ≥ 6. Solution: From the given condition, we have q + 6r = 9 and we need to prove p + 3r ≥ 6, or 2p − q − 3 ≥ 0. If p ≥ 6, then it is trivial since 2p−q −3 ≥ 2·6−9−3 = 0. If 6 ≥ p, by AM-GM inequality, we get p ≥ 3. p(4q − p2 ) And the Schur’s inequality for third degree gives us r ≥ . 9 2 3 2p(4q − p ) 2p + 27 Hence q + ≤ 9, or q ≤ . Thus 2p − q − 3 ≥ 2p − 3 8p + 3 2p3 + 27 2(6 − p)(p − 3)(p + 1) −3= ≥ 0. 8p + 3 8p + 3 9. Let a, b, c ≥ 0 such that a2 + b2 + c2 = 3. Prove that 12 + 9abc ≥ 7(ab + bc + ca). Solution: The given condition can be rewritten as p2 − 2q = 3, and we need to prove 12 + 9r ≥ 7q. By Schur’s inequality for third degree, we have 9r ≥ p(4q − p2 ) = p 2(p2 − 3) − p2 = p(p2 − 6). Hence 12 + 9r − 7q ≥ 12 + p(p2 − 6) − 7q = 12 + p(p2 − 6) − 7 · =

p2 − 3 2

1 (2p + 5)(p − 3)2 ≥ 0. 2

10. If a, b, c are positive reals such that abc = 1. Then 1 + 6 . ab + bc + ca

3 ≥ a+b+c

3 6 − ≥ 0. By AM-GM p q 3 9 3 6 inequality, we have q 2 ≥ 3pr = 3p, hence ≥ 2 . Thus 1 + − ≥ p q p q  2 9 6 3 − 1 ≥ 0. − +1= q2 q q Solution: The inequality can be rewritten as 1 +

11. If a, b, c > 0, abc = 1. Then 2(a2 +b2 +c2 )+12 ≥ 3(a+b+c)+3(ab+bc+ca). Solution: Rewrite the original inequality in the pqr form 2p2 − 3p − 7q + 12 ≥ 0. By Schur’s inequality for third degree, we have p(4q −p2 ) ≤ 9r = 157

9, hence q ≤

p3 +9 4p ,

thus

p3 + 9 (p − 3)(p2 − 9p + 21) + 12 = . 4p 4p √
By AM-GM inequality, we have p ≥ 3 and p2 −9p+21 ≥ 2 21 − 9 p > 0. Our proof is complete. 2p2 − 3p − 7q + 12 ≥ 2p2 − 3p − 7 ·

12. If a, b, c are nonnegative real numbers, no two of which are zero. Deter   3−a  x + y + z a xy + yz + zx 2 mine the least constant a such that ≥ 3 3 (x + y)(y + z)(z + x) . 8 3 ln 3 − 4 ln 2 Solution: Let a = b = 1, c → 0, then we get a ≥ = a0 ≈ 2 ln 2 − ln 3 1.81884 . . . We will show that it is the value which we find, that is 

x+y+z 3

a0 

xy + yz + zx 3

 3−a0 2

≥

(x + y)(y + z)(z + x) . 8

Since this inequality is homogeneous, we can normalize that p = 1, then 8q

3−a0 2

8q

3−a0 2

− q ≥ 0. If 1 ≥ 4q, then we have 3+a0 − q =   3 2 !   a0 −1 3−a0 3−a0 a0 −1 2 8 8 1  = 0. If 1 ≥ q ≥ 2 2  ≥ q q 2 − q − 3 3+a0 3+a0 4 3 2 3 2 4q − 1 1 . It 4 , then by Schur’s inequality for third degree, we have r ≥ 9 3−a0 3−a0 5 4q − 1 8q 2 8q 2 suffices to show that + 3+a0 −q ≥ 0, or f (q) = − q + 3+a0 − 9 9 3 2 3 2 1 2(3 − a )(a − 1) 0 0 ≥ 0. We have f 00 (q) = − < 0. Hence f (q) is concave, 3+a0 1+a0 9 2 ·q 2 3      1 1 thus f (q) ≥ min f ,f = 0. Our proof is complete. 3 4 it becomes r +

3+a0 3 2

13. Let a, b, c ≥ 0 such that a + b + c = 1. Prove that

1 1 1 + + + a+b b+c c+a

247 . 54 247 1+r Solution: The inequality is equivalent to 1+q q−r + 2r ≥ 54 , or q−r + 2r − 193 1+9r 54 ≥ 0. By Schur’s inequality for third degree, we have q ≤ 4 . Hence 2abc ≥

1+r 193 + 2r − q−r 54

≥ =

1+r 193 4(r + 1) 193 + 2r − = + 2r − 54 5r + 1 54 −r (23 − 20r)(1 − 27r) ≥0 54(5r + 1) 1+9r 4

aince by AM-GM inequality, we have 27r ≤ 1. 158

1 (a + b + c)5 . 12 Solution: We normalize that p = 1, then the inequality becomes (5q − 1 1)r + q − 3q 2 ≤ , or 12(5q − 1)r ≤ (1 − 6q)2 . If q ≤ 15 , then it is trivial. 12 1 1 If ≥ q ≥ , then from 3r ≤ q 2 , we have 3 5

14. If a, b, c ≥ 0, prove that a4 (b + c) + b4 (c + a) + c4 (a + b) ≤

12(5q − 1)r − (1 − 6q)2 ≤ 4q 2 (5q − 1) − (1 − 6q)2 = 20q 3 − 40q 2 + 12q − 1   3 28 3 2 − 4q 2 + = 20q − 36q + 12q − 25 25 4 3 = (5q − 1)2 (5q − 7) − 4q 2 + 25 25  2 3 1 3 1 ≤ −4q 2 + = − < 0. ≤ −4· 25 25 5 25 15. If a, b, c are nonnegative real numbers, no two of which are zero. Prove that  2  2  2 a b c 10abc + + + ≥ 2. b+c c+a a+b (a + b)(b + c)(c + a) 2a 2b 2c 1 ,y = ,z = , then we have x+2 + b+c c+a a+b 1 1 y+2 + z+2 = 1, or xy + yz + zx + xyz = 4. Now, put p = x + y + z, q = xy + yz + zx and r = xyz, then we have q + r = 4, and we need to prove x2 + y 2 + z 2 + 5xyz ≥ 8, or p2 − 2q + 5(4 − q) ≥ 8, or p2 − 7q + 12 ≥ 0. If p ≥ 4, then it is trivial since p2 − 7q + 12 ≥ 42 − 7 · 4 + 12 = 0. If 4 ≥ p, then we can easily check by AM-GM inequality that p ≥ 3. p(4q − p2 ) Using Schur’s inequality for third degree, we have r ≥ , hence 9 p(4q − p2 ) p3 + 36 q+ ≤ 4, or q ≤ . It follows that p2 − 7q + 12 ≥ 9 4p + 9 p3 + 36 3(p − 3)(16 − p2 ) p2 − 7 · + 12 = ≥ 0. 4p + 9 4p + 9 Solution: Denote x =

1.28

The tangent line technique and its extensions

The technique of using tangent line is an elegant method that provides nice and simple solutions for many inequalities we solve. But as we see, the original technique can only help us solve the inequalities with only one equality case when all variables are equal. This is really a big disadvantage of this technique. In the latter part of this section, we will expand the technique to a more complicated version which can sometimes solve inequalities that the original tangent line technique cannot tackle. 159

Starting with some examples Example 1. If a, b, c are positive real numbers, then a b c 3 + + ≥ . b+c c+a a+b 2 Solution: Without loss of generality, we may assume that a + b + c = 3, then our inequality becomes a b c 3 + + ≥ . 3−a 3−b 3−c 2 4x 3(x − 1)2 − 3x + 1 = ≥ 0. Hence 3−x 3−x 4a 4b 4c + + ≥ (3a − 1) + (3b − 1) + (3c − 1) = 6. 3−a 3−b 3−c Example 2. Let a, b, c be positive real numbers, then prove that For any x ≤ 3, we have

(2b + c + a)2 (2c + a + b)2 (2a + b + c)2 + + ≤ 8. 2a2 + (b + c)2 2b2 + (c + a)2 2c2 + (a + b)2 Solution: Suppose that a + b + c = 3, then our inequality becomes (3 + a)2 (3 + b)2 (3 + c)2 + + ≤ 8, 2a2 + (3 − a)2 2b2 + (3 − b)2 2c2 + (3 − c)2 or equivalently a2 + 6a + 9 b2 + 6b + 9 c2 + 6c + 9 + + ≤ 24. a2 − 2a + 3 b2 − 2b + 3 c2 − 2c + 3 For any x > 0, we have Hence

x2 + 6x + 9 (4x + 3)(x − 1)2 − (4x + 4) = − ≤ 0. x2 − 2x + 3 x2 − 2x + 3

a2 + 6a + 9 b2 + 6b + 9 c2 + 6c + 9 + + ≤ (4a + 4) + (4b + 4) + (4c + 4) = 24. a2 − 2a + 3 b2 − 2b + 3 c2 − 2c + 3 Example 3. Let a, b, c be positive real numbers such that abc = 1. Prove that p p p √ a2 + 1 + b2 + 1 + c2 + 1 ≤ 2(a + b + c). √ √ 1 Solution: Consider the function f (x) = x2 + 1 − 2x + √ ln x with x > 0, 2 we have √ x 1 x 1 − 2x f 0 (x) = √ +√ − 2= √ + √ . 2 2 2x 2x x +1 x +1 1 1 It is easy to check that f 0 (x) > 0 if x ≤ . And in the case x > , we have 2 2 f 0 (x) =

=

x2 (1 − 2x)2 − 2 x 1 − 2x 2x2 √ + √ = x +1 x 2x − 1 2x x2 + 1 √ + √ 2 2x x +1 (1 − x)(2x3 − 2x2 + 3x − 1)  . x 2x − 1 2 2 2x (x + 1) √ + √ 2x x2 + 1 160

And since 2x3 − 2x2 + 3x − 1 = x3 + x(x − 1)2 + (2x − 1) > 0 (for any x > 1/2), we obtain f 0 (x) = 0 if and only if x = 1. By writing the variation board, we can easily check that f (x) ≤ f (1) = 0 ∀x > 0.Hence p p p √ 1 a2 + 1 + b2 + 1 + c2 + 1 ≤ 2(a + b + c) − √ (ln a + ln b + ln c) 2 √ = 2(a + b + c). Establishing the basic technique While considering the above examples the reader may ask wonder about the origin of the intermediate inequalities used. For example, with Example 1, the √ √ 4x fact is ≥ 3x−1, and with Example 3, the fact is x2 + 1 ≤ 2x− √12 ln x. 3−x The method applies when we have to prove an inequality of the form f (x1 ) + f (x2 )+. . .+f (xn ) ≥ 0 but working directly with f (x) is troublesome . In that case we try to find a function g(x) such that f (x) ≥ g(x) and it is easier to work with g(x), then we will only need to prove the inequality g(x1 ) + g(x2 ) + . . .+g(xn ) ≥ 0. As a matter of fact, g(x) is usually created from what are given in the hypothesis. For example, if the hypothesis is x1 +x2 +. . .+xn = n, then we may predict g(x) = k(x − 1). If the hypothesis is x1 x2 . . . xn = 1, then we may predict g(x) = k ln x. If the hypothesis is xk1 + xk2 + . . . + xkn = n, then we may predict g(x) = k(xk − 1). In general, we will consider inequalities where equality is attained when all variables are equal. Notice that these predictions cannot be mechanical, they must depend on the case of equality (for example, the above predictions of g(x) count on the case x1 = x2 = . . . = xn = 1). How do we find the valid value of k? In the case when we want to compare f with x − 1, and f (x) is differentiable, then we must choose k = f 0 (1) since we need to choose k such that the sign of h(x) = f (x) − g(x) does not change when x gets through the point x = 1. Similarly, if we are subject to x1 . . . xn = 1 so we compare f to ln x we must have k = −f 0 (1) because ln x has derivative −1 at 1. 4x In the above examples, the estimate 3−x ≥ 3x − 1 is obtained in exactly this P 4x P 4x 4x way: we want to prove ≥ 6 i.e. ( 3−x − 2) ≥ 0. As 3−x − 2 has 3−x 4x derivative equal to 3 when evaluated at 1, we propose the inequality 3−x − 2 ≥ 4x 3(x − 1) i.e. 3−x ≥ 3x − 1, which is true. This way is very neat and powerful with simple inequalities. But with sharper ones, it is more difficult to apply useless since the inequality f (x) ≥ g(x) does not always holds. For example, consider with the following Example 4. Let a, b, c be real numbers such that a + b + c = 1. Prove that

a2

b c 9 a + 2 + 2 ≤ . +1 b +1 c +1 10

(Poland 1997) Solution: Using the above way, we can establish the following inequality x 36x + 3 ≤ . But unfortunately, this inequality can only holds when x2 + 1 50 161

3 36x + 3 x (4x + 3)(3x − 1)2 since − 2 = while the original prob4 50 x +1 50(x2 + 1) lem is asking us to prove it in the case a, b, c are arbitrary real numbers. Hence, we need to examine the small cases in order to prove it. 3 Case 1. If min {a, b, c} ≥ − , then we have 4 x≥−

a2

a b c 36 9 9 + 2 + 2 ≤ (a + b + c) + = . +1 b +1 c +1 50 50 10

3 3 Case 2. Assume that one of a, b, c is less than − , for example c < − . Then, 4 4 b 1 by AM-GM inequality, we have 2 ≤ .Hence, our inequality is proved if b +1 2 a 2 1 1 ≤ , or a ≤ ∨ a ≥ 2. It suffices for us to consider the case 2 ≥ a ≥ . 2 a +1 5 2 2 And by the same manner, we only need to prove the inequality in the case 3 c 3 2 ≥ a, b ≥ 12 , then we have − > c = 1 − a − b ≥ −3, hence 2 ≤− . 4 c +1 10 Thus a b c 1 1 3 7 9 + + ≤ + − = < . a2 + 1 b2 + 1 c2 + 1 2 2 10 10 10 In this example, the tangent line technique still works, but we need some casework for when it fails. This is one way to extend the method, but it makes the solutions longer and less elegant. In the next subsection, we will present another extension of the technique. An extension of tangent line technique Example 5. Let a, b, c be nonnegative real numbers such that a + b + c = 3. Prove that 1 1 1 3 + 2 + 2 ≥ . 2 2a + 3 2b + 3 2c + 3 5 Solution: The original technique cannot be applied directly since this inequality attains equality not only when a = b = c but also for a = b, c = 0. Without loss of generality, assume that a ≥ b ≥ c. We will consider 2 cases 1 1 4 9 2(4x − 1)(x − 1)2 Case 1. If c ≥ , then we have + x − = ≥0 4 2x2 + 3 25 25 25(2x2 + 3) 1 ∀x ≥ , hence 4       1 1 1 9 4 9 4 9 4 + + ≥ − a + − b + − c 2a2 + 3 2b2 + 3 2c2 + 3 25 25 25 25 25 25 3 = . 5 Case 2. If c ≤

1 1 8 22 (4x + 1)(2x − 3)2 , then + x− = ≥ 0 ∀x ≥ 0, 2 4 2x + 3 75 75 75(2x2 + 3)

hence 1 1 + 2 + 3 2b + 3

2a2

≥ =

44 8 44 8 − (a + b) = − (3 − c) 75 75 75 75 8 4 c+ . 75 15 162

1 8 1 + c ≥ which is true since + 3 75 3   1 8 1 2c 8c2 + 23c + 12(1 − 4c) + c− − ≥ 0. 2c2 + 3 75 3 75(2c2 + 3)

It suffices to show that

2c2

The idea of this solution is quite similar to the original technique, but is one level higher in complexity. In order to prove the inequality f (x1 )+f (x2 )+. . .+ f (xn ) ≥ 0, we try establish the inequality f (x) ≥ g(x) and find when it holds for all x1 , x2 , . . . , xn . In the other case, there exist xk , xk+1 , . . . , xn such that f (xj ) ≤ g(xj ) for all j = k, . . . , n then will find another function h(x) such that f (x) ≥ h(x) for any x1 , x2 , . . . , xk−1 , then we may reduce our inequality into h(x1 ) + h(x2 ) + . . . + h(xk−1 ) + f (xk ) + . . . + f (xn ) ≥ 0. And from now, we will prove this inequality based on the relationship between x1 , x2 , . . . , xn . There is many ways to establish the function h(x). In the case we considered, after consider the first case (which yields the first equality case a = b = c), in the second case, we have established the function h(x) based on the second 3 equality case a = b = , c = 0, since it is better to choose h(x) as a linear 2 function: h(x) = kx + m and we need to have f (x) ≥ h(x)  for  all x ∈ [0, 3] , 3 3 and it has equality when x = , we can choose k = f 0 , then we can 2 2 easily choose m (notice that this way is quite similar to the first case). Example 6. Let a, b, c, d be nonnegative real numbers such that a+b+c+d = 2. Prove that 1 1 1 1 16 + 2 + 2 + 2 ≥ . 2 3a + 1 3b + 1 3c + 1 3d + 1 7 Solution: Notice that this inequality attains equality when a = b = c = d = 1 2 and a = b = c = , d = 0, we will use the idea above to prove it. Suppose 3 that a ≥ b ≥ c ≥ d, then we have 2 cases 1 1 48 52 3(12x − 1)(2x − 1)2 Case 1. If d ≥ , then we have 2 + x− = ≥0 12 3x + 1 49 49 39(3x2 + 1) 1 ∀x ≥ , hence 12 1 1 1 1 + 2 + 2 + 2 ≥ 1  3d + 1  +1 3c +      + 1 3b 52 48 52 48 52 48 52 48 ≥ − a + − b + − c + − d = 2. 49 49 49 49 49 49 49 49 3a2

Case 2. If d ≤

1 36 45 (12x + 1)(3x − 2)2 1 , then we have 2 + x− = ≥0 12 3x + 1 49 49 49(3x2 + 1)

∀x ≥ 0, hence 1 1 1 + 2 + 2 + 1 3b + 1 3c + 1

3a2

≥ = 163

135 36 − (a + b + c) 49 49 135 36 36 9 − (2 − d) = d + . 49 49 49 7

It suffices to prove that

1 36 + d ≥ 1, which is true since + 1 49

3d2

1 36 3d[36d2 + 95d + 12(1 − 12d)] + d − 1 = ≥ 0. 3d2 + 1 49 49(3d2 + 1) Example 7. Let a, b, c be nonnegative real numbers, not all are zero. Show that b2 c2 2 a2 + + ≤ . 2 2 2 2 2 2 2a + (b + c) 2b + (c + a) 2c + (a + b) 3 Solution: This inequality attains equality only for a = b, c = 0. But we still can use the above idea to prove it. Indeed, without loss of generality, assume that a + b + c = 1 and a ≥ b ≥ c, then our inequality becomes a2 b2 c2 2 + + ≤ . 3a2 − 2a + 1 3b2 − 2b + 1 3c2 − 2c + 1 3 There are 2 cases 1 x2 8 (6x − 1)(2x − 1)2 Case 1. If b ≥ , then we have 2 + 19 − x = − ≤ 6 3x − 2x + 1 9 9(3x2 − 2x + 1) 1 0 ∀x ≥ , hence 6 b2 8 2 8 2 2 8 a2 + ≤ (a + b) − = (1 − c) − = − c. 2 2 3a − 2a + 1 3b − 2b + 1 9 9 9 9 3 9 c2 8 c[(1 − 3c)(17 − 24c) + 7] − c=− ≤ 0. Therefore, our in2 3c − 2c + 1 9 27(3c2 − 2c + 1) equality is proved. 2 1 Case 2. If ≥ b ≥ c, then a = 1 − b − c ≥ , we have 6 3 But

x2 2 x(6x − 1)(2 − x) 1 − x= ≤ 0 ∀x ≤ , 3x2 − 2x + 1 9 9(3x2 − 2x + 1) 6 Hence

b2 c2 2 2 2 + ≤ (b + c) = − a. 2 2 3b − 2b + 1 3c − 2c + 1 9 9 9

It suffices to show that

a2 2 4 − a − ≤ 0, which is true since 2 3a − 2a + 1 9 9

a2 2 4 (3a − 2)(6a2 + 3a − 4) + 4 − a − = − ≤ 0. 3a2 − 2a + 1 9 9 27(3a2 − 2a + 1) Another extension Example 8. Let a, b, c be nonnegative real numbers such that a + b + c = 3. Prove that 1 1 1 3 + + ≤ . 9 − ab 9 − bc 9 − ca 8 Solution: This inequality has the form of f (x1 ) + f (x2 ) + f (x3 ) ≥ 0 which is similar to the above inequalities, that we have considered but the difference is 164

that x1 , x2 , x3 are equal to ab, bc, ca and not the original variables. hypothesis is related to a, b, c not ab, bc, ca. Even more, when we try to establish the inequality f (x) ≤ g(x) where g(x) has the form k(x − 1) then we actually obtain a reversed inequality. Indeed, by this way, we would want to establish 1 x+7 1 x+7 (x − 1)2 the inequality ≤ . But − = ≥ 0 which is 9−x 64 9−x 64 64(9 − x) not what we want. This can still be remedied. Taking notice that ab, bc, ca < 3, it suffices to 1 x + 7 (x − 1)2 x + 7 (x − 1)2 consider x < 3, then we have = + ≤ + = 9−x 64 64(9 − x) 64 64(9 − 3) x2 + 4x + 43 . Then we get 384 1 1 1 1 + + ≤ (a2 b2 + b2 c2 + c2 a2 + 4ab + 4bc + 4ca + 43). 9 − ab 9 − bc 9 − ca 384 It suffices to show that a2 b2 + b2 c2 + c2 a2 + 4(ab + bc + ca) ≤ 15. Put x = ab + bc + ca, then x ≤ 3 and by Schur’s inequality degree  a(a − b)(a − c) + b(b − c)(b − a) + c(c − a)(c − b) ≥ 0, we have abc ≥ max 0, 4x−9 . 3 If 4x ≤ 9, then a2 b2 + b2 c2 + c2 a2 + 4(ab + bc + ca) = x2 + 4x − 6abc 225 ≤ x2 + 4x ≤ < 15. 16 If 4x ≥ 9, then a2 b2 + b2 c2 + c2 a2 + 4(ab + bc + ca) = x2 + 4x − 6abc ≤ x2 + 4x − 2(4x − 9) = (x − 1)(x − 3) + 15 ≤ 15. Example 9. Let a, b, c be positive real number such that a4 + b4 + c4 = 3. Prove that 1 1 1 + + ≤ 1. 4 − ab 4 − bc 4 − ca 3 3 Solution: Notice that ab, bc, ca < , and for all x < , we have 2 2 1 x + 2 (x − 1)2 x+2 (x − 1)2 2x2 + x + 12 = = + ≤ +  . 3 4−x 9 9(4 − x) 9 45 9 4− 2 We have  1 1 1 1  2 2 + + ≤ 2(a b + b2 c2 + c2 a2 ) + ab + bc + ca + 12 . 4 − ab 4 − bc 4 − ca 45 On the other hand, by AM-GM inequality, we have a2 b2 + b2 c2 + c2 a2 ≤ a2 b2 + b2 c2 + c2 a2 + 3 a4 + b4 + c4 = 3 and ab + bc + ca ≤ ≤ 3. From this, we 2 get the result.

165

Section for self study As we saw above, the simple and intuitive tangent line technique can be expanded in many complicated, creative or lucrative ways to solve harder inequalities. Unfortunately, not all these expansions can be categorized. In practice, one needs to start with the original idea and modify it accordingly to suit the needs of every particular problem. In this section, we present some examples that can be solved in this way. Example 10. Let a, b, c be positive real numbers such that abc = 1. Show that a b c 3 + + ≤ . a2 + 3 b2 + 3 c2 + 3 4 Solution: Notice that

x 3x(x + 1) x(3x + 1)(x − 1)2 − = − ≤0 x2 + 3 8(x2 + x + 1) 8(x2 + 3)(x2 + x + 1)

∀x ≥ 0. Hence a b c + 2 + 2 2 a +3 b +3 c +3

≤ =

 2  3 a +a b2 + b c2 + c + + 8 a2 + a + 1 b2 + b + 1 c2 + c + 1   1 3 1 1 3− 2 . − 2 − 2 8 a +a+1 b +b+1 c +c+1

It suffices to prove that a2 Put a =

1 1 1 + 2 + 2 ≥ 1. +a+1 b +b+1 c +c+1

yz zx xy , b = 2 , c = 2 where x, y, z > 0, our inequality becomes 2 x y z

x4 y4 z4 + + ≥ 1. x4 + x2 yz + y 2 z 2 y 4 + y 2 zx + z 2 x2 z 4 + z 2 xy + x2 y 2 which is true since by Cauchy Schwartz inequality and AM-GM inequality, we have

≥ ≥

x4 y4 z4 + + ≥ x4 + x2 yz + y 2 z 2 y 4 + y 2 zx + z 2 x2 z 4 + z 2 xy + x2 y 2 (x2 + y 2 + z 2 )2 x4 + y 4 + z 4 + x2 y 2 + y 2 z 2 + z 2 x2 + xyz(x + y + z) (x2 + y 2 + z 2 )2 = 1. x4 + y 4 + z 4 + 2(x2 y 2 + y 2 z 2 + z 2 x2 )

Example 11. Let a, b, c be positive real number such that a + b + c = 2. Prove that bc ca ab + + ≤ 1. a2 + 1 b2 + 1 c2 + 1 Solution: Our inequality is equivalent to   a b c abc + + + 1 − (ab + bc + ca) ≥ 0. a2 + 1 b2 + 1 c2 + 1 166

1 x(x − 1)2 −1+ x= ≥ 0. Hence 2 x2 + 1       a b c 1 1 1 + + ≥ a 1 − a + b 1 − b + c 1 − c a2 + 1 b2 + 1 c2 + 1 2 2 2 1 = 2 − (a2 + b2 + c2 ) = ab + bc + ca. 2 It suffices to show that abc(ab + bc + ca) + 1 − (ab + bc + ca) ≥ 0. Put x = 4 1 ab + bc + ca ≤ (a + b + c)2 = , then by Schur’s inequality for fourth degree 3 3 (x − 1)(4 − x) 2 2 a (a−b)(a−c)+b (b−c)(b−a)+c2 (c−a)(c−b) ≥ 0, we get r ≥ . 3 Therefore Note that for all x ≥ 0, we have

1 x2 +1

abc(ab + bc + ca) + 1 − (ab + bc + ca) = xabc + 1 − x x(x − 1)(4 − x) ≥ +1−x 3 (3 − x)(x − 1)2 ≥ 0. = 3 Example 12. Let a, b, c be nonnegative real numbers such that a + b + c = 1. Prove that a2 + bc b2 + ca c2 + ab 13 + 2 + 2 ≤ . 2 a +1 b +1 c +1 20 Solution: Notice that for all 0 ≤ x ≤ 1, we have x 12 4 (3x + 4)(2x − 1)2 − x − = − ≤ 0, x2 + 1 25 25 25(x2 + 1) and

1 1 2 x2 (x2 − 1) − 1 + x = ≤ 0. x2 + 1 2 2(x2 + 1)

We have b2 c2 a2 + + a2 + 1 b2 + 1 c2 + 1

and bc ca ab + + a2 + 1 b2 + 1 c2 + 1



12 4 a+ 25 25





12 4 ≤ a +b b+ 25 25 12 2 4 = (a + b2 + c2 ) + , 25 25 



 +c

12 4 c+ 25 25

    1 2 1 2 + ca 1 − b + ab 1 − c 2 2 1 = ab + bc + ca − abc. 2 1 ≤ bc 1 − a2 2



We need to prove 12 2 4 1 13 (a + b2 + c2 ) + + ab + bc + ca − abc ≤ , 25 25 2 20 12 4 Put q = ab + bc + ca, r = abc then the inequality becomes 25 (1 − 2q) + 25 + 1 13 q − 2 r ≤ 20 , or (1 − 4q + 9r) + 41r ≥ 0 which is obviously true by Schur’s inequality for third degree.

Exercises 167



1. Show that if a + b + c = 6 then a4 + b4 + c4 ≥ 2(a3 + b3 + c3 ).   Solution: We have a4 − 2a3 − 8(a − 2) = (a + 1)2 + 3 (a − 2)2 ≥ 0, hence a4 + b4 + c4 − 2(a3 + b3 + c3 ) ≥ 8(a − 2) + 8(b − 2) + 8(c − 2) = 0. a b c 9 + + ≥ . 1 + bc 1 + ca 1 + ab 10 1 81 99 (9x − 1)2 Solution: We have + x− = ≥ 0 ∀x ≥ 0.Hence 1 + x 100 100 100(x + 1)

2. Let a, b, c > 0, a + b + c = 1. Prove that

b c a + + ≥ 1 + bc 1 + ca  1 +ab    99 81 99 81 99 81 ≥ a − bc + b − ca + c − ab 100 100 100 100 100 100   243 99 243 9 99 a+b+c 3 − abc ≥ − · = . = 100 100 100 100 3 10 3. Show that if a, b, c > 0 and abc = 1 then a2 + b2 + c2 + 9(ab + bc + ca) ≥ 10(a + b + c). Solution: The inequality is equivalent to   1 1 1 2 2 2 a +b +c +9 + + − 10(a + b + c) ≥ 0. a b c 9 − 10x + 17 ln x with x > 0, we have x   (x − 1) 2(x − 2)2 + 1 17 9 0 = , f (x) = 2x − 2 − 10 + x x x2

Consider the function f (x) = x2 +

Hence f 0 (x) = 0 if and only if x = 1. Now, by making variation board, we get f (x) ≥ f (1) = 0 ∀x > 0. Therefore   1 1 1 a2 + b2 + c2 + 9 + + − 10(a + b + c) ≥ −17(ln a + ln b + ln c) = 0. a b c 4. Show that if a, b, c > 0 then (c + a − b)2 (a + b − c)2 3 (b + c − a)2 + + ≥ . 2 2 2 2 2 2 a + (b + c) b + (c + a) c + (a + b) 5 Solution: Assume that a + b + c = 1, then our inequality becomes (1 − 2a)2 (1 − 2b)2 (1 − 2c)2 3 + + ≥ . 2 2 2 2a − 2a + 1 2b − 2b + 1 2c − 2c + 1 5 We have

(1 − 2a)2 23 54 2(6a + 1)(1 − 3a)2 − + a = ≥ 0. Hence 2a2 − 2a + 1 25 25 25(2a2 − 2a + 1)

(1 − 2a)2 (1 − 2b)2 (1 − 2c)2 69 54 3 + + ≥ − (a + b + c) = . 2 2 2 2a − 2a + 1 2b − 2b + 1 2c − 2c + 1 25 25 5

168

1 1 1 4 + + + (a + b + c) ≥ 7. a b c 3 a2 + 13 (6 − a)(a − 1)2 1 4 = ≥ 0. Hence Solution: We have + a − a 3 6 6a

5. Let a, b, c > 0, a2 + b2 + c2 = 3. Prove that

1 1 1 4 a2 + 13 b2 + 13 c2 + 13 + + + (a + b + c) ≥ + + = 7. a b c 3 6 6 6 6. For any positive real numbers a, b, c such that a + b + c = 1, then 10(a3 + b3 + c3 ) − 9(a5 + b5 + c5 ) ≥ 1. 16 25 (16 + 21a − 18a2 − 27a3 )(1 − 3a)2 Solution: We have 10a3 −9a5 + − a = . 27 9 27 Hence, assume that a ≥ b ≥ c, and we have 2 cases Case 1. If 16 + 21a − 18a2 − 27a3 ≥ 0, then we observe that 16 + 21b − 18b2 − 27b3 ≥ 0,

16 + 21c − 18c2 − 27c3 ≥ 0.

We have 10(a3 + b3 + c3 ) − 9(a5 + b5 + c5 ) ≥

16 25 (a + b + c) − = 1. 9 9

2 3 3 2 Case 2. If 16 + 21a − 18a √ − 27a ≤ 0, then we have 27a + 18a ≥ 2 30 − 3 2 1 21a + 16 ≥ 37a, or a ≥ > , it follows that b, c < . In this 9 3 3 case, we have

10b3 − 9b5 ≥ 9b3 ,

10c3 − 9c5 ≥ 9c3 .

Hence 9 9 10b3 − 9b5 + 10c3 − 9c5 ≥ 9(b3 + c3 ) ≥ (b + c)3 = (1 − a)3 . 4 4 9 It suffices to show that 10a3 − 9a5 + (1 − a)3 ≥ 1 which is true since 4 (1 − a)(3a − 1)(12a3 + 16a2 + 7a − 5)  4 2 ≥ 0. since a > 3

9 10a3 − 9a5 + (1 − a)3 − 1 = 4

This completes our proof. 7. Let a, b, c be positive real numbers, then a c b √ +√ +√ ≥ c+a b+c a+b

r

3 (a + b + c). 2

Solution: Assume that a + b + c = 1, then our inequality becomes r a b c 3 √ +√ +√ ≥ . 2 1−a 1−c 1−b 169

√   5 6 a 1 √ ≥ We will show that a− . Indeed, put 1 − a = 6t2 8 15 1−a with t ≥ 0, then we have a = 1 − 6t2 , therefore √  √    1 1 − 6t2 5 6 1 a 5 6 √ √ a− = 1 − 6t2 − − − 8 15 8 15 1−a 6t √ 6(5t + 2)(3t − 1)2 = ≥ 0. 12t Hence √   r 1 a b c 5 6 3 √ a+b+c− = +√ +√ ≥ . 8 5 2 1−a 1−c 1−b 8. Let a, b, c be real numbers such that a2 + b2 + c2 = 1. Prove that 1 1 1 9 + + ≤ . 1 − ab 1 − bc 1 − ca 2 Solution: Since 1 1 1 1 1 1 + + ≤ + + , 1 − ab 1 − bc 1 − ca 1 − |ab| 1 − |bc| 1 − |ca| and |a|2 + |b|2 + |c|2 = a2 + b2 + c2 = 1. 1 It suffices to consider the case a, b, c ≥ 0. Now, for all x ≤ , we have 2 1 9x + 3 (1 − 3x)2 9x + 3 (1 − 3x)2 18x2 − 3x + 5 = = + ≤ +  . 1 1−x 4 4(1 − x) 4 4 4 1− 2 1 Now, since ab, bc, ca ≤ , we get 2  1 1 3 2 2 1 + + ≤ 6(a b + b2 c2 + c2 a2 ) − (ab + bc + ca) + 5 . 1 − ab 1 − bc 1 − ca 4 It suffices to show that 6(a2 b2 + b2 c2 + c2 a2 ) − (ab + bc + ca) ≤ 1, or 6(a2 b2 + b2 c2 + c2 a2 ) − (ab + bc + ca)(a2 + b2 + c2 ) ≤ (a2 + b2 + c2 )2 , or X

a2 (a − b)(a − c) + 2ab(a − b)2 + 2bc(b − c)2 + 2ca(c − a)2 ≥ 0.

cyc

which is obviously true by Schur’s inequality for fourth degree. 170

9. Let a, b, c be real numbers such that a2 + b2 + c2 + d2 = 1. Prove that 1 1 1 1 16 + + + ≤ . 1 − ab 1 − bc 1 − cd 1 − da 3 1 Solution: For all x ≤ , we have 2 1 1−x

16x + 8 (1 − 4x)2 16x + 8 (1 − 4x)2  + ≤ +  1 9 9(1 − x) 9 9 1− 2 2 32x + 10 . 9

=

=

1 Now, since ab, bc, cd, da ≤ , we have 2 1 1 1 1 + + + 1 − ab 1 − bc 1 − cd 1 − da

≤ = ≤ =

32(a2 b2 + b2 c2 + c2 d2 + d2 a2 ) + 40 9 2 2 2 32(a + c )(b + d2 ) + 40 9 2 2 8(a + b + c2 + d2 )2 + 40 9 16 . 3

10. Suppose a, b, c are real numbers such that a2 + b2 + c2 = 1. Show that 1 3+

a2

− 2bc

+

1 1 9 + ≤ . 2 3 + − 2ca 3 + c − 2ab 8 b2

Solution: Since 1

1 1 + 2 3 + − 2bc 3 + − 2ca 3 + c − 2ab 1 1 1 + + , 2 2 2 3 + |a| − 2 |bc| 3 + |b| − 2 |ca| 3 + |c| − 2 |ab| a2

≤

+

b2

and |a|2 + |b|2 + |c|2 = a2 + b2 + c2 = 1. It suffices for us to consider the case a, b, c ≥ 0. Now, for all x ∈ [−1, 1] , we have 1 3+x

= =

21 − 9x (1 + 3x)2 21 − 9x (1 + 3x)2 + ≤ + 64 64(3 + x) 64 64(3 − 1) 2 9x − 12x + 43 . 128 171

Since a2 − 2bc, b2 − 2ca, c2 − 2ab ∈ [−1, 1] (it is easy to check), we have 1

1 1 + 2 3 + " − 2bc 3 + − 2ca 3 + c − 2ab # X X 3 3 (a2 − 2bc)2 − 4 (a2 − 2bc) + 43 128 cyc cyc # " X X 3 2 2 3 (a − 2bc) + 8 ab + 39 . 128 cyc cyc a2

≤

=

+

b2

It suffices to show that X X 3 (a2 − 2bc)2 + 8 ab ≤ 9, cyc

cyc

or 3

X (a2 − 2bc)2 + 8(ab + bc + ca)(a2 + b2 + c2 ) ≤ 9(a2 + b2 + c2 )2 , cyc

or X X X X 6 a4 + 6 a2 b2 + 4abc a ≥ 8 ab(a2 + b2 ). cyc

cyc

cyc

cyc

By Schur’s inequality for fourth degree, we have X X X 4 a4 + 4abc a ≥ 4 ab(a2 + b2 ). cyc

cyc

cyc

We have to prove 2

X X X a4 + 6 a2 b2 ≥ 4 ab(a2 + b2 ). cyc

cyc

cyc

which is true since a4 + b4 + 6a2 b2 − 4ab(a2 + b2 ) = (a − b)4 ≥ 0.

11. Let a, b, c be positive real numbers such that abc = 1. Prove that 3a2

1 1 1 + 2 + 2 ≥ 1. 2 2 + (a − 1) 3b + (b − 1) 3c + (c − 1)2

1 Solution: Assume that a ≥ b ≥ c, then if c ≤ , we have 2 3c2

1 1 1 = 2 = ≥ 1. 2 + (c − 1) 4c − 2c + 1 2c(2c − 1) + 1

1 1 In the case a ≥ b ≥ c ≥ , then consider the function f (x) = 2 − 2 3x + (x − 1)2 1 2 1 2(x − 1)(16x3 − 1) + ln x with x ≥ , we have f 0 (x) = , hence 3 3 2 3x(4x2 − 2x + 1)2 172

1 ). Now, by writing the 2 1 variation board, we have f (x) ≥ f (1) = 0 ∀x ≥ . Therefore 2 f 0 (x) = 0 if and only if x = 1 (since x ≥

3a2

1 1 2 1 + 2 + 2 ≥ 1− (ln a+ln b+ln c) = 1. 2 2 2 + (a − 1) 3b + (b − 1) 3c + (c − 1) 3

Remark. There is another approach for this inequality: we have 1 1 a(a + 2)(a − 1)2 − = ≥ 0. 3a2 + (a − 1)2 a4 + a2 + 1 (4a2 − 2a + 1)(a4 + a2 + 1) Hence, it suffices to show that a4

1 1 1 + 4 + 4 ≥ 1. 2 2 + a + 1 b + b + 1 c + c2 + 1

This inequality has been proved in Example 10. 12. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that b2 c2 1 a2 + + ≤ . 2 2 2 2 2 2 5a + (b + c) 5b + (c + a) 5c + (a + b) 3 Solution: Normalize by declaring a + b + c = 1 and assume a ≥ b ≥ c, then our inequality becomes

6a2

a2 b2 c2 1 + 2 + 2 ≤ . − 2a + 1 6b − 2b + 1 6c − 2c + 1 3

Consider 2 cases 1 x2 12x − 1 (8x − 1)(3x − 1)2 Case 1. If c ≥ , then 2 − =− ≤ 8 6x − 2x + 1 27 27(6x2 − 2x + 1) 1 0 ∀x ≥ . Hence 8 6a2

a2 b2 c2 + 2 + 2 − 2a + 1 6b − 2b + 1 6c − 2c + 1

≤ =

12a − 1 12b − 1 12c − 1 + + 27 27 27 1 . 3

1 x2 4x + 1 (6x + 1)(2x − 1)2 , then − = − ≤ 8 6x2 − 2x + 1 18 18(6x2 − 2x + 1) 0 ∀x ≥ 0. Hence

Case 2. If c ≤

6a2

a2 b2 4(a + b) + 2 4(1 − c) + 2 1 2 + 2 ≤ = = − c. − 2a + 1 6b − 2b + 1 18 18 3 9

It suffices to show that

c2 6c2 −2c+1

− 29 c ≤ 0 which is true since

c2 2 c[12c2 + 3c + 2(1 − 8c)] − c = − ≤ 0. 6c2 − 2c + 1 9 9(6c2 − 2c + 1) 173

13. Let a, b, c, d be positive real numbers such that a + b + c + d = 4. Prove that (a2 + 1)(b2 + 1)(c2 + 1)(d2 + 1) ≥ (a + 1)(b + 1)(c + 1)(d + 1). Solution: Assume that a ≥ b ≥ c ≥ d. Then our inequality is equivalent to X [ln(a2 + 1) − ln(a + 1)] ≥ 0. cyc

We have 2 cases Case 1. If a ≤ 2, consider the function f (x) = ln(x2 + 1) − ln(x + 1) − 1 (x − 1) with x ≤ 2, we have 2 f 0 (x) =

(x − 1)(3 − x2 ) 2(x2 + 1)(x + 1)

√ From this, we get f 0 (x) = 0 if and only if x = 1 or x = 3 and by writing the variation board, we have f (x) ≥ min {f (1), f (2)} = 0 ∀x ≤ 2. Hence X

[ln(a2 + 1) − ln(a + 1)] ≥

cyc

1X (a − 1) = 0. 2 cyc

Case 2. If a ≥ 2, then 2 ≥ b ≥ c ≥ d, consider the function g(x) = 7 13 ln(x2 + 1) − ln(x + 1) − (3x − 2) − ln with x ≤ 2, we have 65 15 g 0 (x) =

(3x − 2)(43 + 10x − 7x2 ) . 65(x2 + 1)(x + 1)

2 Since x ≤ 2, we have g 0 (x) = 0 if and only if x = and by writing the 3 variation board, we get   2 = 0 ∀x ≤ 2. g(x) ≥ g 3 Hence [ln(b2 + 1) − ln(b + 1)] + [ln(c2 + 1) − ln(c + 1)] + [ln(d2 + 1) − ln(d + 1)] 7 13 21 13 ≥ (3b + 3c + 3d − 6) + 3 ln = (2 − a) + 3 ln . 65 15 65 15 We need to prove h(a) = ln(a2 + 1) − ln(a + 1) + We have h0 (a) =

21 13 (2 − a) + 3 ln ≥ 0. 65 15

(3a − 2)(43 + 10a − 7a2 ) , 65(a2 + 1)(a + 1) √

and since a ≥ 2, we get h0 (a) = 0 if and only if a = 5+ 7 326 . Writing the variation board again, we have h(a) ≥ min {h(2), h(4)} > 0. 174

14. Let a, b, c be nonnegative real numbers, no two of which are zero. Prove that r r r 48a 48b 48c 1+ + 1+ + 1+ ≥ 15. b+c c+a a+b Solution: Suppose that a + b + c = 1 and a ≥ b ≥ c, then our inequality becomes r r r 1 + 47a 1 + 47b 1 + 47c + + ≥ 15. 1−a 1−b 1−c We have 2 cases 2 Case 1. If c ≥ , then 27   1 + 47x 54 7 2 12(27x − 2)(3x − 1)2 2 = − x+ ≥ 0 ∀1 ≥ x ≥ . 1−x 5 5 25(1 − x) 27 Hence r

1 + 47a + 1−a

r

1 + 47b + 1−b

r

1 + 47c 54 21 ≥ (a + b + c) + = 15. 1−c 5 5

2 , then 27   96 1 2 48(48x + 1)(2x − 1)2 1 + 47x − x+ = ≥ 0 ∀1 ≥ x ≥ 0. 1−x 7 7 49(1 − x)

Case 2. If c ≤

Hence r 96 2 96 1 + 47a 1 + 47b + ≥ (a + b) + = 14 − c. 1−a 1−b 7 7 7 q 96 1+47c 2 We need to prove 1+47c 1−c ≥ 1 + 7 c. Put 1−c = t (t ≥ 0) , then we r

have c =

t2 −1 t2 +47

and since 0 ≤ c ≤

becomes t ≥ 1 + since

11 5

96(t2 −1) 7(t2 +47)

, or (t −

2 27 ,

we get

1)(7t2

11 5

≥ t ≥ 1, our inequality

− 96t + 233) ≥ 0 which is true

≥ t ≥ 1.

15. Let a, b, c be nonnegative real numbers such that a + b + c = 1. Prove that 1 − 4a2 1 − 4b2 1 − 4c2 + + ≤ 1. 1 + 3a − 3a2 1 + 3b − 3b2 1 + 3c − 3c2 Solution: Without loss of generality, assume that a ≥ b ≥ c. We have 2 cases 1 Case 1. If c ≥ , then 9 1 − 4x2 27x − 14 (1 − 9x)(1 − 3x)2 1 + = ≤ 0 ∀1 ≥ x ≥ . 2 1 + 3x − 3x 15 15(1 + 3x − 3x2 ) 9 Hence 1 − 4a2 1 − 4b2 1 − 4c2 + + 2 2 1 + 3a − 3a 1 + 3b − 3b 1 + 3c − 3c2

175

42 − 27(a + b + c) 15 = 1.

≤

1 Case 2. If c ≤ , then 9 8(1 − 2x) 1 − 4x2 (12x + 1)(2x − 1)2 − =− ≤ 0 ∀1 ≥ x ≥ 0. 2 1 + 3x − 3x 7 7(1 + 3x − 3x2 ) Hence 1 − 4b2 8 1 − 4a2 16 + ≤ (2 − 2a − 2b) = c. 2 2 1 + 3a − 3a 1 + 3b − 3b 7 7 16 7 c

1−4c2 1+3c−3c2

≤ 1 which is true since   16 c(48c2 − 41c + 5) 1 − 4c2 1 −1=− . c+ ≤ 0 since c ≤ 7 1 + 3c − 3c2 7(1 + 3c − 3c2 ) 9

We need to prove

+

16. Let a, b, c, d be nonnegative real numbers such that a + b + c + d = 2. Prove that a2 b2 c2 d2 16 + + + ≤ . (a2 + 1)2 (b2 + 1)2 (c2 + 1)2 (d2 + 1)2 25 Solution: Without loss of generality, assume that a ≥ b ≥ c ≥ d. We have 2 cases Case 1. If 12d3 + 11d2 + 32d ≥ 4, then 48x − 4 (12x3 + 11x2 + 32x − 4)(2x − 1)2 x2 − = − ≤0 (x2 + 1)2 125 125(x2 + 1)2 for all x such that 12x3 + 11x2 + 32x ≥ 4. Hence

≤

b2 c2 d2 a2 + + + ≤ (a2 + 1)2 (b2 + 1)2 (c2 + 1)2 (d2 + 1)2 16 48a − 4 48b − 4 48c − 4 48d − 4 + + + = . 125 125 125 125 25

1 Case 2. If 4 ≥ 12d3 + 11d2 + 32d ≥ 32d, it follows that d ≤ , we have 8 x2 108(5x + 1) (60x3 + 92x2 + 216x + 27)(3x − 2)2 − = − ≤ 0 ∀x ≥ 0. (x2 + 1)2 2197 2197(x2 + 1)2 Hence a2 b2 c2 + + (a2 + 1)2 (b2 + 1)2 (c2 + 1)2

≤ = =

108 [5(a + b + c) + 3] 2197 108 (5(2 − d) + 3) 2197 108 540 − d. 169 2197

It suffices to prove that d2 540 108 16 − a+ ≤ , 2 2 (d + 1) 2197 169 25 176

or

540 169d2 4 − d≤ . 2 2 (d + 1) 13 25

We have 169d2 540 4 − d− 2 2 (d + 1) 13 25

169d2 540 4 − a− 2 2 (d + 1) 15 25 2 169d 4 − 36a − (d2 + 1)2 25 4 + 900d − 4217a2 + 1800d3 + 4d4 + 900d5 − 25(d2 + 1)2 4 + 8 · 900d2 − 4217d2 + 1800d3 + 4d4 + 900d5 − 25(d2 + 1)2 4 + 2983d2 + 1800d3 + 4d4 + 900d5 < 0. − 25(d2 + 1)2

≤ = = ≤ =

17. Let a1 , a2 , . . . , an (n ≥ 2) be nonnegative real numbers such that a1 + a2 + . . . + an = n. Prove that (n − 1)(a31 + a32 + . . . + a3n ) + n2 ≥ (2n − 1)(a21 + a22 + . . . + a2n ). Solution: If n = 2, then the inequality becomes equality. If n = 3, put q = a1 a2 + a2 a3 + a3 a1 , r = a1 a2 a3 , then the inequality becomes 2(27 − 9q + 3r) + 9 ≥ 5(9 − 2q), or 3r + 9 − 4q ≥ 0 which is just Schur’s inequality for third degree. Consider the case n ≥ 4, assume that a1 ≥ a2 ≥ . . . ≥ an . We have 2 cases 1 Case 1. If an ≥ , then we have n−1 (n−1)x3 −(2n−1)x2 +(n+1)x−1 = (x−1)2 [(n−1)x−1] ≥ 0 ∀x ≥ Hence

1 . n−1

n n X X (n − 1) a3i ≥ [(2n − 1)a2i − (n + 1)ai + 1], i=1

or

i=1

n n X X (n − 1) a3i + n2 ≥ (2n − 1) a2i . i=1

i=1

1 , then we have Case 2. If an ≤ n−1 n(n − 1)(n − 2)x + n2 = (n − 1)2 [(n − 1)x − n]2 [(n − 1)x + 1] ≥ 0 ∀x ≥ 0. (n − 1)2

(n − 1)x3 − (2n − 1)x2 + = Hence n−1 X i=1

(n −

1)a3i

− (2n −

1)a2i

 n(n − 1)(n − 2)ai + n2 + ≥ 0, (n − 1)2

177

or

n−1 X

n−1 X

i=1

i=1

a3i − (2n − 1)

(n − 1)

a2i + n2 ≥

n(n − 2) an . n−1

We need to prove (n − 1)a3n − (2n − 1)a2n +

n(n − 2) an ≥ 0, n−1

or

an [(n − 1)(2 − an )[1 − (n − 1)an ] + n2 − 4n + 2] ≥0 n−1 1 . which is true since n ≥ 4 and an ≤ n−1

1.29

Using identities to prove inequalities

This is another collection of loosely related methods which provide short solutions to hard problems. The idea is to find special identities which help us solve the problem. Let us consider some examples: Example 1. Let x, y be real numbers such that x 6= −y, show that   1 + xy 2 2 2 x +y + ≥ 2. x+y Solution: Put z = − 1+xy x+y then we have xy + yz + zx = −1 and our original inequality is equivalent to x2 + y 2 + z 2 ≥ 2, or x2 + y 2 + z 2 ≥ −2(xy + yz + zx), which is just simply to (x + y + z)2 ≥ 0. Example 2. If a, b, c are distinct real numbers, then 1 + a2 b2 1 + b2 c2 1 + c2 a2 3 + + ≥ . 2 2 2 (a − b) (b − c) (c − a) 2 Solution: Notice that     1 − ab 1 − bc 1 − ca −1 −1 −1 = a−b b−c c−a (1 − a)(1 + b)(1 − b)(1 + c)(1 − c)(1 + a) = (a − b)(b − c)(c − a) (1 + a)(1 − b) (1 + b)(1 − c) (1 + c)(1 − a) = · · b − c  c − a  a − b  1 − ab 1 − bc 1 − ca = +1 +1 +1 , a−b b−c c−a and 

   1 + ab 1 + bc 1 + ca −i −i −i = a−b b−c c−a (i + a)(b − i)(i + b)(c − i)(i + c)(a − i) = (a − b)(b − c)(c − a) (a − i)(i + b) (b − i)(i + c) (c − i)(i + a) = · · b − c  c−a  a − b  1 + ab 1 + bc 1 + ca = +i +i +i , a−b b−c c−a 178

Hence 1 − ab a−b 1 + ab a−b

1 − bc 1 − bc + b−c b−c 1 + bc 1 + bc · + b−c b−c ·

1 − ca 1 − ca + c−a c−a 1 + ca 1 + ca · + c−a c−a ·

1 − ab a−b 1 + ab · a−b ·

= −1, = 1.

Furthermore, for any x, y, z ∈ R, we always have (x−y)2 +(y−z)2 +(z−x)2 ≥ 0, and (x + y + z)2 ≥ 0. Hence x2 + y 2 + z 2 ≥ xy + yz + zx, and x2 + y 2 + z 2 ≥ −2(xy + yz + zx). Therefore 

1 + ab a−b

2

 +

1 + bc b−c

2

 +

1 + ca c−a

2 ≥

X 1 + ab 1 + bc · = 1. a−b b−c cyc

and 

1 − ab a−b

2

 +

1 − bc b−c

2

 +

1 − ca c−a

2 ≥ −2

X 1 − ab 1 − bc · = 2. a−b b−c cyc

Thus 2(1 + a2 b2 ) 2(1 + b2 c2 ) 2(1 + c2 a2 ) + + (a − b)2 (b − c)2 (c − a)2

=

X (1 + ab)2 + (1 − ab)2

(a − b)2 X  1 + ab 2 X  1 − ab 2 + = a−b a−b cyc cyc cyc

≥ 3. Example 3. Let a, b, c be positive real numbers such that abc = 1. Prove that   a b c 16 4 + + ≤1+ . 2 2 2 (a + 1) (b + 1) (c + 1) (a + 1)(b + 1)(c + 1) Solution: Put x =

1−a 1−b 1−c ,y = ,z = , then we have 1+a 1+b 1+c

−1 ≤ x, y, z ≤ 1,

a=

1−x 1−y 1−z ,b = ,c = . 1+x 1+y 1+z

It is easy to verify that (1 − x)(1 − y)(1 − z) = (1 + x)(1 + y)(1 + z), hence x + y + z + xyz = 0. But 4 · 1−x 4a 1+x 2 =  2 = 1 − x , (a + 1)2 1−x 1+ 1+x

2 = a+1

2 = 1 + x. 1−x 1+ 1+x

Therefore, our inequality is equivalent to 1 − x2 + 1 − y 2 + 1 − z 2 ≤ 1 + 2(1 + x)(1 + y)(1 + z), or x2 + y 2 + z 2 + 2(xy + yz + zx) + 2(x + y + z + xyz) ≥ 0. 179

or (x + y + z)2 ≥ 0. Example 4. Let a, b, c be positive real numbers, then show that a b c 4abc + + + ≥ 2. b + c c + a a + b (a + b)(b + c)(c + a) Solution: Put x =

b c 1 1 a ,y = ,z = , then we have + + b+c c+a a+b x+1 y+1

1 = 2, or xy+yz+zx+2xyz = 1. And we need to prove x+y+z+4xyz ≥ 2, z+1 or x+y+z ≥ 2(xy+yz+zx). Put p = x+y+z, q = xy+yz+zx, r = xyz then we have q+2r = 1 and we need to prove p ≥ 2q. If p ≥ 2, then it is trivial since p ≥ 3 2 ≥ 2q. If p ≤ 2, then it is easy to verify that 2 ≥ p ≥ . By Schur’s inequality 2 for third degree, we have x(x − y)(x − z) + y(y − z)(y − x) + z(z − x)(z − y) ≥ 0, 2p(4q − p2 ) 2p3 + 9 p(4q − p2 ) . Hence q + ≤ 1, which implies us q ≤ . or r ≥ 9 9 8p + 9 2 3 (4p − 9)(2 − p) 2(2p + 9) = ≥ 0. Therefore p − 2q ≥ p − 8p + 9 8p + 9 Remark. Since (a+b)(b+c)(c+a) ≥ 8abc, this inequality implies us a stronger version of the famous Nesbitt inequality a b c 3 + + ≥ . b+c c+a a+b 2 Exercises 1. Let a, b, c be positive real numbers such that abc = 1. Prove that 1 1 1 2 + + + ≥ 1. 2 2 2 (1 + a) (1 + b) (1 + c) (1 + a)(1 + b)(1 + c) Solution: Using the substitution as Example 3, we need to prove (1 + x)2 + (1 + y)2 + (1 + z)2 + (1 + x)(1 + y)(1 + z) ≥ 4, or x2 + y 2 + z 2 + x2 y 2 z 2 ≥ 4xyz. By AM-GM inequality, we have p x2 + y 2 + z 2 + x2 y 2 z 2 ≥ 4 4 x4 y 4 z 4 = 4 |xyz| ≥ 4xyz. 2. If a, b, c are distinct real numbers, then a2 + ab + b2 b2 + bc + c2 c2 + ca + a2 9 + + ≥ . 2 2 2 (a − b) (b − c) (c − a) 4 180

3 1 Solution: We have a2 + ab + b2 = (a + b)2 + (a − b)2 , hence 4 4 " #   3 X a+b 2 a2 + ab + b2 b2 + bc + c2 c2 + ca + a2 + + = +1 . (a − b)2 (b − c)2 (c − a)2 4 cyc a − b Moreover, it is easy to verify that     a+b b+c c+a +1 +1 +1 = a−b b−c c−a     a+b b+c c+a = −1 −1 −1 , a−b b−c c−a Hence

a+b b+c b+c c+a c+a a+b · + · + · = −1 a−b b−c b−c c−a c−a a−b

Therefore       Xa + b b + c b+c 2 c+a 2 a+b 2 + + ≥ −2 · = 2. a−b b−c c−a a−b b−c cyc This inequality implies a2 + ab + b2 b2 + bc + c2 c2 + ca + a2 3 9 + + ≥ (2 + 1) = . 2 2 2 (a − b) (b − c) (c − a) 4 4 3. If a, b, c are distinct real numbers, then  
1 1 1 9 2 2 2 + + ≥ . a +b +c 2 2 2 (a − b) (b − c) (c − a) 2 Solution: According to the proof above, we have       a+b 2 b+c 2 c+a 2 + + ≥ 2. a−b b−c c−a Hence 2(a2 + b2 ) 2(b2 + c2 ) 2(c2 + a2 ) + + (a − b)2 (b − c)2 (c − a)2

=

X (a + b)2 + (a − b)2

(a − b)2 X  a + b 2 = + 3 ≥ 5. a−b cyc cyc

Also, we have 

   a b c +1 +1 +1 = b−c c−a a−b     a b c = −1 −1 −1 , b−c c−a a−b Hence

a b b c c a · + · + · = −1. b−c c−a c−a a−b a−b b−c 181

Thus 

a b−c

2

 +

b c−a

2

 +

c a−b

2

X a b ≥ −2 · = 2. b − c c − a cyc

Hence 

1 1 1 + + a +b +c 2 2 (a − b) (b − c) (c − a)2   2 X a2 + b2 X a 5 9 = + ≥ +2= . 2 (a − b) b−c 2 2 cyc cyc 2

2

2




 =

4. Let a, b, c be positive real numbers, show that 2  2  2  a b c 10abc ≥ 2. + + + b+c c+a a+b (a + b)(b + c)(c + a) Solution: By the same substitution as Example 4, we need to prove x2 + y 2 + z 2 + 10xyz ≥ 2. Put p = x + y + z, q = xy + yz + zx, r = xyz then we have q + 2r = 1 and we need to prove p2 − 2q + 10r ≥ 2, or p2 − 7q + 3 ≥ 0. If p ≥ 2, then it is trivial since p2 + 3 ≥ 7 ≥ 7q. If 2 ≥ p, then we can easily check 3 2p3 + 9 that 2 ≥ p ≥ . And from the proof of Example 4, we get q ≤ . 2 8p + 9 Hence p2 − 7q + 3 ≥ p2 + 3 −

3(2p − 3)(4 − p2 ) 7(2p3 + 9) = ≥ 0. 8p + 9 8p + 9

5. Let a, b, c be distinct real numbers. Prove that (a − 2b)2 + (a − 2c)2 (b − 2c)2 + (b − 2a)2 (c − 2a)2 + (c − 2b)2 + + ≥ 22. (b − c)2 (c − a)2 (a − b)2 Solution: We have (a − 2b)2 + (a − 2c)2 −2=2 (b − c)2



b+c−a b−c

2 ,

Hence, our inequality is equivalent to       c+a−b 2 a+b−c 2 b+c−a 2 + + ≥ 8. b−c c−a a−b We have 

   b+c−a c+a−b a+b−c −2 −2 −2 = b−c c−a a−b (3c − a − b)(3a − b − c)(3b − c − a) = (a − b)(b − c)(c − a) 3b − c − a 3c − a − b 3a − b − c = · · c−a a − b   b−c  b+c−a c+a−b a+b−c = +2 +2 +2 . b−c c−a a−b 182

Hence b+c−a c+a−b c+a−b a+b−c a+b−c b+c−a · + · + · = −4. b−c c−a c−a a−b a−b b−c Thus, our inequality is equivalent to x2 + y 2 + z 2 ≥ 8, or x2 + y 2 + z 2 ≥ −2(xy + yz + zx), which is just simply to (x + y + z)2 ≥ 0. 6. Let a, b, c be distinct real numbers. Prove that (1 − a2 )(1 − b2 ) (1 − b2 )(1 − c2 ) (1 − c2 )(1 − a2 ) + + ≥ −1. (a − b)2 (b − c)2 (c − a)2 Solution: Since (1 − a2 )(1 − b2 ) +1= (a − b)2



1 − ab a−b

2 ,

Our inequality is equivalent to       1 − ab 2 1 − bc 2 1 − ca 2 + + ≥ 2. a−b b−c c−a This inequality has been already proved in Example 2. 7. Let a, b, c be positive real numbers such that abc = 1. Prove that b+3 c+3 a+3 + + ≥ 3. (a + 1)2 (b + 1)2 (c + 1)2 Solution: Using the substitution as Example 3, we have 1−x +3 a+3 1 2 1+x = 2 = (x + 3x + 2).  2 (a + 1) 2 1−x +1 1+x Hence, our inequality is equivalent to x2 + y 2 + z 2 + 3(x + y + z) ≥ 0, or x2 + y 2 + z 2 ≥ 3xyz. By AM-GM inequality, we have p x2 + y 2 + z 2 ≥ 3 3 x2 y 2 z 2 ≥ 3 |xyz| (since |xyz| ≤ 1) ≥ 3xyz. 8. Let a, b, c be real numbers such that a, b, c 6= 1 and abc = 1. Prove that a2 b2 c2 + + ≥ 1. 2 2 (a − 1) (b − 1) (c − 1)2 Solution: By the same subsitution as Example 3, put   a 1 1 1 1 = = = 1− , 1 1+x a−1 2 x 1− 1− a 1−x 183

Hence, our inequality is equivalent to 

or

1 1− x

2

    1 2 1 2 + 1− + 1− ≥ 4, y z

1 1 1 + 2 + 2 −2 2 x y z



1 1 1 + + x y z

 − 1 ≥ 0.

Since x + y + z + xyz = 0 and x, y, z 6= 0 (since a, b, c 6= 1), we have 1 1 1 + + = −1. Hence xy yz zx 1 1 1 + 2+ 2 2 x y z

 =  =

1 1 1 + + x y z

2

1 1 1 + + x y z

2

 −2

1 1 1 + + xy yz zx

+ 2.

It suffices to show that     1 1 1 2 1 1 1 + + −2 + + + 1 ≥ 0. x y z x y z which is just simply to 

1 1 1 + + −1 x y z

184

2 ≥ 0.



Chapter 2

Problems 83.1. Let a, b, c be the sidelengths of a triangle. Prove that a2 b(a − b) + b2 c(b − c) + c2 a(c − a) ≥ 0. (IMO 1983) 83.2. Show that for any positive reals a, b, c, d, e, f , we have cd ef (a + c + e)(b + d + f ) ab + + ≤ . a+b c+d e+f a+b+c+d+e+f (United Kingdom 1983) 84.1. Let a1 , a2 , . . . , an > 0, n ≥ 2. Prove that a2 a21 a22 a2 + + · · · + n−1 + n ≥ a1 + a2 + · · · + an . a2 a3 an a1 (China 1984) 84.2. Let x, y, z be nonnegative real numbers such that x + y + z = 1. Prove that 7 0 ≤ xy + yz + zx − 2xyz ≤ . 27 (IMO 1984) 84.3. Prove that for any a, b > 0, we have that √ √ (a + b)2 a + b + ≥ a b + b a. 2 4 (Russia 1984) 85.1. Let x1 , x2 , . . . , xn be real numbers from the interval [0, 2]. Prove that n X

|xi − xj | ≤ n2 .

i,j=1

When do we have equality? (United Kingdom 1985) 86.1. Find the maximum value of the constant c such that for any x1 , x2 , . . . , xn > 0 for which xk+1 ≥ x1 + x2 + · · · + xk for any k, the inequality √ √ √ √ x1 + x2 + · · · + xn ≤ c x1 + x2 + · · · + xn 185

also holds for any n. (IMO Shorlist 1986) 86.2. Let n be a positive integer. Prove that 8 |sin 1| + |sin 2| + · · · + |sin 3n| > n. 5 (Russia 1986) 86.3. Show that for all positive real numbers a1 , a2 , . . . , an , we have   1 2 n 1 1 1 + + ··· + ≤4 + + ··· + . a1 a1 + a2 a1 + a2 + · · · + an a1 a2 an (Russia 1986) 86.4. Find the maximum value of x2 y + y 2 z + z 2 x for reals x, y, z with x + y + z = 0 and x2 + y 2 + z 2 = 6. (United Kingdom 1986) 87.1. Prove that if x, y, z are real numbers such that x2 + y 2 + z 2 = 2, then x + y + z ≤ xyz + 2. (IMO Shorlist 1987) 88.1. Show that (1 + x)n ≥ (1 − x)n + 2nx(1 − x2 )

n−1 2

for all 0 ≤ x ≤ 1 and all positive integer n. (Ireland 1988) 88.2. Given a triangle ABC. Prove that         sin A sin B sin C 1 1 1 1 1 1 2 + + ≤ + sin A+ + sin B+ + sin C. A B C B C C A A B (Russia 1988) 89.1. Let x1 , x2 , . . . , xn be positive real numbers, and let S = x1 +x2 +· · ·+xn . Prove that (1 + x1 )(1 + x2 ) · · · (1 + xn ) ≤ 1 + S +

S2 Sn + ··· + . 2! n!

(APMO 1989) π 89.2. Let x, y, z be real numbers such that 0 < x < y < z < . Prove that 2 π + 2 sin x cos y + 2 sin y cos z > sin 2x + sin 2y + sin 2z. 2 (Iberoamerica 1989) 89.3. Let a, b, c be the sidelengths of a triangle. Prove that a − b b − c c − a 1 a + b + b + c + c + a < 16 . 186

(Iberoamerica 1989) 89.4. Find the least possible value of (x + y)(y + z) for positive reals x, y, z satisfying xyz(x + y + z) = 1. (Russia 1989) 90.1. Let a, b, c, d be positive real numbers such that ab + bc + cd + da = 1. Prove that a3 b3 c3 d3 1 + + + ≥ . b+c+d c+d+a d+a+b a+b+c 3 (IMO Shortlist 1990) 90.2. Show that x4 > x −

1 2

for all real x. (Russia 1990) 90.3. Let x1 , x2 , . . . , xn be positive reals with sum 1. Show that x22 x2n 1 x21 + + ··· + ≥ . x1 + x2 x2 + x3 xn + x1 2 (Russia 1990) 90.4. Show that p p p x2 − xy + y 2 + y 2 − yz + z 2 ≥ z 2 + zx + x2 for any positive real numbers x, y, z. (United Kingdom 1990) 91.1. Let a1 , a2 , . . . , an and b1 , b2 , . . . , bn be positive real numbers such that a1 + a2 + · · · + an = b1 + b2 + · · · + bn . Prove that a21 a22 a2n a1 + a2 + · · · + an + + ··· + ≥ . a1 + b1 a2 + b2 an + bn 2 (APMO 1991) 91.2. Given positive real numbers a, b, c satisfying a + b + c = 1, show that (1 + a) (1 + b) (1 + c) ≥ 8 (1 − a) (1 − b) (1 − c) . (Russia 1991) 91.3. Show that √ (x + y + z)2 √ √ ≥ x yz + y zx + z xy 3 for all nonnegative reals x, y, z. (Russia 1991) 91.4. The real numbers x1 , x2 , . . . , x1991 satisfy |x1 − x2 | + |x2 − x3 | + · · · + |x1990 − x1991 | = 1991. 187

Denote sn =

x1 + x2 + . . . + xn . What is the maximum possible value of n |s1 − s2 | + |s2 − s3 | + . . . + |s1990 − s1991 |? (Russia 1991)

91.5. A triangle has sides a, b, c with sum 2. Show that a2 + b2 + c2 + 2abc < 2. (United Kingdom 1991) 91.6. Prove the inequality x2 y y 2 z z 2 x + + ≥ x2 + y 2 + z 2 z x y for any positive real numbers x, y, z with x ≥ y ≥ z. (Vietnam 1991) 92.1. For every integer n ≥ 2, find the smallest positive number λ = λ(n) 1 such that if 0 ≤ a1 , a2 , . . . , an ≤ , b1 , b2 , . . . , bn > 0, and a1 + a2 + · · · + an = 2 b1 + b2 + · · · + bn = 1, then b1 b2 · · · bn ≤ λ(a1 b1 + a2 b2 + · · · + an bn ). (China 1992) 92.2. Show that

√ x4 + y 4 + z 2 ≥ 2 2xyz

for all positive reals x, y, z. (Russia 1992) 92.3. Show that for any real numbers x, y > 1, we have y2 x2 + ≥ 8. y−1 x−1 (Russia 1992) 92.4. Positive real numbers a, b, c satisfy a ≥ b ≥ c. Prove that a2 − b2 c2 − b2 a2 − c2 + + ≥ 3a − 4b + c. c a b (Ukraine 1992) 92.5. Let x, y, z, w be positive real numbers. Prove that   X 1 12 3 1 1 1 1 ≤ ≤ + + + . x + y + z + w sym x + y 4 x y z w (United Kingdom 1992) 93.1. Let a, b, c, d be positive real numbers. Prove that a b c d 2 + + + ≥ . b + 2c + 3d c + 2d + 3a d + a + 3b a + 2b + 3c 3 (IMO Shortlist 1993) 188

93.2. Let a, b, c ∈ [0, 1]. Prove that a2 + b2 + c2 ≤ a2 b + b2 c + c2 a + 1. (Italia 1993) 93.3. Let x, y, u, v be positive real numbers. Prove that xy uv xy + xv + yu + uv ≥ + . x+y+u+v x+y u+v 93.4. If the equation then a2 + b2 ≥ 8.

x4

+

ax3

+

2x2

(Poland 1993) + bx + 1 = 0 has at least one real root,

(Tournament of the Towns 1993) 93.5. Let x1 , x2 , x3 , x4 be real numbers such that 1 ≤ x21 + x22 + x23 + x24 ≤ 1. 2 Find the largest and the smallest values of the expression A = (x1 − 2x2 + x3 )2 + (x2 − 2x3 + x4 )2 + (x2 − 2x1 )2 + (x3 − 2x4 )2 . (Vietnam 1993) 94.1. Let a1 , a√ 2 , . . . , an be a sequence of positive real numbers satisfying a1 + · · · + ak ≥ k for all k = 1, 2, . . . , n. Prove that   1 1 1 2 2 2 1 + + ··· + . a1 + a2 + · · · + an > 4 2 n (USA 1994) 95.1. Prove that for any positive real numbers x, y and any positive integers m, n,

(m−1)(n−1) xm+n + y m+n +(m + n − 1) (xm y n + xn y m ) ≥ mn xm+n−1 y + xy m+n−1 . (Austrian-Polish Competition 1995) 95.2. Let a, b, c, d be positive real numbers. Prove that a+c b+d c+a d+b + + + ≥ 4. a+b b+c c+d d+a (Baltic Way 1995) 95.3. Let x, y, z be positive real numbers. Prove that xx y y z z ≥ (xyz)

x+y+z 3

.

(Canada 1995) 95.4. If a, b, c are positive real numbers such that abc = 1, then 1 1 1 3 + 3 + 3 ≥ . + c) b (c + a) c (a + b) 2

a3 (b

(IMO 1995) 189

95.5. Suppose that x1 , x2 , . . . , xn are real numbers satisfying |xi − xi+1 | < 1 and xi ≥ 1 for i = 1, 2, . . . , n (xn+1 = x1 ). Prove the following inequality x1 x2 xn + + ··· + < 2n − 1. x2 x3 x1 (India 1995) 95.6. (a) Find the maximum value of the expression x2 y−y 2 x when 0 ≤ x ≤ 1 and 0 ≤ y ≤ 1. (b) Find the maximum value of the expression x2 y +y 2 z +z 2 x−y 2 x−z 2 y −x2 z when 0 ≤ x ≤ 1, 0 ≤ y ≤ 1 and 0 ≤ z ≤ 1. (United Kingdom 1995) 95.7. Let a, b, c be real numbers satisfying a < b < c, a + b + c = 6 and ab + bc + ca = 9. Prove that 0 < a < 1 < b < 3 < c < 4. (United Kingdom 1995) 95.8. Find the greatest constant k such that for any positive integer n which is not a square, √
√
1 + n sin π n > k. (Vietnam (IMO training camp) 1995) 96.1. Let m and n be positive integers such that n ≤ m. Prove that 2n · n! ≤

(m + n)! ≤ (m2 + m)n . (m − n)!

(APMO 1996) 96.2. Let a, b, c be the lengths of the sides of a triangle. Prove that √ √ √ √ √ √ a + b − c + b + c − a + c + a − b ≤ a + b + c, and determine when equality occurs. (APMO 1996) 96.3. The real numbers x, y, z, t satisfy the equalities x + y + z + t = 0 and x2 + y 2 + z 2 + t2 = 1. Prove that −1 ≤ xy + yz + zt + tx ≤ 0. (Austrian-Polish Competition 1996) √ 96.4. Let a, b, c be positive real numbers, such that a + b + c = abc. Prove that ab + bc + ca ≥ 9(a + b + c). (Belarus 1996) 96.5. Suppose n ∈ N, x0 = 0, xi > 0 for i = 1, 2, . . . , n, and x1 +x2 +· · ·+xn = 1. Prove that 1≤

n X i=1

√

xi π √ < . 2 1 + x0 + · · · + xi−1 · xi + · · · + xn (China 1996) 190

96.6. (a) Find the minimum value of xx for x a positive real number. (b) If x and y are positive real numbers, show that xy + y x > 1. (France 1996) 96.7. Let a, b, c be positive real numbers such that abc = 1. Prove that ab bc ca + + ≤ 1. a5 + b5 + ab b5 + c5 + bc c5 + a5 + ca (IMO Shortlist 1996) 96.8. Let a and b be positive real numbers with a + b = 1. Prove that a2 b2 1 + ≥ . a+1 b+1 3 (Hungary 1996) 96.9. Prove the following inequality for positive real numbers x, y, z,   1 1 9 1 + + ≥ . (xy + yz + zx) (y + z)2 (z + x)2 (x + y)2 4 (Iran 1996) 96.10. Let n ≥ 2 be a fixed natural number and let a1 , a2 , . . . , an be positive numbers whose sum is 1. Prove that for any positive numbers x1 , x2 , . . . , xn whose sum is 1, n X n − 2 X ai x2i + , 2 xi xj ≤ n−1 1 − ai i=1

1≤i 0, y > 0, z > 0 and + + ≤ 1; x y z (ii) for every quadrilateral with sides a, b, c, d, a2 x + b2 y + c2 z > d2 . (Romania 1996) 96.14. Let a, b, c be positive real numbers. (a) Prove that 4(a3 + b3 ) ≥ (a + b)3 . (b) Prove that 9(a3 + b3 + c3 ) ≥ (a + b + c)3 . 191

(United Kingdom 1996) 96.15. Let a, b, c, d be positive real numbers such that 2(ab + ac + ad + bc + bd + cd) + abc + bcd + cda + dab = 16. Prove that 2 a + b + c + d ≥ (ab + ac + ad + bc + bd + cd). 3 (Vietnam 1996) 96.16. Prove that for any real numbers a, b, c, 4 (a + b)4 + (b + c)4 + (c + a)4 ≥ (a4 + b4 + c4 ). 7 (Vietnam 1996) 97.1. a, b, c be positive numbers such that abc = 1. Prove that 1 1 1 1 1 1 + + ≤ + + . 1+a+b 1+b+c 1+c+a a+2 b+2 c+2 (Bulgaria 1997) 97.2. Prove that

1 1 3 1997 1 < · ··· < . 1999 2 4 1998 44

(Canada 1997) 97.3. Let x1 , x2 , . . . , x1997 be real numbers satisfying the following conditions √ 1 (a) − √ ≤ xi ≤ 3 for i = 1, 2, . . . , 1997; 3 √ (b) x1 + x2 + · · · + x1997 = −318 3. 12 12 Determine the maximum value of x12 1 + x2 + · · · + x1997 . (China 1997) 97.4. For each natural number n ≥ 2, determine the largest possible value of the expression Vn = sin x1 cos x2 + sin x2 cos x3 + · · · + sin xn cos x1 , where x1 , x2 , . . . , xn are arbitrary real numbers. (Czech-Slovak 1997) 97.5. Let x, y, z be positive real numbers. Prove that   p √ xyz x + y + z + x2 + y 2 + z 2 3+ 3 ≤ . (x2 + y 2 + z 2 ) (xy + yz + zx) 9 (Hong Kong 1997) 97.6. Given x1 , x2 , x3 , x4 are positive real numbers such that x1 x2 x3 x4 = 1. Prove that   1 1 1 1 3 3 3 3 + + + . x1 + x2 + x3 + x4 ≥ max x1 + x2 + x3 + x4 , x1 x2 x3 x4 (Iran 1997) 192

97.7. Let a, b, c be nonnegative real numbers such that a + b + c ≥ abc. Prove that a2 + b2 + c2 ≥ abc. (Ireland 1997) 97.8. Let a, b, c be positive real numbers. Prove that (b + c − a)2 (c + a − b)2 (a + b − c)2 3 + + ≥ . 2 2 2 2 2 2 (b + c) + a (c + a) + b (a + b) + c 5 (Japan 1997) 97.9. Let a1 , . . . , an be positive numbers, and define A=

a1 + · · · + an , n

G=

√ n

a1 · · · an ,

H=

a−1 1

n . + · · · + a−1 n

 n A A (a) If n is even, show that ≤ −1 + 2 . H G   n − 2 2(n − 1) A n A ≤− + . (b) If n is odd, show that H n n G (Korea 1997) 97.10. For any positive real numbers x, y, z such that xyz = 1, prove the inequality x9 + y 9 y9 + z9 z 9 + x9 + + ≥ 2. x6 + x3 y 3 + y 6 y 6 + y 3 z 3 + z 6 z 6 + z 3 z 3 + x6 (Romania 1997) 97.11. Let a, b, c be positive real numbers. Prove that b2 c2 bc ca ab a2 + + ≥1≥ 2 + 2 + 2 . 2 2 2 a + 2bc b + 2ca c + 2ab a + 2bc b + 2ca c + 2ab (Romania 1997) 97.12. Show that if 1 < a < b < c, then loga (loga b) + logb (logb c) + logc (logc a) > 0. (Russia 1997) 97.13. Prove that for x, y, z ≥ 2, (y 3 + x)(z 3 + y)(x3 + z) ≥ 125xyz. (Saint Petersburg 1997) 97.14. Given an integer n ≥ 2, find the minimal value of x51 x52 x5n + + ··· + , x2 + x3 + · · · + xn x3 + · · · + xn + x1 x1 + x2 + · · · + xn−1 for positive real numbers x1 , . . . , xn subject to the condition x21 + · · · + x2n = 1. (Turkey 1997) 97.15. Let x, y and z be positive real numbers. 193

1 1 1 + + ≤ 3? x y z 1 1 1 (b) If x + y + z ≤ 3, is it necessarily true that + + ≥ 3? x y z (United Kingdom 1997) 97.16. Prove that, for all positive real numbers a, b, c, (a) If x + y + z ≥ 3, is it necessarily true that

a3

1 1 1 1 + 3 + 3 ≤ . 3 3 3 + b + abc b + c + abc c + a + abc abc (USA 1997)

98.1. Let a, b, c be positive real numbers. Prove that      a b  c a+b+c 1+ 1+ . 1+ ≥2 1+ √ 3 b c a abc (APMO 1998) 98.2. Let x1 , x2 , y1 , y2 be real numbers such that x21 + x22 ≤ 1. Prove the inequality (x1 y1 + x2 y2 − 1)2 ≥ (x21 + x22 − 1)(y12 + y22 − 1). (Austrian-Polish Competition 1998) 98.3. If n ≥ 2 is an integer and 0 < a1 < a2 < . . . < a2n+1 are real numbers, prove the inequality √ n

a1 −

√ n

a2 +

√ n

a3 −· · ·−

√ n

a2n +

√ n

a2n+1 <

√ n

a1 − a2 + a3 − . . . − a2n + a2n+1 . (Balkan 1998)

98.4. Let a, b, c be positive real numbers. Prove that a b c a+b b+c + + ≥ + + 1. b c a b+c a+b (Belarus 1998) 98.5. Let n be a natural number such that n ≥ 2. Prove that     1 1 1 1 1 1 1 1 + + ... + > + + ... + . n+1 3 2n − 1 n 2 4 2n (Canada 1998) 98.6. Let n ≥ 2 be a positive integer and let x1 , x2 , . . . , xn be real numbers such that n n−1 X X x2i + xi xi+1 = 1. i=1

i=1

For every positive integer k, 1 ≤ k ≤ n, determine the maximum value of |xk | . (China 1998) 98.7. Let a, b, c ≥ 1. Prove that p √ √ √ a − 1 + b − 1 + c − 1 ≤ c (ab + 1). (Hong Kong 1998) 194

98.8. Let a1 , a2 , . . . , an > 0 such that a1 + a2 + · · · + an < 1. Prove that a1 a2 · · · an (1 − a1 − a2 − · · · − an ) 1 ≤ n+1 . (a1 + a2 + · · · + an ) (1 − a1 ) (1 − a2 ) · · · (1 − an ) n (IMO Shortlist 1998) 98.9. Let a, b, c be positive real numbers such that abc = 1. Prove that a3 b3 c3 3 + + ≥ . (1 + b) (1 + c) (1 + a) (1 + c) (1 + a) (1 + b) 4 (IMO Shortlist 1998) 1 1 1 98.10. Let x, y, z > 1 such that + + = 2. Prove that x y z p √ √ √ x + y + z ≥ x − 1 + y − 1 + z − 1. (Iran 1998) 98.11. Suppose that a1 < a2 < · · · < an are real numbers. Prove that a1 a42 + a2 a43 + · · · + an−1 a4n + an a41 ≥ a2 a41 + a3 a42 + · · · + an a4n−1 + a1 a4n . (Iran 1998) 98.12. Show that if x is a nonzero real number, then x8 − x5 −

1 1 + ≥ 0. x x4 (Ireland 1998)

98.13. Prove that if a, b, c are positive real numbers, then   9 1 1 1 ≤2 + + , a+b+c a+b b+c c+a and

1 1 1 1 + + ≤ a+b b+c c+a 2



1 1 1 + + a b c

 .

(Ireland 1998) 98.14. If x, y, z are positive real numbers such that x + y + z = xyz, then √

1 1 1 3 +p +√ ≤ . 2 1 + x2 1 + z2 1 + y2 (Korea 1998)

98.15. Let a, b, c, d, e, f be positive real numbers such that a + b + c + d + e + f = 1,

ace + bdf ≥

Prove that abc + bcd + cde + def + ef a + f ab ≤

1 . 108

1 . 36 (Poland 1998)

195

98.16. Let a be a positive real numbers and let x1 , x2 , . . . , xn be positive real numbers such that x1 + x2 + · · · + xn = 1. Prove that ax2 −x3 axn −x1 n2 ax1 −x2 + + ··· + ≥ . x1 + x2 x2 + x3 xn + x1 2 (Serbia 1998) 98.17. Find the minimum of the expression p p p F (x, y) = (x + 1)2 + (y − 1)2 + (x − 1)2 + (y + 1)2 + (x + 2)2 + (y + 2)2 , where x, y are real numbers. (Vietnam 1998) 98.18. Let n ≥ 2 and x1 , x2 , . . . , xn be positive real numbers satisfying 1 1 1 1 + + ··· + = . x1 + 1998 x2 + 1998 xn + 1998 1998 Prove that

√ n

x1 x2 · · · xn ≥ 1998. n−1

99.1. Let {an } be a sequence of real numbers such that ai+j i, j. Prove that the following inequality holds a1 +

(Vietnam 1998) ≤ ai + aj for all

a2 an + ··· + ≥ an . 2 n a2

99.2. Let a, b, c be positive real numbers such that + that 1 1 1 3 + + ≥ . 1 + ab 1 + bc 1 + ca 2

b2

(APMO 1999) + c2 = 3. Prove

(Belarus 1999) 99.3. For any nonnegative real numbers x, y, z such that x + y + z = 1, prove the following inequality x2 y + y 2 z + z 2 x ≤

4 . 27 (Canada 1999)

99.4. Let a, b, c be positive real numbers. Prove that a b c + + ≥ 1. b + 2c c + 2a a + 2b (Czech-Slovak 1999) 99.5. Let n ≥ 2 be a fixed integer. Find the least constant C such that the inequality X xi xj (x2i + x2j ) ≤ C(x1 + x2 + · · · + xn )4 1≤i 3 and a1 , a2 , . . . , an be real numbers such that a1 +a2 +· · ·+an ≥ n and a21 + a22 + · · · + a2n ≥ n2 . Prove that max {a1 , a2 , . . . , an } ≥ 2. (USA 1999)  π 99.9. Let a0 , a1 , . . . , an be numbers from the interval 0, such that 2    π π π + tan a1 − + · · · + tan an − ≥ n − 1. tan a0 − 4 4 4 Prove that tan a0 tan a1 · · · tan an ≥ nn+1 . (USA 1999) 99.10. Let x, y, z > 1. Prove that xx

2 +2yz

yy

2 +2zx

zz

2 +2xy

≥ (xyz)xy+yz+zx . (USA (Shortlist) 1999)

00.1. Let a, b be real numbers and a 6= 0. Prove that a2 + b2 +

1 b √ + ≥ 3. 2 a a

(Austria 2000) 00.2. For all real numbers a, b, c ≥ 0 such that a + b + c = 1, prove that 2 ≤ (1 − a2 )2 + (1 − b2 )2 + (1 − c2 )2 ≤ (1 + a)(1 + b)(1 + c). (Austrian-Polish Competition 2000) 00.3. Let a, b, c, x, y, z be positive real numbers. Prove that a3 b3 c3 (a + b + c)3 + + ≥ . x y z 3 (x + y + z) (Belarus 2000) 00.4. Suppose that the real numbers a1 , a2 , . . . , a100 satisfy (i) a1 ≥ a2 ≥ · · · ≥ a100 ≥ 0; (ii) a1 + a2 ≤ 100; (iii) a3 + a4 + · · · + a100 ≤ 100. Determine the maximum possible value of a21 + a22 + · · · + a2100 , and find all possible sequences a1 , a2 , . . . , a100 which achieve this maximum. (Canada 2000) 197

00.5. Show that

s 3



2 (a + b)

1 1 + a b

r

 ≥

3

a + b

r 3

b a

for all positive real numbers a and b, and determine when the equality occurs. (Czech-Slovak 2000) 00.6. Let a, b, c be positive real numbers such that abc = 1. Prove that 1 + ab2 1 + bc2 1 + ca2 18 + + ≥ 3 . 3 3 3 c a b a + b3 + c3 (Hong Kong 2000) 00.7. Let x, y, and z denote positive real numbers, each less than 4. Prove 1 1 1 1 1 1 , + , and + is greater that at least one of the numbers + x 4−y y 4−z z 4−x than or equal to 1. (Hungary 2000) 00.8. Let a, b, c be positive real numbers such that abc = 1. Prove the inequality     1 1 1 b+ −1 c + − 1 ≤ 1. a+ −1 b c a (IMO 2000) 00.9. Let x, y ≥ 0 with x + y = 2. Prove that x2 y 2 (x2 + y 2 ) ≤ 2. (Ireland 2000) 00.10. The real numbers a, b, c, x, y, z satisfy a ≥ b ≥ c > 0 and x ≥ y ≥ z > 0. Prove that a2 x2 b2 y 2 c2 z 2 3 + + ≥ . (by + cz)(bz + cy) (cz + ax)(cx + az) (ax + by)(ay + bx) 4 (Korea 2000) 00.11. Let x, y, z be real numbers. Prove that √ x2 + y 2 + z 2 ≥ 2 (xy + yz) . (Macedonia 2000) 00.12. Let a, b, x, y, z be positive real numbers. Prove that y z 3 x + + ≥ . ay + bz az + bx ax + by a+b (MOSP 2000) 00.13. Let ABC be an acute triangle. Prove that 

cos A cos B

2

 +

cos B cos C

2

 +

cos C cos A

2 + 8 cos A cos B cos C ≥ 4. (MOSP 2000)

198

00.14. Let a, b, c be positive real numbers such that min {a, b} ≥ c. Prove that p p √ c(a − c) + c(b − c) ≤ ab. (MOSP 2000) 00.15. Let a1 , a2 , . . . , an be nonnegative real numbers. Prove that √ a1 + a2 + · · · + an − n a1 a2 · · · an ≥ n n√ √ 2 √ √ 2 √ √ 2o 1 ≥ min ( a1 − a2 ) , ( a2 − a3 ) , . . . , ( an − a1 ) . 2 (MOSP 2000) 00.16. Let (an ) be an infinite sequence of positive numbers such that a 2 a1 a4 + + ··· + k ≤ 1 1 2 k for all k. Prove that

a1 a2 an + + ··· + < 2, 1 2 n

for all n. (MOSP 2000) 00.17. Let a, b, c be nonnegative real numbers such that ab + bc + ca = 1. Prove that 1 1 5 1 + + ≥ . b+c c+a a+b 2 (MOSP 2000) 00.18. Let a1 , a2 , . . . , an be positive real numbers such that 1 1 1 + + ··· + ≤ 1. a1 a2 an Prove that for any positive integer k, (ak1 − 1)(ak2 − 1) · · · (akn − 1) ≥ (nk − 1)n . (MOSP 2000) 00.19. Let a, b, c be positive real numbers. Prove that  2 √ 3 a + b + c + abc 1 1 1 1 + + + √ ≥ . 3 a + b b + c c + a 2 abc (a + b)(b + c)(c + a) (MOSP 2000) 00.20. For any integer n ≥ 2, consider n−1 positive real numbers a1 , a2 , . . . , an−1 having sum 1, and n real numbers b1 , b2 , . . . , bn . Prove that b21 +

b22 b2 b2 + 3 + . . . + n ≥ 2b1 (b2 + b3 + . . . + bn ). a1 a2 an−1

(Romania 2000) 00.21. Positive real numbers x, y, z satisfy xyz = 1. Prove that the following inequality holds x2 + y 2 + z 2 + x + y + z ≥ 2 (xy + yz + zx) . 199

(Russia 2000) 00.22. Let x1 , x2 , . . . , xn be real numbers (n ≥ 2), satisfying the conditions −1 < x1 < x2 < · · · < xn < 1 and 13 13 x13 1 + x2 + · · · + xn = x1 + x2 + · · · + xn .

Prove that 13 13 x13 1 y1 + x2 y2 + · · · + xn yn < x1 y1 + x2 y2 + · · · + xn yn

for any real numbers y1 < y2 < · · · < yn . (Russia 2000) 00.23. Show that for all n ∈ N and x ∈ R, sinn 2x + (sinn x − cosn x)2 ≤ 1. (Russia 2000) 00.24. Let n ≥ 3 be a positive integer. Prove that for all positive real numbers a1 , a2 , . . . , an , we have a1 + a2 a2 + a3 an + a1 a1 + a2 + a3 an + a1 + a2 √ √ · ··· ≤ ··· . 2 2 2 2 2 2 2 (Saint Petersburg 2000) 00.25. Let n ≥ 3 be an integer. Prove that for positive numbers x1 ≤ x2 ≤ · · · ≤ xn , xn x1 x1 x2 xn−1 xn + + ··· + ≥ x1 + x2 + · · · + xn . x2 x3 x1 (Saint Petersburg 2000)
3 00.26. Let a, b, c, d be positive real numbers such that c2 + d2 = a2 + b2 . Prove that a3 b3 + ≥ 1. c d (Singapore 2000) 00.27. Given that x, y, z are positive real numbers satisfying xyz = 32, find the minimum value of x2 + 4xy + 4y 2 + 2z 2 . (United Kingdom 2000) 00.28. Prove that for any nonnegative real numbers a, b, c, the following inequality holds   √ 2 √ √ √ 2 √ √
2 a+b+c √ 3 − abc ≤ max a− b , b− c , c− a . 3 (USA 2000) 01.1. Let a, b, c ≥ 0 such that a + b + c ≥ abc. Prove that the following inequality holds √ a2 + b2 + c2 ≥ 3abc. (Balkan 2001) 200

01.2. Let x1 , x2 , x3 be real numbers in [−1, 1], and let y1 , y2 , y3 be real numbers in [0, 1). Find the maximum possible value of the expression 1 − x1 1 − x2 1 − x3 · · . 1 − x2 y3 1 − x3 y1 1 − x1 y2 (Belarus 2001) 01.3. Let x and y be any two real numbers. Prove that 3(x + y + 1)2 + 1 ≥ 3xy. Under what conditions does equality hold? (Colombia 2001) 01.4. Let n (n ≥ 2) be an integer and let a1 , a2 , . . . , an be positive real numbers. Prove the inequality (a31 + 1)(a32 + 1) · · · (a3n + 1) ≥ (a21 a2 + 1) · · · (a2n a1 + 1). (Czech-Slovak-Polish 2001) 01.5. Prove that for any positive real numbers a, b, c, the following inequality holds a b c √ +√ +√ ≥ 1. a2 + 8bc b2 + 8ca c2 + 8ab (IMO 2001) 01.6. Let a, b, c be positive real numbers such that abc = 1. Prove that ab+c bc+a ca+b ≤ 1. (India 2001) 01.7. Let x, y, z be positive real numbers such that xyz ≥ xy + yz + zx. Prove that xyz ≥ 3 (x + y + z) . (India 2001) 01.8. Prove that for any real numbers a, b, c the following inequality holds (b + c − a)2 (c + a − b)2 (a + b − c)2 ≥ (b2 + c2 − a2 )(c2 + a2 − b2 )(a2 + b2 − c2 ). (Japan 2001) 01.9. Let a, b, c be the sidelengths of an acute-angled triangle. Prove that (a + b + c)(a2 + b2 + c2 )(a3 + b3 + c3 ) ≥ 4(a6 + b6 + c6 ). (Japan 2001) 01.10. Prove that for any real numbers x1 , x2 , . . . , xn , y1 , y2 , . . . , yn such that x21 + x22 + · · · + x2n = y12 + y22 + · · · + yn2 = 1, ! n X 2 (x1 y2 − x2 y1 ) ≤ 2 1 − xk yk . k=1

(Korea 2001) 201

01.11. Prove that if a, b, c are positive real numbers, then p p p p a4 + b4 + c4 + a2 b2 + b2 c2 + c2 a2 ≥ a3 b + b3 c + c3 a + ab3 + bc3 + ca3 . (Korea 2001) 01.12. Prove that for all a, b, c > 0, p p (a2 b + b2 c + c2 a) (ab2 + bc2 + ca2 ) ≥ abc+ 3 (a3 + abc) (b3 + abc) (c3 + abc). (Korea 2001) 01.13. Prove that if a, b, c > 0 have product 1, then (a + b)(b + c)(c + a) ≥ 4(a + b + c − 1). (MOSP 2001) 01.14. Show that the inequality n X i=1

  X n n ixi ≤ + xii 2 i=1

holds for every integer n ≥ 2 and all real numbers x1 , x2 , . . . , xN ≥ 0. (Poland 2001) 01.15. Let a and b be positive real numbers in the interval (0, 1]. Prove that √

1 2 1 +√ ≤√ . 2 2 1 + ab 1+a 1+b

(Russia 2001) 01.16. Let a, b, c, x, y, z be positive real numbers such that x + y + z = 1. Prove that p ax + by + cz + 2 (xy + yz + zx)(ab + bc + ca) ≤ a + b + c. (Ukraine 2001) 01.17. Let a, b, c be positive real numbers such that a + b + c ≥ abc. Prove that at least two of the inequalities 2 3 6 + + ≥ 6, a b c

2 3 6 + + ≥ 6, b c a

2 3 6 + + ≥6 c a b

are true. (USA 2001) 01.18. Prove that for any nonnegative real numbers a, b, c such that a2 + b2 + c2 + abc = 4, we have 0 ≤ ab + bc + ca − abc ≤ 2. (USA 2001) 01.19. Let x, y, z be positive real numbers satisfying  √ √ 1 1 (i) √ ≤ z ≤ min x 2, y 3 ; 2 √ 2√ (ii) x + ≥ 6; √ √z 3 √ (iii) y 3 + z 10 ≥ 2 5. 202

Find the maximum of P (x, y, z) =

1 2 3 + 2 + 2. 2 x y z (Vietnam 2001)

01.20. Let x, y, z be positive real numbers such that 2 (i) ≤ z ≤ min {x, y} ; 5 4 (ii) xz ≥ ; 15 1 (iii) yz ≥ . 5 Determine the maximum possible value of P (x, y, z) =

1 2 3 + + . x y z

(Vietnam 2001) 1 2 3 01.21. Find the minimum value of the expression + + where a, b, c are a b c positive real numbers such that 21ab + 2bc + 8ca ≤ 12. (Vietnam 2001) 02.1. Let x, y, z be positive real numbers such that 1 1 1 + + = 1. x y z Prove that √

x + yz +

√

y + zx +

√

z + xy ≥

√

xyz +

√

x+

√

y+

√

z.

(APMO 2002) 02.2. Let a, b, c be positive real numbers. Prove that a3 b3 c3 a2 b2 c2 + + ≥ + + . b2 c2 a2 b c a (Balkan (Shortslit) 2002) 02.3. If a, b, c are positive real numbers such that abc = 2, then √ √ √ a3 + b3 + c3 ≥ a b + c + b c + a + c a + b. (Balkan (Shortlist) 2002) 02.4. Let a, b, c be real numbers such that a2 +b2 +c2 = 1. Prove the inequality a2 b2 c2 3 + + ≥ . 1 + 2bc 1 + 2ca 1 + 2ab 5 (Bosnia and Herzegovina 2002) 02.5. Show that for any positive real numbers a, b, c, we have a3 b3 c3 + + ≥ a + b + c. bc ca ab (Canada 2002) 203

02.6. Assume (P1 , P2 , . . . , Pn ) (n ≥ 2) is an arbitrary permutation of (1, 2, . . . , n). Prove that 1 1 1 n−1 + + ... + > . P1 + P2 P2 + P3 Pn−1 + Pn n+2 (China 2002) 02.7. Let x, y be positive real numbers such that x + y = 2. Prove that x3 y 3 (x3 + y 3 ) ≤ 2. (India 2002) 02.8. For any positive real numbers a, b, c, show that the following inequality holds a b c c+a a+b b+c + + ≥ + + . b c a c+b a+c b+a (India 2002) 02.9. Let x1 , x2 , . . . , xn be positive real numbers. Prove that √ x1 x2 xn + + ··· + < n. 2 2 2 2 2 2 1 + x1 1 + x1 + x2 1 + x1 + x2 + · · · + xn (India 2002) 02.10. Let a, b, c, d be the positive real numbers such that 1 1 1 1 + + + = 1. 4 4 4 1+a 1+b 1+c 1 + d4 Prove that abcd ≥ 3. (Latvia 2002) 02.11. Prove that for any positive real numbers a, b, c, we have b c a + + ≤ 1. 2a + b 2b + c 2c + a 02.12. Positive numbers α, β, x1 , x2 , . . . , xn x1 + x2 + · · · + xn = 1. Prove that

(Moldova 2002) (n ≥ 1) satisfy the condition

x31 x32 x3n 1 + + ··· + ≥ . αx1 + βx2 αx2 + βx3 αxn + βx1 n(α + β) (Moldova 2002) 02.13. Let a, b, c be positive real numbers. Prove that 

2a b+c

2



3

+

2b c+a

2



3

+

2c a+b

2 3

≥ 3. (MOSP 2002)

02.14. If a, b, c ∈ (0, 1), prove that p √ abc + (1 − a)(1 − b)(1 − c) < 1. (Romania 2002) 204

02.15. Given positive real numbers a, b, c and x, y, z, for which a+x = b+y = c + z = 1. Prove that   1 1 1 ≥ 3. (abc + xyz) + + ay bz cx (Russia 2002) 02.16. Let x, y, z be positive real numbers with sum 3. Prove that √ √ √ x + y + z ≥ xy + yz + zx. (Russia 2002) 02.17. Let a1 , a2 , . . . , an and b1 , b2 , . . . , bn be real numbers between 1001 and 2002 inclusive. Suppose a21 + · · · + a2n = b21 + · · · + b2n . Prove that n X a3

n

17 X 2 ≤ ai . bi 10 i

i=1

i=1

Determine when equality holds. (Singapore   2002) 1 02.18. Let a, b, c, d be real numbers contained in the interval 0, . Prove 2 that a4 + b4 + c4 + d4 (1 − a)4 + (1 − b)4 + (1 − c)4 + (1 − d)4 ≥ . abcd (1 − a)(1 − b)(1 − c)(1 − d) 02.19. Let x, y, z be positive real numbers such that that 1 x2 yz + y 2 zx + z 2 xy ≤ . 3 a2

02.20. Let a, b, c be real numbers satisfying inequality 2(a + b + c) − abc ≤ 10.

x2

+

y2

(Taiwan 2002) + z 2 = 1. Prove

(United Kingdom 2002) + b2 + c2 = 9. Prove the

(Vietnam 2002) 03.1. If a, b, c > −1, then 1 + a2 1 + b2 1 + c2 + + ≥ 2. 2 2 1+b+c 1+c+a 1 + a + b2 (Laurentiu Panaitopol, Balkan 2003) 03.2. Prove that if a, b and c are positive real numbers with sum 3, then a b c 3 + + ≥ . b2 + 1 c2 + 1 a2 + 1 2 (Bulgaria 2003) 03.3. Let a, b, c, d be positive real numbers such that ab + cd = 1 and let x1 , x2 , x3 , x4 , y1 , y2 , y3 , y4 be real numbers such that x21 + y12 = x22 + y22 = x23 + y32 = x24 + y42 = 1. Prove that the following inequality holds  2  a + b2 c2 + d2 2 2 (ay1 + by2 + cy3 + dy4 ) + (ax4 + bx3 + cx2 + dx1 ) ≤ 2 + . ab cd 205

(China 2003) 03.4. Let a1 , a2 , . . . , a2n be real numbers such that 2n−1 X

(ai − ai+1 )2 = 1.

i=1

Determine the maximum value of (an+1 + an+2 + · · · + a2n ) − (a1 + a2 + . . . + an ). (China 2003)  3 , 5 . Prove that 03.5. Suppose x be a real number in the interval 2 √ √ √ √ 2 x + 1 + 2x − 3 + 15 − 3x < 2 19. 

(China 2003) 03.6. Let x1 , x2 , . . . , x5 be nonnegative real numbers such that 5 X i=1

Prove that

1 = 1. 1 + xi

5 X

xi 2 + 4 ≤ 1. x i=1 i

(China 2003) 03.7. Find the greatest real number k such that, for any positive a, b, c with a2 > bc, (a2 − bc)2 > k(b2 − ca)(c2 − ab). (Japan 2003) 03.8. Prove that in any acute triangle ABC, cot3 A + cot3 B + cot3 C + 6 cot A cot B cot C ≥ cot A + cot B + cot C. (MOSP 2003) 03.9. Let ai be positive real numbers, for all i = 1, 2, . . . , n, satisfying a1 + a2 + · · · + an =

1 1 1 + + ··· + . a1 a2 an

Prove that 1 1 1 + + ··· + ≤ 1. n − 1 + a1 n − 1 + a2 n − 1 + an (Vasile Cirtoaje, MOSP 2003) 03.10. Let a, b, c be nonnegative real numbers satisfying a2 + b2 + c2 = 1. Prove that √ b c a 1≤ + + ≤ 2. 1 + bc 1 + ca 1 + ab (Faruk Zejnulahi, MOSP 2003) 206

03.11. Let a, b, c be positive real numbers so that abc = 1. Prove that 1+

3 6 ≥ . a+b+c ab + bc + ca (Romania 2003)

03.12. Let a, b, c be positive real numbers. Prove that (2a + b + c)2 (2b + c + a)2 (2c + a + b)2 + + ≤ 8. 2a2 + (b + c)2 2b2 + (c + a)2 2c2 + (a + b)2 (USA 2003) π , the following inequality holds 03.13. Prove that for any a, b, c ∈ 0, 2 

X sin a · sin(a − b) · sin(a − c) sin(b + c)

≥ 0. (USA 2003)

04.1. Prove that (a2 + 2)(b2 + 2)(c2 + 2) ≥ 9(ab + bc + ca) for any positive real numbers a, b, c. (APMO 2004) 04.2. Let x, y, z, t be positive real numbers such that xyzt = 1. Prove that 1 1 1 1 + + + ≥ 1. 2 2 2 (1 + x) (1 + y) (1 + z) (1 + t)2 (China 2004) 04.3. If a, b, c are positive real numbers, then √ a b c 3 2 1< √ . +√ +√ ≤ 2 a2 + b2 b2 + c2 c2 + a2 (China 2004) 04.4. Let n be a positive integer with n greater than one, and let a1 , a2 , . . . , an be positive integers such that a1 < a2 < · · · < an and 1 1 1 + + ··· + ≤ 1. a1 a2 an Prove that, for any real number x, the following inequality holds  2 1 1 1 1 1 + 2 + ··· + 2 ≤ · . 2 2 2 2 an + x 2 a1 (a1 − 1) + x2 a1 + x a2 + x (China 2004) 04.5. Find all real numbers k such that the following inequality a3 + b3 + c3 + d3 + 1 ≥ k(a + b + c + d) holds for all real numbers a, b, c, d ≥ −1. 207

(China 2004) 04.6. Determine the maximum constant λ such that x + y + z ≥ λ, √ √ √ where x, y, z are positive real numbers with x yz + y zx + z xy ≥ 1. (China 2004) 04.7. Let a, b, c be positive real numbers. Determine the minimal value of the following expression a + 3c 4b 8c + − . a + 2b + c a + b + 2c a + b + 3c (China 2004) 04.8. Determine the largest constant M such that the following inequality holds for all real numbers x, y, z, x4 + y 4 + z 4 + xyz (x + y + z) ≥ M (xy + yz + zx)2 . (Hellenic 2004) 04.9. Let n ≥ 3 be an integer and let t1 , t2 , . . . , tn be positive real numbers such that   1 1 1 2 n + 1 > (t1 + t2 + · · · + tn ) + + ··· + . t1 t2 tn Show that ti , tj , tk are the side lengths of a triangle for all i, j, k with 1 ≤ i < j < k ≤ n. (IMO 2004) 04.10. For a, b, c positive reals, find the minimum value of b2 + c2 c2 + a2 a2 + b2 + + . a2 + bc b2 + ca c2 + ab   (India 2004) 1 04.11. Let x1 , x2 , . . . , xn be real numbers in the interval 0, . Prove that 2 n Y xi i=1 n X

xi

i=1

!n ≤

n Y

" i=1 n X

(1 − xi ) #n .

(1 − xi )

i=1

(India 2004) 04.12. Prove that for all positive real numbers a, b, the following inequality holds p p 2a(a + b)3 + b 2(a2 + b2 ) ≤ 3(a2 + b2 ). (Ireland 2004) 04.13. If a, b, c are positive reals such that a + b + c = 1, prove that   b c a 1+a 1+b 1+c + + ≤2 + + . 1−a 1−b 1−c a b c 208

(Japan 2004) 04.14. Let a, b be real numbers in the interval [0, 1]. Prove that √

a b 2 +√ ≤√ . 2 2 7 2b + 5 2a + 5

(Lithuania 2004) 04.15. Prove that for any positive real numbers a, b, c, 3 a − b3 b3 − c3 c3 − a3 (a − b)2 + (b − c)2 + (c − a)2 . a+b + b+c + c+a ≤ 4 (Moldova 2004) 04.16. Prove that if n > 3 and x1 , x2 , . . . , xn > 0 have product 1, then 1 1 1 + + ··· + > 1. 1 + x1 + x1 x2 1 + x2 + x2 x3 1 + xn + xn x1 (Russia 2004) 04.17. (a) Given real numbers a, b, c with a + b + c = 0, prove that a3 + b3 + c3 > 0 if and only if a5 + b5 + c5 > 0. (b) Given real numbers a, b, c, d with a + b + c + d = 0, prove that a3 + b3 + c3 + d3 > 0

if and only if a5 + b5 + c5 + d5 > 0.

(United Kingdom 2004) 04.18. Let a, b, c be positive real numbers. Prove that (a5 − a2 + 3)(b5 − b2 + 3)(c5 − c2 + 3) ≥ (a + b + c)3 . (USA 2004) 04.19. Let x, y, z > 0 such that (x + y + z)3 = 32xyz. Find the minimum and x4 + y 4 + z 4 maximum of . (x + y + z)4 (Vietnam 2004) 05.1. For any positive real numbers a, b, c such that abc = 8, then a2 b2 c2 4 p +p +p ≥ . 3 (a3 + 1) (b3 + 1) (b3 + 1) (c3 + 1) (c3 + 1) (a3 + 1) (APMO 2005) 05.2. Let a, b, c, d be positive real numbers. Show that 1 1 1 1 a+b+c+d + + + ≥ . a3 b3 c3 d3 abcd (Austria 2005) 05.3. Let a, b, c be positive real numbers. Prove that 4(a − b)2 a2 b2 c2 + + ≥a+b+c+ . b c a a+b+c (Balkan 2005) 209

05.4. If a, b, c are positive real numbers such that abc = 1, then the inequality holds a b c + + ≤ 1. a2 + 2 b2 + 2 c2 + 2 (Baltic Way 2005) 05.5. Let a, b, c be positive real numbers. Prove that       3 1 1 3 2 2 b +a+ ≥ 2a + 2b + . a +b+ 4 4 2 2 (Belarus 2005) 05.6. Given positive real numbers a, b, c such that a + b + c = 1. Prove that √ √ √ 1 a b+b c+c a≤ √ . 3 (Bosnia and Hercegovina 2005) 05.7. Let x, y be positive real numbers such that x3 + y 3 = x − y. Prove that x2 + 4y 2 < 1. (China 2005) 1 05.8. Let a, b, c be positive real numbers such that ab + bc + ca = . Prove 3 that 1 1 1 + 2 + 2 ≤ 3. 2 a − bc + 1 b − ca + 1 c − ab + 1 (China 2005) 05.9. Given positive real numbers a, b, c such that a + b + c = 1. Prove that the following inequality holds 10(a3 + b3 + c3 ) − 9(a5 + b5 + c5 ) ≥ 1. (China 2005) 05.10. Let ABC be an acute triangle. Determine the least value of the following expression P =

cos2 B cos2 C cos2 A + + . cos A + 1 cos B + 1 cos C + 1

(China 2005) 1 1 1 05.11. Let a, b, c be positive real numbers such that + + = 1. Prove a b c that (a − 1) (b − 1) (c − 1) ≥ 8. (Croatia 2005) 05.12. Let a, b, c > 0 such that abc = 1. Prove that a a a 3 + + ≥ . (a + 1) (b + 1) (b + 1) (c + 1) (c + 1) (a + 1) 4 (Czech-Slovak 2005) 210

05.13. If a, b, c are three positive real numbers such that ab + bc + ca = 1, prove that r r r 1 3 1 3 1 3 1 + 6b + + 6c + + 6a ≤ . a b c abc (Germany 2005) 05.14. Given positive real numbers x, y, z such that xyz ≥ 1. Prove that x5 − x2 y5 − y2 z5 − z2 + + ≥ 0. x5 + y 2 + z 2 y 5 + z 2 + x2 z 5 + x2 + y 2 (IMO 2005) 05.15. Let a1 ≤ a2 ≤ · · · ≤ an be positive real numbers such that a21 + a22 + · · · + a2n = 1, n

a1 + a2 + · · · + an = m, n

where 1 ≥ m > 0. Prove that for all i satisfying ai ≤ m, we have n − i ≥ n(m − ai )2 . (Iran 2005) 05.16. If three nonnegative real numbers a, b, c satisfy the condition a2

1 1 1 + 2 + 2 = 2, +1 b +1 c +1

prove that 3 ab + bc + ca ≤ . 2 (Iran 2005) 05.17. If x, y, z are real numbers satisfying xyz = −1, prove that x4 + y 4 + z 4 + 3(x + y + z) ≥

y 2 + z 2 z 2 + x2 x2 + y 2 + + . x y z

(Iran 2005) 05.18. Let a, b, c be positive real numbers such that a + b + c = 1. Prove that √ √ √ 3 3 a 1 + b − c + b 3 1 + c − a + c 1 + a − b ≤ 1.

05.19. For any real numbers x1 , x2 , . . . , xn with that

x21

+ x22

(Japan 2005) + · · · + x2n = 1, prove

x1 x2 xn + + ··· + < 2 2 2 2 1 + x1 1 + x1 + x2 1 + x1 + x22 + · · · + x2n

r

n . 2

(Korea 2005) 05.20. Given positive real numbers a, b, c such that a4 + b4 + c4 = 3. Prove that 1 1 1 + + ≤ 1. 4 − ab 4 − bc 4 − ca (Moldova 2005) 211

05.21. Let a, b, c ∈ [0, 1]. Prove that a b c + + ≤ 2. bc + 1 ca + 1 ab + 1 (Poland 2005) 05.22. Let a, b, c be positive real numbers such that a + b + c = 1. Prove that r p p p 2 ab (1 − c) + bc (1 − a) + ca (1 − b) ≤ . 3 (Republic of Srpska 2005) 05.23. Let a, b, c be positive real numbers such that abc ≥ 1. Prove that 1 1 1 + + ≤ 1. 1+a+b 1+b+c 1+c+a (Romania 2005) 05.24. Let n is a positive integer. Prove that if x is a positive real numbers, then (2x)n . 1 + xn+1 ≥ (1 + x)n−1 05.25. Given positive real numbers x, y, z such that the following inequality

x2

+

y2

(Russia 2005) + z 2 = 1. Prove

x y z + + > 3. x3 + yz y 3 + zx z 3 + xy (Russia 2005) 05.26. Let a, b, c be positive real numbers. Prove that r a 3 b c √ (a + b + c). +√ +√ ≥ 2 c + a b+c a+b (Serbia and Montenegro 2005) 05.27. If a, b, c are positive real numbers such that ab + bc + ca = 1. Prove that r √ 3 1 3 3 3 + 6 (a + b + c) ≤ . abc abc (Slovenia 2005) 05.28. Let a1 , a2 , . . . , a95 be positive real numbers. Prove that 95 95 X Y ak ≤ 94 + max {1, ak } . k=1

k=1

(Taiwan 2005) 05.29. Let a, b, c be positive real numbers. Prove that 

a b c + + b c a

2

 ≥ (a + b + c) 212

1 1 1 + + a b c

 .

(United Kingdom 2005) 05.30. Let a, b, c be positive real numbers. Prove that 

a a+b

3

 +

b b+c

3

 +

c c+a

3

3 ≥ . 8

(Vietnam 2005) 06.1. Let x1 , x2 , . . . , xn be positive real numbers such that x1 +x2 +· · ·+xn = 1. Prove that ! n ! n X X √ 1 n2 √ xi ≤√ . 1 + xi n+1 i=1 i=1 (China 2006) 06.2. Let a, b, c be positive real numbers satisfying a + b + c = 1. Prove that √ bc ca 2 ab √ +√ +√ ≤ . 2 ab + bc bc + ca ca + ab (China 2006) 06.3. Suppose that a1 , a2 , . . . , an are real numbers with sum 0. Prove that the following inequality holds n−1

max a2i 1≤i≤n

nX ≤ (ai − ai+1 )2 . 3 i=1

(China 2006) 06.4. Let a, b, c be the sidelengths of a triangle. Prove that √ √ √ b+c−a c+a−b a+b−c √ √ +√ √ √ √ +√ √ √ ≤ 3. b+ c− a c+ a− b a+ b− c (IMO Shortlist 2006) 06.5. Determine the least real number M such that the inequality ab(a2 − b2 ) + bc(b2 − c2 ) + ca(c2 − a2 ) ≤ M (a2 + b2 + c2 )2 holds for all real numbers a, b, c. (IMO 2006) 06.6. Let x1 , x2 , x3 , y1 , y2 , y3 , z1 , z2 , z3 be positive real numbers. Find the maximum value of real number A if M = (x31 + x32 + x33 + 1)(y13 + y23 + y33 + 1)(z13 + z23 + z33 + 1), and N = A(x1 + y1 + z1 )(x2 + y2 + z2 )(x3 + y3 + z3 ), then M ≥ N always holds, and find the condition that the equality holds. (Japan 2006) 06.7. Let a, b, c, d be real numbers with sum 0. Prove the inequality (ab + ac + ad + bc + bd + cd)2 + 12 ≥ 6(abc + abd + acd + bcd). 213

(Kazakhstan 2006) 06.8. Let a, b, c be sides of the triangle. Prove that   c  a  b a2 − 1 + b2 − 1 + c2 − 1 ≥ 0. c a b (Moldova 2006) 06.9. Let a, b, c be positive real numbers with ab + bc + ca = abc. Prove that a4 + b4 b4 + c4 c4 + a4 + + ≥ 1. ab(a3 + b3 ) bc(b3 + c3 ) ca(c3 + a3 ) (Poland 2006) 06.10. Find the maximum value of (x3 + 1)(y 3 + 1), for all real numbers x, y, satisfying the condition that x + y = 1. (Romania 2006) 06.11. Let a, b, c be three positive real numbers with sum 3. Prove that 1 1 1 + + ≥ a2 + b2 + c2 . a2 b2 c2 (Romania   2006) 1 06.12. Consider real numbers a, b, c contained in the interval , 1 . Prove 2 that a+b b+c c+a 2≤ + + ≤ 3. 1+c 1+a 1+b (Romania 2006) 06.13. Let a, b be positive real numbers. Determine the largest constant M such that for all k ∈ [0, π] , we have 1 1 M + ≥ . ka + b kb + a a+b (Thailand 2006) 06.14. If x, y, z are positive numbers satisfying the condition xy +yz +zx = 1, show that √
2 √ √ √ 27 (x + y)(y + z)(z + x) ≥ x + y + y + z + z + x ≥ 6 3. 4 (Turkey 2006) 06.15. Let a, b, c be real numbers. Prove that the following inequality holds Xp (a2 − ab + b2 )(b2 − bc + c2 ) ≥ a2 + b2 + c2 .

07.1. Let x, y, z be positive real numbers such that that the following inequality holds

√

(VMEO 2006) √ √ x + y + z = 1. Prove

x2 + yz y 2 + zx z 2 + xy p +p +p ≥ 1. 2x2 (y + z) 2y 2 (z + x) 2z 2 (x + y) 214

(APMO 2007) 07.2. If a, b, c ∈ R such that abc = 1, then     1 1 1 b+c c+a a+b 1 1 1 2 2 2 ≥ 6+2 . + + a +b +c + 2 + 2 + 2 +2 a + b + c + + + a b c a b c a b c (Brazil 2007) 07.3. Given an integer n ≥ 2, find the largest constant C(n) for which the inequality n X i=1

xi ≥ C(n)

X

2xi xj +

√

xi xj




1≤j 0 such that a + b + c = 1. Prove that a2 b2 c2 + + ≥ 3(a2 + b2 + c2 ). b c a (Croatia 2007) 07.8. Let a, b, c, d be positive real numbers such that a + b + c + d = 1. Prove that 1 6(a3 + b3 + c3 + d3 ) ≥ a2 + b2 + c2 + d2 + . 8 (France 2007) 07.9. Let a, b, c be sides of a triangle, show that (c + a − b)4 (a + b − c)4 b(b + c − a) + + ≥ ab + bc + ca. a(a + b − c) b(b + c − a) c(c + a − b) (Greece 2007) 215

07.10. Let a, b, c, d be real numbers such that a2 ≤ 1,

a2 + b2 ≤ 5,

a2 + b2 + c2 ≤ 14,

a2 + b2 + c2 + d2 ≤ 30.

Prove that a + b + c + d ≤ 10. 07.11. Let a1 , a2 , . . . , a100 · · · + a2100 = 1. Prove that

(Hungary-Isarel 2007) be nonnegative eral numbers such that a21 + a22 +

a21 a2 + a22 a3 + · · · + a2100 a1 <

12 . 25

(IMO Shortlist 2007) 07.12. Let n be a positive integer, and let x and y be positive real numbers such that xn + y n = 1. Prove that ! n ! n X 1 + y 2k X 1 + x2k 1 < . 4k 4k (1 − x)(1 − y) 1+x 1+y k=1

k=1

(IMO Shortlist 2007) 07.13. Real numbers a1 , a2 , . . . , an are given. For each i (1 ≤ i ≤ n) define di = max {aj : 1 ≤ j ≤ i} − min {aj : i ≤ j ≤ n}, and let d = max {di : 1 ≤ i ≤ n}. (a) Prove that for any real numbers x1 ≤ x2 ≤ · · · ≤ xn , we have max {|xi − ai | : 1 ≤ i ≤ n} ≥

d . 2

(b) Show that there are real numbers x1 ≤ x2 ≤ · · · ≤ xn such that we have equality in (a). (IMO 2007) 07.14. If x, y, z are positive real numbers, prove that the following inequality holds (x + y + z)2 (yz + zx + xy)2 ≤ 3(y 2 + yz + z 2 )(z 2 + zx + x2 )(x2 + xy + y 2 ). (India 2007) 07.15. Let a, b, c be distinct positive real numbers. Show that a + b b + c c + a a − b + b − c + c − a > 1. (Iran 2007) 07.16. Find the largest constant T such that for all nonnegative real numbers a, b, c, d, e satisfying a + b = c + d + e, we have √ p √ √ √ √ 2 a2 + b2 + c2 + d2 + e2 ≥ T a+ b+ c+ d+ e . 216

(Iran 2007) 07.17. Prove that for any positive real numbers a, b, c, we have r   a+b+c a2 + b2 + c2 1 bc ca ab . ≤ ≤ + + 3 3 3 a b c (Ireland 2007) 07.18. Let n ≥ 2 be a given integer. Determine (a) the largest real cn such that 1 1 1 + + ··· + ≥ cn 1 + a1 1 + a2 1 + an holds for any positive numbers a1 , a2 , . . . , an with a1 a2 · · · an = 1. (b) the largest real dn such that 1 1 1 + + ··· + ≥ dn 1 + 2a1 1 + 2a2 1 + 2an holds for any positive numbers a1 , a2 , . . . , an with a1 a2 · · · an = 1. (Italy 2007) 07.19. If a, b are positive real numbers such that ab ≥ 1, then 1 2 1 + ≥ . 2 2 (2a + 3) (2b + 3) 5(2ab + 3) (Kiev 2007) 07.20. For all positive real numbers a, b, c, find all values of positive number k such that the following inequality holds a b c 1 + + ≥ . c + kb a + kc b + ka 2007 (Korea 2007) 07.21. Let a, b, c be positive real numbers. Prove that 1+

3 6 ≥ . ab + bc + ca a+b+c

(Macedonia 2007) 07.22. Let a, b, c, d be nonnegative real numbers such that a + b + c + d = 4. Prove that a2 bc + b2 cd + c2 da + d2 ab ≤ 4. (Middle 2007)  Europe  1 07.23. Let a, b, c, d be positive real numbers in the interval , 2 and abcd = 2 1. Find the maximum value of      1 1 1 1 a+ b+ c+ d+ . b c d a (Middle Europe 2007) 217

1 07.24. Let a1 , a2 , . . . , an be positive real numbers such that ai ≥ for all i i = 1, 2, . . . , n. Prove the inequality     1 1 2n (a1 + 1) a2 + · · · an + ≥ (1 + a1 + 2a2 + · · · + nan ). 2 n (n + 1)! (Moldova 2007) 07.25. Let a1 , a2 , . . . , an ∈ [0, 1]. If S = a31 + a32 + · · · + a3n , then prove that a1 a2 an 1 + + ··· + ≤ . 3 3 3 2n + 1 + S − an 3 2n + 1 + S − a1 2n + 1 + S − a2 (Moldova 2007) 07.26. Let a, b, c be positive real numbers such that a+b+c≥ Prove that a+b+c≥

1 1 1 + + . a b c

3 2 + . a + b + c abc

(Peru 2007) 07.27. a, b, c, d are positive real numbers satisfying the following condition 1 1 1 1 + + + = 4. a b c d Prove that r r r r 3 3 3 3 3 3 3 3 3 b + c 3 c + d 3 d + a 3 a + b + + + ≤ 2(a + b + c + d) − 4. 2 2 2 2 (Poland 2007) 07.28. Let x, y, z be nonnegative real numbers. Prove that the following inequality holds x3 + y 3 + z 3 3 ≥ xyz + |(x − y)(y − z)(z − x)| . 3 4 (Romania 2007) 07.29. For n ∈ N, n ≥ 2, determine max

n Y

(1 − xi ),

for xi ∈ R+ ,

1 ≤ i ≤ n,

i=1

n X

x2i = 1.

i=1

(Romania 2007) 07.30. Let a, b, c be positive real numbers such that 1 1 1 + + ≥ 1. a+b+1 b+c+1 c+a+1 Show that a + b + c ≥ ab + bc + ca. 218

(Romania 2007) 07.31. For n ∈ N, n ≥ 2, ai , bi ∈ R, 1 ≤ i ≤ n, such that n X

a2i = 1,

n X

i=1

b2i = 1,

and

i=1

n X

ai bi = 0,

i=1

prove that n X i=1

!2 ai

+

n X

!2 bi

≤ n.

i=1

(C. Lupu and T. Lupu, Romania 2007) 07.32. Positive real numbers a, b, c satisfy a + b + c = 1. Show that 1 1 1 1 + + ≥ . 2 2 2 ab + 2c + 2c bc + 2a + 2a ca + 2b + 2b ab + bc + ca (Turkey 2007) 07.33. Let a, b, c be positive real numbers such that abc ≥ 1. Prove that     1 1 27 1 b+ c+ ≥ , a+ a+1 b+1 c+1 8 and 27(a3 +a2 +a+1)(b3 +b2 +b+1)(c3 +c2 +c+1) ≥ 64(a2 +a+1)(b2 +b+1)(c2 +c+1). (Ukraine 2007) 07.34. Show that for any real numbers a, b, c, then (a2 + b2 )2 ≥ (a + b + c)(b + c − a)(c + a − b)(a + b − c). (United Kingdom 2007) 07.35. Given a triangle ABC. Determine the minimum value of A B B C C A cos2 cos2 cos2 cos2 cos2 2 2 + 2 2 + 2 2. C A B cos2 cos2 cos2 2 2 2

cos2

(Vietnam 2007) 08.1. Let a, b, c be real numbers such that a2 + b2 + c2 = 3. Prove that a2 b2 c2 (a + b + c)2 + + ≥ . 2 + b + c2 2 + c + a2 2 + a + b2 12 (Baltic Way 2008) 08.2. Suppose that a, b, c are positive real numbers with a2 + b2 + c2 = 1. Prove that a5 + b5 b5 + c5 c5 + a5 + + ≥ 3(ab + bc + ca) − 2. ab(a + b) bc(b + c) ca(a + b) (Bosnia 2008) 219

08.3. Let x, y, z be real numbers. Show that the following inequality holds 2

2



2

x + y + z − xy − yz − zx ≥ max

3(x − y)2 3(y − z)2 3(z − x)2 , , 4 4 4

 .

(Bosnia 2008) 08.4. Let a, b, c be positive real numbers. Prove that     4a 4b 4c 1+ 1+ 1+ > 25. b+c a+c a+b (Bosnia 2008) 08.5. Let x, y, z be real numbers such that x+y +z = xy +yz +zx. Determine the least value of the following expression P =

x2

x y z + 2 + 2 . +1 y +1 z +1

(Brazil 2008) 08.6. Let a, b, c be positive real numbers such that a + b + c = 1. Prove that 3 a − bc b − ca c − ab + + ≤ . a + bc b + ca c + ab 2 (Canada 2008) 08.7. Let a, b, c be nonnegative real numbers such that a + b + c = 1. Prove that r √ √ (b − c)2 √ a+ + b + c ≤ 3. 4 (China 2008) 08.8. Let x, y, z be positive numbers. Find the minimal value of x2 + y 2 + z 2 ; (a) xy + yz x2 + y 2 + 2z 2 (b) . xy + yz (Croatia 2008) 08.9. Determine the smallest constant C such that the following inequality 1 + (x + y)2 ≤ C(1 + x2 )(1 + y 2 ) holds for any real numbers x, y. (Germany 2008) 08.10. Prove that if a1 , a2 , . . . , an are positive integers, then the following inequality  kn  2 a1 + a22 + · · · + a2n t ≥ a1 a2 · · · an a1 + a2 + · · · + an holds for k = max {a1 , a2 , . . . , an } and t = min {a1 , a2 , . . . , an } . When do we have the equality? (Greece 2008) 220

08.11. Let x, y, z be positive real numbers such that x2 + y 2 + z 2 = 3. Prove the inequality 3 1 + y 2 1 + z 2 1 + x2 < + + < 3. 2 2+x 2+y 2+z (Greece 2008) 08.12. Prove that for any positive real numbers a, b, c, d, we have the following inequality (a − b)(a − c) (b − c)(b − d) (c − d)(c − a) (d − a)(d − b) + + + ≥ 0. a+b+c b+c+d c+d+a d+a+b (Darij Grinberg, IMO Shortlist 2008) 08.13. (i) If x, y, z are three real numbers, all different from 1, such that xyz = 1, then prove that y2 z2 x2 + + ≥ 1. (x − 1)2 (y − 1)2 (z − 1)2 (ii) Prove that equality is achieved for infinitely many triples of rational numbers x, y, z. (IMO 2008) 08.14. Let n ≥ 3 is an integer and let x1 , x2 , . . . , xn be real numbers such that xi > 1 for all i. Prove the following inequality x1 x2 xn−1 xn xn x1 + ··· + + ≥ 4n. x3 − 1 x1 − 1 x2 − 1 (Indonesia 2008) 08.15. Let a, b, c be nonnegative real numbers, from which at least two are nonzero and satisfying the condition ab + bc + ca = 1. Prove that p p p √ a3 + a + b3 + b + c3 + c ≥ 2 a + b + c. (Iran 2008) 08.16. Let x, y, z be positive real numbers such that x + y + z = 3. Prove that x3 y3 z3 1 2 + + ≥ + (xy + xz + yz). y 3 + 8 z 3 + 8 x3 + 8 9 27 (Iran 2008) 08.17. Find the smallest real k such that for each x, y, z > 0, we have the inequality p √ √ √ x y + y z + z x ≤ k (x + y)(y + z)(z + x). 08.18. For any positive real numbers a, b, c, d such that prove the inequality

a2

+ b2

(Iran 2008) + c2 + d2 = 1,

a2 b2 cd + b2 c2 da + c2 d2 ab + d2 a2 bc + c2 a2 db + d2 b2 ac ≤

3 . 32

(Ireland 2008) 221

08.19. Let a, b, c be positive real numbers satisfying abc = 1. Prove that 1 1 1 3 + + ≥ . b(a + b) c(b + c) a(c + a) 2 (Kazakhstan 2008) 08.20. Let a, b, c be positive real numbers such that (a + b) (b + c) (c + a) = 8. Prove the inequality r 3 3 3 a+b+c 27 a + b + c ≥ . 3 3 (Macedonia 2008) 08.21. Let a, b, c, d be positive real numbers such that a ≥ b ≥ c ≥ d and abcd = 1. Prove that a3

1 1 1 3 + 3 + 3 ≥ . abc + 1 +1 b +1 c +1

(MathLinks Contest 2008) 08.22. Let a, b, c be nonnegative real numbers satisfying ab + bc + ca = 3. Prove that 1 1 1 3 + + ≤ . 1 + a2 (b + c) 1 + b2 (c + a) 1 + c2 (a + b) 1 + 2abc (MathLinks Contest 2008) 08.23. Determine the least value of the expression P = abc +

1 abc

3 where a, b, c are positive real numbers satisfying a + b + c ≤ . 2 (Moldova 2008) 08.24. Let a1 , a2 , . . . , an be positive real numbers such that a1 +a2 +· · ·+an ≤ n . Determine the smallest value of the following expression 2 s s s 1 1 1 A = a21 + 2 + a22 + 2 + · · · + a2n + 2 . a2 a3 a1 (Moldova 2008) 08.25. Find the maximum value of constant C such that the inequality x3 + y 3 + z 3 + C(xy 2 + yz 2 + zx2 ) ≥ (C + 1)(x2 y + y 2 z + z 2 x) holds for any nonnegative real numbers x, y, z. (Mongolia 2008) 08.26. If a, b, c are nonnegative real numbers, then the inequality holds  √ √ √ 4 a3 b3 + b3 c3 + c3 a3 ≤ 4c3 + (a + b)3 . (Poland 2008) 222

08.27. For real numbers xi > 1, 1 ≤ i ≤ n, n ≥ 2, such that n

X x2i ≥S= xj , xi − 1

for all i = 1, 2, . . . , n,

j=1

find, with proof, sup S. (Romania 2008) 08.28. Show that for all integers n ≥ 1, we have     1 1 1 1 1 ≥ (n + 1) . + + ··· + n 1 + + ··· + 2 n 2 3 n+1 (Romania 2008) 08.29. Let a, b, c be positive real numbers with ab + bc + ca = 3. Prove that 1 1+

a2 (b

+ c)

+

1 1+

b2 (c

+ a)

+

1 1+

c2 (a

+ b)

≤

1 . abc

(Romania 2008) 08.30. Determine the maximum value of real number k such that   1 1 1 (a + b + c) + + −k ≥k a+b b+c c+a for all real numbers a, b, c ≥ 0 with a + b + c = ab + bc + ca. (Romania 2008) 08.31. Let a, b ∈ [0, 1]. Prove the inequality 1 a + b ab ≤1− + . 1+a+b 2 3 (Romania 2008) 08.32. Let n ≥ 3 is an odd integer. Determine the maximum value of the cyclic sum, for 0 ≤ xi ≤ 1, i = 1, 2, . . . , n, p p p E = |x1 − x2 | + |x2 − x3 | + · · · + |xn − x1 |. (Romania 2008) 08.33. Let a, b, c be three positive real numbers satisfying abc = 8. Prove that a−2 b−2 c−2 + + ≤ 0. a+1 b+1 c+1 (Romania 2008) 08.34. Let a1 , a2 , . . . , an be positive real numbers satisfying the condition that a1 + a2 + · · · + an = 1. Prove that n X j=1

aj 1 0, it becomes x3 z + y 3 x + z 3 y ≥ x2 yz + xy 2 z + xyz 2

or

x2 y 2 z 2 + + ≥ x + y + z. y z x

However, an application of the Cauchy Schwarz Inequality gives  2  x y2 z2 (y + z + x) + + ≥ (x + y + z)2 . y z x Note that the equality holds if and only if a = b = c. Second solution: Without loss of generality, we may assume that b is the smallest number between a and c. Then we have two cases to consider: a ≥ c ≥ b and c ≥ a ≥ b. For the former case, we have a2 b(a − b) = b(a − b)(a − c)2 + bc(2a − c)(a − b) ≥ bc(2a − c)(a − b). Therefore, it suffices to prove that b(2a − c)(a − b) + b2 (b − c) + ca(c − a) ≥ 0, or (2b − c)a2 − (2b − c)(b + c)a + b3 ≥ 0. If c ≥ 2b then we are done because (2b − c)a2 − (2b − c)(b + c)a + b3 = a(c − 2b)(b + c − a) + b3 ≥ 0. Conversely, if 2b ≥ c, then we have (2b − c)a2 = (2b − c)(a − c)2 + c(2b − c)(2a − c) ≥ c(2b − c)(2a − c). 225

It now remains to check that c(2b − c)(2a − c) − (2b − c)(b + c)a + b3 ≥ 0. However, this is true because c(2b − c)(2a − c) − (2b − c)(b + c)a + b3 = (2b − c)(c − b)a + c3 − 2bc2 + b3 ≥ (2b − c)(c − b)c + c3 − 2bc2 + b3 = b(b − c)2 ≥ 0. For the latter case, we have a3 b + b3 c + c3 a − (ab3 + bc3 + ca3 ) = (a − b)(c − b)(c − a)(a + b + c) ≥ 0, and hence 2(a3 b+b3 c+c3 a) ≥ (a3 b+ab3 )+(b3 c+bc3 )+(c3 a+ca3 ) ≥ 2(a2 b2 +b2 c2 +c2 a2 ), from which we deduce that a2 b(a − b) + b2 c(b − c) + c2 a(c − a) ≥ 0, as desired. ?F? 83.2. Show that for any positive reals a, b, c, d, e, f , we have ab cd ef (a + c + e)(b + d + f ) + + ≤ . a+b c+d e+f a+b+c+d+e+f (United Kingdom 1983) First solution: The inequality is equivalent to 

or

a+b ab − 4 a+b



 +

   c+d cd e+f ef − + − ≥ 4 c+d 4 e+f (a + c + e)(b + d + f ) a+c+e+b+d+f ≥ − , 4 a+b+c+d+e+f

(a − b)2 (c − d)2 (e − f )2 (a + c + e − b − d − f )2 + + ≥ . a+b c+d e+f a+b+c+d+e+f

Now, using the Cauchy Schwarz Inequality, we get (a − b)2 (c − d)2 (e − f )2 [(a − b) + (c − d) + (e − f )]2 + + ≥ a+b c+d e+f (a + b) + (c + d) + (e + f ) (a + c + e − b − d − f )2 = . a+b+c+d+e+f Therefore, the last inequality is valid and so, the problem is solved. 226

Second solution: According to the Cauchy Schwarz Inequality, we have h i h √ √ i2 (b + d + f )2 a + (a + c + e)2 b (b + a) ≥ (b + d + f ) ab + (a + c + e) ba = ab(a + b + c + d + e + f )2 . This implies that ab (b + d + f )2 a + (a + c + e)2 b . ≤ a+b (a + b + c + d + e + f )2 Similarly, we have cd (b + d + f )2 c + (a + c + e)2 d , ≤ c+d (a + b + c + d + e + f )2 and

ef (b + d + f )2 e + (a + c + e)2 f ≤ . e+f (a + b + c + d + e + f )2

Adding up these three inequalities, we get ab cd ef (b + d + f )2 (a + c + e) + (a + c + e)2 (b + d + e) + + ≤ a+b c+d e+f (a + b + c + d + e + f )2 =

(a + c + e)(b + d + f ) , a+b+c+d+e+f

as desired. ?F? 84.1. Let a1 , a2 , . . . , an > 0, n ≥ 2. Prove that a2 a21 a22 a2 + + · · · + n−1 + n ≥ a1 + a2 + · · · + an . a2 a3 an a1 (China 1984) Solution: From the AM-GM Inequality, for each i (an+1 = a1 ), we get s a2i a2i + ai+1 ≥ 2 · ai+1 = 2ai . ai+1 ai+1 Setting i = 1, 2, . . . , n, we get the n similar inequalities and summing them up, we can get the desired result. Note that the equality holds if and only if a1 = a2 = · · · = an . ?F? 84.2. Let x, y, z be nonnegative real numbers such that x + y + z = 1. Prove that 7 0 ≤ xy + yz + zx − 2xyz ≤ . 27 (IMO 1984) 227

Solution: Let f (x, y, z) = xy + yz + zx − 2xyz. Without loss of generality, we may assume that 0 ≤ x ≤ y ≤ z ≤ 1. Since x + y + z = 1, this implies 1 that x ≤ . It follows that f (x, y, z) = (1 − 3x)yz + xyz + zx + xy ≥ 0, which 3 proves the left inequality. For the right part, applying the AM-GM Inequality,     1−x 2 y+z 2 we obtain yz ≤ = . Since 1 − 2x ≥ 0, this implies that 2 2   1−x 2 −2x3 + x2 + 1 f (x, y, z) = x(y+z)+yz(1−2x) ≤ x(1−x)+ . (1−2x) = 2 4 1 Our job is now to maximize a one-variable function F (x) = (−2x3 + x2 + 1),  4     3 1 1 1 0 − x ≥ 0 on 0, , we conclude that where x ∈ 0, . Since F (x) = x 3  3 2 3 1 7 1 F (x) ≤ F = for all x ∈ 0, . 3 27 3 ?F? 84.3. Prove that for any a, b > 0, we have that √ √ (a + b)2 a + b + ≥ a b + b a. 2 4 (Russia 1984) Solution: The AM-GM Inequality gives us    √ √ b a 2 2a + + 2b2 + ≥ 2a b + 2b a, 2 2 and

   √ √ b a 2ab + + 2ab + ≥ 2b a + 2a b. 2 2

By summing up these two inequalities, we get √ √ 2(a + b)2 + (a + b) ≥ 4a b + 4b a, and then, we deduce that √ √ (a + b)2 a + b + ≥ a b + b a, 2 4 1 as desired. Note that the equality holds if and only if a = b = . 4 ?F? 85.1. Let x1 , x2 , . . . , xn be real numbers from the interval [0, 2]. Prove that n X

|xi − xj | ≤ n2 .

i,j=1

When do we have equality? 228

(United Kingdom 1985) Solution: Since the inequality is symmetric, we can assume that 0 ≤ x1 ≤ x2 ≤ · · · ≤ xn ≤ 2. Accordingly, we find that n X

|xi − xj | = 2

i,j=1

n X

|xi − xj | = 2

1≤i n. 5 (Russia 1986) Solution: We will prove first that the following inequality holds for any x ≥ 0 f (x) = |sin x| + |sin(x + 1)| + |sin(x + 2)| −

8 > 0. 5

Since f (x) = f (x + π), it suffices to prove it for x ∈ [0, π], then for x > π we can put x = kπ + r where k ≥ 1, k ∈ N and 0 < r < π, and thus f (x) = f (kπ + r) = f (r) > 0. Now, from the assumption that x ∈ [0, π], we will consider two cases 230

The first case is when 0 ≤ x ≤ π−2. In this case, we have sin x ≥ 0, sin(x+2) ≥ 0 and sin(x + 1) ≥ sin 1 > 0. Therefore |sin x| + |sin(x + 1)| + |sin(x + 2)| = sin x + sin(x + 1) + sin(x + 2) 8 = (1 + 2 cos 1) sin(x + 1) ≥ (1 + 2 cos 1) sin 1 > . 5 The second one is when π−2 ≤ x ≤ π. In this case, sin x ≥ 0 and sin(x+2) ≤ 0, our inequality is equivalent to 8 sin x + |sin(x + 1)| − sin(x + 2) > . 5 Denote t = cos(x + 1) then −1 ≤ t ≤ − cos 1. We have sin x + |sin(x + 1)| − sin(x + 2) = |sin(x + 1)| − 2 sin 1 cos(x + 1) p = 1 − t2 − 2t sin 1 = g(t) Since g(t) is concave, we have 8 g(t) ≥ min {g(−1), g (− cos 1)} = min {2 sin 1, sin 1 + 2 sin 1 cos 1} > . 5 Now, we will prove the original inequality by induction on n. For n = 1, it is clear. Suppose that the inequality holds for n and let us prove it for n + 1, that is to prove 8 |sin 1|+|sin 2|+· · ·+|sin 3n|+| sin(3n+1)|+| sin(3n+2)|+| sin(3n+3)| > (n+1). 5 However, this is trivial since from the inductive hypothesis, we have 8 |sin 1| + |sin 2| + · · · + |sin 3n| > n, 5 and from what we have shown above, 8 | sin(3n + 1)| + | sin[(3n + 1) + 1]| + | sin[(3n + 1) + 2]| ≥ . 5 This completes our proof. ?F? 86.3. Show that for all positive real numbers a1 , a2 , . . . , an , we have   1 2 n 1 1 1 + + ··· + ≤4 + + ··· + . a1 a1 + a2 a1 + a2 + · · · + an a1 a2 an (Russia 1986) Solution: We will show that the stronger inequality holds n X

k k X k=1

n X 1 ≤2 . ak

aj

j=1

231

k=1

Indeed, by the Cauchy Schwarz Inequality, we have   2 b2k b1 + ··· + ≥ (b1 + · · · + bk )2 , (a1 + · · · + ak ) a1 ak where bi are arbitrary positive real numbers for all i = 1, 2, . . . , n. Therefore  2  b2k k b1 k + ··· + , ≤ k (b1 + · · · + bk )2 a1 ak X aj j=1

from which we deduce that n X

n

X ci k ≤ , k a X i=1 i k=1 aj j=1

where ck =

(k + 1)b2k nb2k kb2k + + · · · + , (b1 + · · · + bk )2 (b1 + · · · + bk+1 )2 (b1 + · · · + bn )2

for all k = 1, 2, . . . , n. Choosing now bk = k, we get ck =

k3 k X

!2 +

i

i=1 n X 2

k 2 (k + 1) !2 + · · · + k+1 X i

k2 n !2 n X i i=1

i=1 n  X

 1 1 1 2 = 4k − j(j + 1)2 j(j + 1) (j + 1)2 j=k j=k  n  X 1 1 1 2 ≤ 4k + − 2j 2 2(j + 1)2 (j + 1)2 j=k    n  X 1 1 1 2 2 1 = 2k − = 2k − < 2, j 2 (j + 1)2 k 2 (n + 1)2 = 4k

j=k

for all k = 1, 2, . . . , n. In this case, it follows that n X

k k X k=1

n X 1 ≤2 , ai

aj

i=1

j=1

and so our proof is completed. ?F? 86.4. Find the maximum value of x2 y + y 2 z + z 2 x 232

for reals x, y, z with x + y + z = 0 and x2 + y 2 + z 2 = 6. (United Kingdom 1986) First solution: From the given condition, we have xy + yz + zx = −3, and then, it follows that x2 y 2 + y 2 z 2 + z 2 x2 = (xy + yz + zx)2 − 2xyz(x + y + z) = 9. Now, let us expand the following nonnegative expression (x − xy − 1)2 + (y − yz − 1)2 + (z − zx − 1)2 . The work is very simple, we find that it is equal to X X X X X 3+ x2 + x2 y 2 + 2 xy − 2 x−2 x2 y. X X X X Replacing x2 by 6, x2 y 2 by 9, xy by −3 and x by 0, respectively, we conclude that (x − xy − 1)2 + (y − yz − 1)2 + (z − zx − 1)2 = 12 − 2(x2 y + y 2 z + z 2 x). Accordingly, we infer that x2 y + y 2 z + z 2 x ≤ 6. 2π 4π Note that the equality can occur, for example when x = 2 cos ,y = 2 cos ,z = 9 9 8π 2 2 2 2 cos . This allows us to conclude that the maximum of x y + y z + z x is 9 6. Second solution: We will give another way to prove that x2 y +y 2 z +z 2 x ≤ 6. Denote with p, q, r the values of x + y + z, xy + yz + zx and xyz, respectively. By this substitution, one can easily check that (x − y)2 (y − z)2 (z − x)2 = p2 q 2 − 4q 3 + 2(9q − 2p2 )pr − 27r2 , and (x2 y + y 2 z + z 2 x) + (xy 2 + yz 2 + zx2 ) = pq − 3r. On the other hand, from the given hypothesis, we have p = 0 and q = −3. Therefore, the above identities imply (x − y)2 (y − z)2 (z − x)2 = −4 · (−3)3 − 27r2 = 27(4 − r2 ), and (x2 y + y 2 z + z 2 x) + (xy 2 + yz 2 + zx2 ) = −3r. Accordingly, we have X  X X X X 2 x2 y = x2 y + xy 2 + x2 y − xy 2 p = −3r + (x − y)(x − z)(y − z) ≤ −3r + (x − y)2 (y − z)2 (z − x)2 s √ 2   p 2  p 2 2 2 2 = −3r + 3 3(4 − r ) ≤ 3 1 + 3 (−r) + 4−r = 12, 233

from which it deduces that x2 y + y 2 z + z 2 x ≤ 6, as desired. Remark: In addition, we can prove that the following inequality holds (in the same manner with the two solutions above) 2 − 9



3n2 − m2 2

3/2

m3 2 + ≤ x2 y + y 2 z + z 2 x ≤ 9 9



3n2 − m2 2

3/2 +

m3 9

for all real numbers x, y, z satisfying x + y + z = m and x2 + y 2 + z 2 = n2 (3n2 > m2 ). ?F? 87.1. Prove that if x, y, z are real numbers such that x2 + y 2 + z 2 = 2, then x + y + z ≤ xyz + 2. (IMO Shorlist 1987) First solution: If one of x, y, z is negative, for example z < 0. Then, we rewrite the inequality in the form −z(1 − xy) + 2 − x − y ≥ 0, which is true because 2−x−y ≥2− and

p p 2(x2 + y 2 ) ≥ 2 − 2(x2 + y 2 + z 2 ) = 0,

  x2 + y 2 z3 −z(1 − xy) ≥ −z 1 − = − ≥ 0. 2 2

Let us consider the now the case x, y, z ≥ 0. Without loss of generality, we 1 may assume that z = max {x, y, z} . By this assumption, we have z > √ . 3 Denote t = x + y, we find that xy =

t2 − x2 − y 2 t2 + z 2 − 2 = . 2 2

Therefore, our inequality is equivalent to 2+z·

t2 + z 2 − 2 ≥ t + z, 2

or f (t) = zt2 − 2t + z 3 − 4z + 4 ≥ 0. We see that f (t) is a quadratic function of t with the largest coefficient is z > 0. In addtion, its discrimimant is   2 1 + √ − 1 (1−z)2 ≤ 0. ∆0f = 1−z(z 3 −4z+4) = −(z 2 +2z−1)(1−z)2 < − 3 3 Therefore f (t) ≥ 0, and this ends our proof. Note that the equality holds if and only if x = y = 1, z = 0 or y = z = 1, x = 0 or z = x = 1, y = 0. 234

Second solution: Using the Cauchy Schwarz Inequality, we find that p x + y + z − xyz = x(1 − yz) + (y + z) · 1 ≤ [x2 + (y + z)2 ][(1 − yz)2 + 1]. So, it is enough to prove that this last quantity is at most 2, which is equivalent to the inequality (2 + 2yz)(2 − 2yz + y 2 z 2 ) ≤ 4, or 2y 3 z 3 ≤ 2y 2 z 2 , which is clearly true, because 2 ≥ y 2 + z 2 ≥ 2yz. Third solution: By the same arguments with the first solution, we see that it suffices to prove the inequality for x, y, z ≥ 0. Because of symmetry, we may assume that 0 ≤ x ≤ y ≤ z. If z ≤ 1, then 2 + xyz − x − y − z = (1 − z)(1 − xy) + (1 − x)(1 − y) ≥ 0. Now, if z > 1, we have p p z + (x + y) ≤ 2[z 2 + (x + y)2 ] = 2 1 + xy ≤ 1 + (1 + xy) ≤ 2 + xyz. This ends the proof. ?F? 88.1. Show that (1 + x)n ≥ (1 − x)n + 2nx(1 − x2 )

n−1 2

for all 0 ≤ x ≤ 1 and all positive integer n. (Ireland 1988) Solution: For n = 1, the inequality becomes equality. Suppose that n ≥ 2, n−1 then since (1 − x2 ) 2 ≤ 1, it suffices to prove that f (x) = (1 + x)n − (1 − x)n − 2nx ≥ 0.   We have f 0 (x) = n (1 + x)n−1 + (1 − x)n−1 − 2 . Since n−1 ≥ 1, the Bernoulli’s Inequality implies that (1 + x)n−1 + (1 − x)n−1 ≥ 1 + (n − 1)x + 1 + (n − 1)(−x) = 2. This means that f 0 (x) ≥ 0. Therefore f (x) is increasing on [0, 1], and we deduce that f (x) ≥ f (0) = 0, as claimed. Note that the equality holds if and only if n = 1 or x = 0. ?F? 88.2. Given a triangle ABC. Prove that         1 1 1 1 1 1 sin A sin B sin C + + ≤ + sin A+ + sin B+ + sin C. 2 A B C B C C A A B (Russia 1988) Solution: Rewrite the inequality as       1 1 2 1 1 2 1 1 2 + − sin A + + − sin B + + − sin C ≥ 0. B C A C A B A B C 235

Now, since this inequality is symmetric, we can assume that A ≥ B ≥ C. By this assumption, we have a ≥ b ≥ c, and since a = 2R sin A, we deduce that sin A ≥ sin B ≥ sin C. Also, it is clear from the assumption that 1 1 2 1 1 2 1 1 2 + − ≥ + − ≥ + − . B C A C A B A B C Thefore, by applying the Chebyshev’s Inequality, we deduce that      1 1 1 1 2 1 2 1 2 sin A + sin B + sin C ≥ + − + − + − B C A C A B A B C   1 1 1 2 1 1 2 1 1 2 (sin A+sin B+sin C) = 0. ≥ + − + + − + + − 3 B C A C A B A B C 

Note that the equality holds if and only if ABC is a equilateral triangle. ?F? 89.1. Let x1 , x2 , . . . , xn be positive real numbers, and let S = x1 +x2 +· · ·+xn . Prove that (1 + x1 )(1 + x2 ) · · · (1 + xn ) ≤ 1 + S +

S2 Sn + ··· + . 2! n! (APMO 1989)

Solution: By the AM-GM Inequality, we have   (1 + x1 ) + (1 + x2 ) + · · · + (1 + xn ) n (1 + x1 )(1 + x2 ) . . . (1 + xn ) ≤ n  n S = 1+ . n Therefore, it suffices to prove that   S2 Sn S n ≤1+S+ + ... + , 1+ n 2! n!   n  n n X S k =1+ S k , we can rewrite this inequality as and since 1 + n nk k=1

  n n X 1   − k  S k ≥ 0.  k! k n  

k=1

This is true since for any k = 1, . . . , n, we have   n n! n! k n − k! = nk − k! · = nk − k (n − k)!k! (n − k)! = nk − (n − k + 1) · · · n ≥ nk − n · · n} = 0. | ·{z k numbers

236

?F? 89.2. Let x, y, z be real numbers such that 0 < x < y < z <

π . Prove that 2

π + 2 sin x cos y + 2 sin y cos z > sin 2x + sin 2y + sin 2z. 2 (Iberoamerica 1989) Solution: Put a = sin x, b = sin y, c = sin z, then 1 > c > b > a > 0. inequality becomes p p p p p π + a 1 − b2 + b 1 − c2 > a 1 − a2 + b 1 − b2 + c 1 − c2 , 4 or p p p π > a 1 − a2 + (b − a) 1 − b2 + (c − b) 1 − c2 . 4 By the AM-GM Inequality, we get p p p a 1 − a2 + (b − a) 1 − b2 + (c − b) 1 − c2 ≤      1 1 ≤ a 1 − a2 + + (b − a) 1 − b2 + + (c − b) 1 − c2 + 4 4 1 = − (4a3 + 4b3 + 4c3 − 4ab2 − 4bc2 − 5c). 4

Our

1 4



Therefore, it suffices to prove that 4a3 + 4b3 + 4c3 − 4ab2 − 4bc2 − 5c + π > 0. 500 3 π π c ≥ 5c, and hence Using the AM-GM Inequality again, we have + + 2 2 27π 2 500 3 π − 5c ≥ − c > −2c3 . Consequently, it is enough to check that 27π 2 2a3 + 2b3 + c3 − 2ab2 − 2bc2 ≥ 0, or

    32 3 22 3 3 2 3 2 c − 2bc + b + 2a − 2ab + b ≥ 0. 27 27

This is true since by the AM-GM Inequality, we have c3 − 2bc2 +

32 3 c3 c3 32 3 b = + + b − 2bc2 27 2r 2 27 3 c3 32 c 3 ≥3 · · b3 − 2bc2 = 0, 2 2 27

and 2a3 − 2ab2 +

22 3 11 11 b = 2a3 + b3 + b3 − 2ab2 27 27 27 r 11 11 3 ≥ 3 2a3 · b3 · b3 − 2ab2 27 27 ! √ 3 242 = − 2 ab2 ≥ 0. 3 237

?F? 89.3. Let a, b, c be the sidelengths of a triangle. Prove that a − b b − c c − a 1 + + a + b b + c c + a < 16 . (Iberoamerica 1989) Solution: Using the identity (a − b)(b − c)(a − c) a−b b−c c−a + + = , a+b b+c c+a (a + b)(b + c)(c + a) we can rewrite the original inequality as (a + b)(b + c)(c + a) > 16 |(a − b)(b − c)(a − c)| . Now, since a, b, c are the sidelengths of a triangle, we may put a = y + z, b = z + x, c = x + y where x, y, z > 0, and hence, our inequality becomes (2x + y + z)(2y + z + x)(2z + x + y) > 16 |(x − y)(y − z)(z − x)| . Without loss of generality, we may assume that x ≥ y ≥ z, then we have (2x + y + z)(2y + z + x)(2z + x + y) > (2x + y)(x + 2y)(x + y), and |(x − y)(y − z)(z − x)| = (x − y)(x − z)(y − z) < xy(x − y). Therefore, it suffices to prove that (2x + y)(x + 2y)(x + y) > 16xy(x − y). Now, we apply the AM-GM Inequality to get p p (2x + y)(x + 2y) = 2(x − y)2 + 9xy ≥ 2 2(x − y)2 · 9xy = 6 2xy(x − y), and

√ x + y ≥ 2 xy.

Multiplying these two inequalities, we deduce that √ (2x + y)(x + 2y)(x + y) ≥ 12 2xy(x − y) ≥ 16xy(x − y). However, it is clear that the equality cannot occur, so we must have (2x + y)(x + 2y)(x + y) > 16xy(x − y). The proof is completed. ?F? 89.4. Find the least possible value of (x + y)(y + z) 238

for positive reals x, y, z satisfying xyz(x + y + z) = 1. (Russia 1989) Solution: From the AM-GM Inequality, we have p (x + y)(y + z) = y(x + y + z) + xz ≥ 2 xyz(x + y + z) = 2. The equality holds iff y(x + y +√ z) = xz and xyz(x + y + z) = 1, which can be attained for x = z = 1, y = 2 − 1. This allows us to conclude that the minimum value of (x + y)(y + z) is 2. ?F? 90.1. Let a, b, c, d be positive real numbers such that ab + bc + cd + da = 1. Prove that a3 b3 c3 d3 1 + + + ≥ . b+c+d c+d+a d+a+b a+b+c 3 (IMO Shortlist 1990) Solution: From the AM-GM Inequality, we get
2 (ab + ac + ad + bc + bd + cd) ≤ 3 a2 + b2 + c2 + d2 . Furthermore, the Cauchy Schwarz Inequality gives us p p a2 + b2 + c2 + d2 = a2 + b2 + c2 + d2 b2 + c2 + d2 + a2 ≥ ab + bc + cd + da = 1. Therefore, from these two inequalities, we deduce that 2 (a2 + b2 + c2 + d2 )2 ≥ (ab + ac + ad + bc + bd + cd). 3 Now, using the Cauchy Schwarz Inequality and this inequality, we obtain a3 b3 c3 d3 + + + ≥ b+c+d c+d+a d+a+b a+b+c
2 a2 + b2 + c2 + d2 1 ≥ ≥ , 2 (ab + ac + ad + bc + bc + cd) 3 1 as desired. Note that the equality holds if and only if a = b = c = d = . 2 ?F? 90.2. Show that x4 > x −

1 2

for all real x. (Russia 1990) Solution: According to the AM-GM Inequality, we have r 4 1 1 1 1 4 4 x 4 4 x + =x + + + ≥4 = √ |x| ≥ |x| ≥ x. 4 3 3 2 6 6 6 6 6 239

It is easy to see that the equality cannot be attained. Therefore 1 x4 > x − , 2 and our proof is completed. ?F? 90.3. Let x1 , x2 , . . . , xn be positive reals with sum 1. Show that x22 x2n 1 x21 + + ··· + ≥ . x1 + x2 x2 + x3 xn + x1 2 (Russia 1990) Solution: According to the Cauchy Schwarz Inequality, we have x21 x22 x2n (x1 + x2 + · · · + xn )2 + + ··· + ≥ x1 + x2 x2 + x3 xn + x1 (x1 + x2 ) + (x2 + x3 ) + · · · + (xn + x1 ) 1 x1 + x2 + · · · + xn = , = 2 2 1 as desired. Note that the equality holds if and only if x1 = x2 = · = xn = . n ?F? 90.4. Show that p p p x2 − xy + y 2 + y 2 − yz + z 2 ≥ z 2 + zx + x2 for any positive real numbers x, y, z. (United Kingdom 1990) Solution: By applying the Minkowsky’s Inequality, we have p p x2 − xy + y 2 + y 2 − yz + z 2 ≥ v v u √ !2 u √ !2    u u 2 3 x 3 z 2 t = x + −y +t z + y− 2 2 2 2 v u √ √ !2  u x z 2 p 2 3 3 ≥t x+ z + −y+y− = x + xz + z 2 , 2 2 2 2 as desired. ?F? 91.1. Let a1 , a2 , . . . , an and b1 , b2 , . . . , bn be positive real numbers such that a1 + a2 + · · · + an = b1 + b2 + · · · + bn . Prove that a21 a22 a2n a1 + a2 + · · · + an + + ··· + ≥ . a1 + b1 a2 + b2 an + bn 2 (APMO 1991) 240

First solution: By the Cauchy Schwarz Inequality, we have a21 a22 a2n (a1 + a2 + · · · + an )2 + + ··· + ≥ a1 + b1 a2 + b2 an + bn (a1 + b1 ) + (a2 + b2 ) + · · · + (an + bn ) (a1 + a2 + · · · + an )2 = (a1 + a2 + · · · + an ) + (b1 + b2 + · · · + bn ) (a1 + a2 + · · · + an )2 = (a1 + a2 + · · · + an ) + (a1 + a2 + · · · + an ) a1 + a2 + · · · + an , = 2 as desired. Note that the equality holds if and only if ai = bi for all i = 1, 2, . . . , n. Second solution: From the AM-GM Inequality, we see that for all i = 1, 2, . . . , n a2i ai bi (ai + bi )2 3ai − bi = ai − ≥ ai − = . ai + bi ai + bi 4(ai + bi ) 4 Therefore a22 a2n 3a1 − b1 3a2 − b2 3an − bn a21 + + ··· + ≥ + + ··· + a1 + b1 a2 + b2 an + bn 4 4 4 3(a1 + a2 + · · · + an ) − (b1 + b2 + · · · + bn ) = 4 a1 + a2 + · · · + an = , 2 as desired. ?F? 91.2. Given positive real numbers a, b, c satisfying a + b + c = 1, show that (1 + a) (1 + b) (1 + c) ≥ 8 (1 − a) (1 − b) (1 − c) . (Russia 1991) Solution: By applying the AM-GM Inequality, we have that p 1 + a = (1 − b) + (1 − c) ≥ 2 (1 − b) (1 − c). Similarly, we also have p 1 + b ≥ 2 (1 − c) (1 − a),

p 1 + c ≥ 2 (1 − a) (1 − b).

Multiplying these three inequalities, we get the desired result. Note that the 1 equality holds if and only if a = b = c = . 3 ?F? 91.3. Show that √ (x + y + z)2 √ √ ≥ x yz + y zx + z xy 3 241

for all nonnegative reals x, y, z. (Russia 1991) Solution: By the AM-GM Inequality, we get √ y+z z+x x+y √ √ x yz + y zx + z xy ≤ x · +y· +z· 2 2 2 (x + y + z)2 = xy + yz + zx ≤ , 3 as desired. The equality occurs iff x = y = z. ?F? 91.4. The real numbers x1 , x2 , . . . , x1991 satisfy |x1 − x2 | + |x2 − x3 | + · · · + |x1990 − x1991 | = 1991. Denote sn =

x1 + x2 + . . . + xn . What is the maximum possible value of n |s1 − s2 | + |s2 − s3 | + . . . + |s1990 − s1991 |? (Russia 1991)

Solution: We denote with E the desired expression. For all 1 ≤ i ≤ 1990, we have i+1 X

si+1 − si =

i X xk

xk

k=1

i+1

−

k=1

i

ixi+1 − =

i X xk

i X

k=1

k=1

=

i(i + 1)

k(xk+1 − xk ) i(i + 1)

.

This implies that i X

|si+1 − si | ≤

k |xk+1 − xk |

k=1

.

i(i + 1)

According to this inequality, we deduce that i X

E=

=

1990 X

|si+1 − si | ≤

i=1 1990 X i=1

i 1− 1991

1990 X i=1

k |xk+1 − xk |

k=1

=

i(i + 1)

 |xi+1 − xi | ≤

1990 X i=1

1990 X

1990 X

i=1

k=i

i

1 1− 1991

1 k(k + 1)

! |xi+1 − xi |

 |xi+1 − xi |

1990

=

1990 X |xi+1 − xi | = 1990. 1991 i=1

On ther other hand, let x1 = 1991, x2 = x3 = · · · = x1991 = 0, we have |x1 − x2 | + |x2 − x3 | + · · · + |x1990 − x1991 | = 1991 and E = 1990. Therefore, the maximum value of E is 1990. 242

?F? 91.5. A triangle has sides a, b, c with sum 2. Show that a2 + b2 + c2 + 2abc < 2. (United Kingdom 1991) Solution: Since a, b, c are the sidelengths of a triangle and a + b + c = 2, we see that 1 > a, b, c > 0. Hence (1 − a)(1 − b)(1 − c) > 0, which implies that abc < 1 − (a + b + c) + ab + bc + ca = ab + bc + ca − 1. And thus, a2 + b2 + c2 + 2abc < a2 + b2 + c2 + 2(ab + bc + ca − 1) = (a + b + c)2 − 2 = 2, as desired. ?F? 91.6. Prove the inequality x2 y y 2 z z 2 x + + ≥ x2 + y 2 + z 2 z x y for any positive real numbers x, y, z with x ≥ y ≥ z. (Vietnam 1991) First solution: By expanding, we see that the original inequality is equivalent to x3 y 2 + y 3 z 2 + z 3 x2 ≥ x3 yz + y 3 zx + z 3 xy, or x3 y(y − z) + y 2 z 2 (y − z) + z 3 (y 2 − 2yx + x2 ) − xyz(y 2 − z 2 ) ≥ 0, that is (y − z)(x − z)[x2 y + yz(x − y)] + z 3 (x − y)2 ≥ 0. The last inequality is obviously true since x ≥ y ≥ z. So, our problem is solved. Note that the equality holds if and only if x = y = z. Second solution: From the Cauchy Schwarz Inequality, we have s s x2 y y 2 z z 2 x x2 z y 2 x z 2 y + + · + + ≥ x2 + y 2 + z 2 . z x y y z x Hence, it suffices to prove that x2 y y 2 z z 2 x x2 z y 2 x z 2 y + + ≥ + + . z x y y z x However, we find that  2   2  x y y2z z2x x z y2x z2y (x − y)(y − z)(x − z)(xy + yz + zx) + + − + + = , z x y y z x xyz 243

which is obviously nonnegative because x ≥ y ≥ z > 0. Thus, the last inequality is true and our proof is completed. ?F? 92.1. For every integer n ≥ 2, find the smallest positive number λ = λ(n) 1 such that if 0 ≤ a1 , a2 , . . . , an ≤ , b1 , b2 , . . . , bn > 0, and a1 + a2 + · · · + an = 2 b1 + b2 + · · · + bn = 1, then b1 b2 · · · bn ≤ λ(a1 b1 + a2 b2 + · · · + an bn ). (China 1992) Solution: If n = 2, then from 0 ≤ a1 , a2 ≤ 12 and a1 + a2 = 1, we deduce that 1 a1 = a2 = . Therefore 2   b1 + b2 1 1 1 2 a1 b1 + a2 b2 = = + ≥ = 2, b1 b2 2b1 b2 2 b1 b2 b1 + b2 1 and the equality holds for b1 = b2 = . This allows us to conclude that 2 1 1 1 λ(2) = = . Now, suppose that n ≥ 3, and let b1 = b2 = , 2 2(2 − 1)2−1 2(n − 1) 1 1 , a1 = a2 = and a3 = · · · = an = 0, we get b3 = · · · = bn = n−1 2 λ(n) ≥

1 . 2(n − 1)n−1

We will show that it is the value we need to find, i.e. a1 b1 + a2 b2 + · · · + an bn ≥ 2(n − 1)n−1 b1 b2 · · · bn , or a1 b1 + a2 b2 + · · · + an bn ≥ 2(n − 1)n−1 b1 b2 · · · bn . a1 + a2 + · · · + an Without loss of generality, we may assume that b1 ≤ b2 ≤ · · · ≤ bn . By this assumption, we have a1 b1 + a2 b2 + · · · + an bn (a2 + · · · + an )b1 + a2 b2 + · · · + an bn − = a1 + a2 + · · · + an 2(a2 + · · · + an ) (a2 + · · · + an − a1 ) [a2 b2 + · · · + an bn − (a2 + · · · + an )b1 ] = 2(a2 + · · · + an )(a1 + a2 + · · · + an ) (a2 + · · · + an − a1 ) [a2 (b2 − b1 ) + · · · + an (bn − b1 )] = . 2(a2 + · · · + an )(a1 + a2 + · · · + an ) The last quantity is nonnegative since a2 (b2 − b1 ) + · · · + an (bn − b1 ) ≥ 0 from the assumption and a2 + · · · + an − a1 = 1 − 2a1 ≥ 0. Therefore, from this, we 244

deduce that (a2 + · · · + an )b1 + a2 b2 + · · · + an bn a1 b1 + a2 b2 + · · · + an bn ≥ a1 + a2 + · · · + an 2(a2 + · · · + an ) (b2 + b1 )a2 + · · · + (bn + b1 )an = 2(a2 + · · · + an ) (b2 + b1 )a2 + · · · + (b2 + b1 )an ≥ 2(a2 + · · · + an ) (b2 + b1 )(a2 + · · · + an ) = 2(a2 + · · · + an ) b1 + b2 p ≥ b1 b2 . = 2 The inequality is reduced to proving that p b1 b2 ≥ 2(n − 1)n−1 b1 b2 · · · bn , or equivalently, b1 b2 b23 · · · b2n ≤

1 . 4(n − 1)2(n−1)

This is true since by the AM-GM Inequality, we have 1 · (2b1 ) · (2b2 ) · b3 · b3 · · · bn · bn 4   1 2b1 + 2b2 + b3 + b3 + · · · + bn + bn 2(n−1) ≤ 4 2(n − 1)   1 b1 + b2 + · · · + bn 2(n−1) 1 = = . 4 n−1 4(n − 1)2(n−1)

b1 b2 b23 · · · b2n =

Our statement is proved. And thus, we conclude that λ(n) =

1 for 2(n − 1)n−1

every integer n. ?F? 92.2. Show that

√ x4 + y 4 + z 2 ≥ 2 2xyz

for all positive reals x, y, z. (Russia 1992) Solution: Using the AM-GM Inequality, we get p √ x4 + y 4 + z 2 ≥ 2x2 y 2 + z 2 ≥ 2 2x2 y 2 · z 2 = 2 2xyz, as desired. Note that the equality holds if and only if x = y and z = ?F? 92.3. Show that for any real numbers x, y > 1, we have y2 x2 + ≥ 8. y−1 x−1 245

√

2x2 .

(Russia 1992) Solution: From the AM-GM Inequality, we have 4y 2 4x2 4y 2 x2 y2 4x2 + = + ≥ + 2 ≥ 8, y−1 x−1 y2 x [1 + (y − 1)]2 [1 + (x − 1)]2 as desired. Note that the equality holds if and only if x = y = 2. ?F? 92.4. Positive real numbers a, b, c satisfy a ≥ b ≥ c. Prove that a2 − b2 c2 − b2 a2 − c2 + + ≥ 3a − 4b + c. c a b (Ukraine 1992) c+b a+c a+b ≥ 2, ≤ 2, and ≥ 1. c a b Now, we get (with noting that a − b ≥ 0, c − b ≤ 0 and a − c ≥ 0)

Solution: From a ≥ b ≥ c > 0, we have

c2 − b2 ≥ 2(c − b), a

a2 − b2 ≥ 2(a − b), c

a2 − c2 ≥ a − c. b

After addition of these inequalities, we have a2 − b2 c2 − b2 a2 − c2 + + ≥ 2(a − b) + 2(c − b) + (a − c) = 3a − 4b + c, c a b as claimed. Note that the equality holds if and only if a = b = c. ?F? 92.5. Let x, y, z, w be positive real numbers. Prove that   X 1 3 1 1 1 1 12 ≤ ≤ + + + . x + y + z + w sym x + y 4 x y z w (United Kingdom 1992) Solution: By the Cauchy Schwarz Inequality, we have 1 62 12 ≥X = , x+y x+y+z+w (x + y) sym X

sym

and

X 1 ≤ x + y sym sym X



1 1 + 4x 4y



3 = 4



1 1 1 1 + + + x y z w

 .

From these two inequalities, we infer that X 1 12 3 ≤ ≤ x + y + z + w sym x + y 4 246



1 1 1 1 + + + x y z w

 ,

as desired. Note that the equality holds (in both parts) if and only if x = y = z = w. ?F? 93.1. Let a, b, c, d be positive real numbers. Prove that a b c d 2 + + + ≥ . b + 2c + 3d c + 2d + 3a d + a + 3b a + 2b + 3c 3 (IMO Shortlist 1993) Solution: From the AM-GM Inequality, we have
2 (ab + ac + ad + bc + bd + cd) ≤ 3 a2 + b2 + c2 + d2 , and hence, it follows that (a + b + c + d)2 = a2 + b2 + c2 + d2 + 2 (ab + ac + ad + bc + bd + cd) 8 ≥ (ab + ac + ad + bc + bd + cd) . 3 Now, using the Cauchy Schwarz Inequality in combination with this inequality, we get a b c d + + + ≥ b + 2c + 3d c + 2d + 3a d + a + 3b a + 2b + 3c (a + b + c + d)2 2 ≥ ≥ , 4 (ab + ac + ad + bc + bd + cd) 3 as desired. The equality holds if and only if a = b = c = d. ?F? 93.2. Let a, b, c ∈ [0, 1]. Prove that a2 + b2 + c2 ≤ a2 b + b2 c + c2 a + 1. (Italia 1993) Solution: The desired inequality is equivalent to a2 (1 − b) + b2 (1 − c) + c2 (1 − a) ≤ 1. Since a, b, c ∈ [0, 1], we have a2 (1 − b) + b2 (1 − c) + c2 (1 − a) ≤ a (1 − b) + b (1 − c) + c (1 − a) = (a − 1)(b − 1)(c − 1) + 1 − abc ≤ 1. ?F? 93.3. Let x, y, u, v be positive real numbers. Prove that xy uv xy + xv + yu + uv ≥ + . x+y+u+v x+y u+v 247

(Poland 1993) Solution: Put a = x + y, b = u + v. Then we have to prove that xy + xv + uy + uv xy uv ≥ + , a+b a b or equivalently, ab (xy + xv + uy + uv) − (a + b) (bxy + auv) ≥ 0. The left hand side of the last inequality is equal to (au − bx)(by − av) which is clearly nonnegative because (au − bx) (by − av) = [(x + y) u − (u + v) x] [(u + v) y − (x + y) v] = (yu − xv)2 ≥ 0. The proof is completed. ?F? 93.4. If the equation x4 + ax3 + 2x2 + bx + 1 = 0 has at least one real root, then a2 + b2 ≥ 8. (Tournament of the Towns 1993) First solution: Let x be the real root of the equation. Using the Cauchy Schwarz Inequality, we infer that a2 + b2 ≥

(x4 + 2x2 + 1)2 ≥ 8, x6 + x2

because the last inequality is equivalent to (x2 − 1)2 ≥ 0. Second solution: Assume that x0 is the root of the desired equation, then x40 + ax30 + 2x20 + bx0 + 1 = 0. According to the two trivial inequalities x20 (a + 2x0 )2 ≥ 0 and (bx0 + 2)2 ≥ 0, we find that 1 x40 + ax30 ≥ − a20 x20 , 4

1 and b0 x0 + 1 ≥ − b20 x20 . 4

Therefore 1 1 1 0 = x40 + ax30 + 2x20 + bx0 + 1 ≥ − a20 x20 + 2x20 − b20 x20 = x20 (8 − a2 − b2 ). 4 4 4 On the other hand, it is easy to see that x0 6= 0. Thus, from the above inequality, we can deduce that a2 + b2 ≥ 8, as desired. ?F? 248

93.5. Let x1 , x2 , x3 , x4 be real numbers such that 1 ≤ x21 + x22 + x23 + x24 ≤ 1. 2 Find the largest and the smallest values of the expression A = (x1 − 2x2 + x3 )2 + (x2 − 2x3 + x4 )2 + (x2 − 2x1 )2 + (x3 − 2x4 )2 . (Vietnam 1993) Solution: (1) To find the max A, we rewrite A = 5(x21 + x24 ) + 6(x22 + x2 − 3) − 8(x1 x2 + x2 x3 + x3 x4 ) + 2x1 x3 + 2x2 x4 . Now, we try to replace xi xj by a sum of squares, so that at the end all squares x2i appear with the same coefficient. This can be done using the following remark: For any positive number r, the AM-GM Inequality implies that 16 • −8x1 x2 ≤ rx21 + x22 with equality iff rx1 = −4x2 ; r • −8x2 x3 ≤ 4(x22 + x23 ) with equality iff x2 = −x3 ; 16 • −8x3 x4 ≤ rx24 + x23 with equality iff rx4 = −4x3 ; r r 2 4 2 • 2x1 x3 ≤ x1 + x3 with equality iff rx1 = 4x3 ; 4 r 4 2 r 2 • 2x2 x4 ≤ x2 + x4 with equality iff rx4 = 4x2 . r 4 Therefore     20 5r 2 2 A≤ 5+ (x1 + x4 ) + 10 + (x22 + x23 ). 4 r 5r 20 Now, by choosing r > 0 such that 5 + = 10 + , or equivalently r = 4 r √
2 1 + 5 , we get √
5 3+ 5 A≤ . 2 The equality holds if and only if the above conditions on xi can be simultaneously satisfied and x21 + x22 + x23 + x24 = 1. One can verify that A reaches its maximum at two points s √ √ √ 1 5− 5 1+ 5 1+ 5 x1 = ± , x2 = − x1 , x3 = x1 , x4 = −x1 . 2 5 2 2

(2) For finding the minimum, we follow the same idea (but more complicated). Fix a number r and rewrites A = (x1 − rx2 + x3 )2 + (x2 − rx3 + x4 )2 − 2(4 − r)x1 x2 − 4(2 − r)x2 x3 − 2(4 − r)x3 x4 + 4(x21 + x24 ) + (5 − r2 )(x22 + x23 ). By the freedom, we may assume  r < 2. Then for any s > 0, we have 2 x • −2(4 − r)x1 x2 ≥ −(4 − r) sx21 + 2 with equality iff sx1 = x2 ; s 249

  x23 2 • −2(4 − r)x3 x4 ≥ −(4 − r) sx4 + with equality iff sx4 = x3 ; s • −4(2 − r)x2 x3 ≥ −2(2 − r)(x22 + x23 ) with equality iff x2 = x3 . Hence A ≥ −2(4 − r)x1 x2 − 4(2 − r)x2 x3 − 2(4 − r)x3 x4 + 4(x21 + x24 ) + (5 − r2 )(x22 + x23 )   4−r 2 2 2 ≥ [4 − (4 − r)s](x1 + x4 ) + 1 + 2r − − r (x22 + x23 ). s Choosing s such that 4 − (4 − r)s = 1 + 2r −

4−r − r2 , s

(∗)

we get A≥

4 − (4 − r)s . 2

The equality holds if and only if   sx1 = x2 x2 = x3         sx = x 4 3   x1 = x4      x2 = x3  x = (r − 1)x 1 2 ⇔ . e x − rx + x = 0 1 2 3   (e = ±1) x1 = √   2   x2 − rx3 + x4 = 0 2 1+s          x2 + x2 + x2 + x2 = 1  r−1= 1 1 2 3 4 s 2 1 From (∗) and r − 1 = , we get s     1 1 1 1 −3 s=2− 2 + −3 , 4+ s s s s √ 1+ 5 2 which is equivalent to s − s − 1 = 0, i.e. s = . Then 2 √ e 1+ 5 x1 = p x1 . √ , x2 = 2 10 + 2 5 √ 7−3 5 and it is attained at Thus min A = 4 √ 1 1+ 5 x1 = x4 = ± p √ , x2 = x3 = ± p √ . 10 + 2 5 2 10 + 2 5 ?F? 94.1. Let a1 , a√ 2 , . . . , an be a sequence of positive real numbers satisfying a1 + · · · + ak ≥ k for all k = 1, 2, . . . , n. Prove that   1 1 1 2 2 2 a1 + a2 + · · · + an > 1 + + ··· + . 4 2 n 250

(USA 1994) √ √ Solution: Denote bi = i − i − 1 for i = 1, 2, . . . , n, then we have a1 + · · · + ak ≥ b1 + · · · + bk for every k = 1, 2, . . . , n. We claim that a21 + a22 + · · · + a2n ≥ b21 + b22 + · · · + b2n . Indeed, from the Cauchy Schwarz Inequality, we have n X a2i

n X

!

i=1

n X ai bi

! b2i

≥

i=1

!2 .

i=1

Therefore, in order to prove our claim, it suffices to prove that n X

n X

ai bi ≥

i=1

b2i .

i=1

Now, using the Abel’s Summation formula, we have n X

ai bi −

i=1

n X

b2i =

i=1

n X

bi (ai − bi ) =

i=1

n−1 X

i X

i=1

k=1

(bi − bi+1 )

(ak − bk ) + bn

n X

(ai − bi ).

i=1

i X Since (ak − bk ) ≥ 0 for all i = 1, 2, . . . , n, and k=1

√  p √ √ √ i + 1 + i − 1 ≥ 2 i − (1 + 1)(i + 1 + i − 1) = 0, bi − bi+1 = 2 i − we see that each element of the above sum is nonnegative and hence, it is clear that n n X X ai bi ≥ b2i . i=1

i=1

This proves the claim that stated n X

a2i ≥

i=1

n X

b2i .

i=1

On the other hand, we have n X

b2i

=

i=1

>

n  X √ i=1 n X i=1

i−

√

i−1

2

=

n X

√

1 √


2 i−1   1 1 1 1 1 + + ··· + . √ √
2 = 4 2 n i+ i i=1

i+

Therefore, from this and the above inequality, we deduce that   1 1 1 2 2 2 a1 + a2 + · · · + an > 1 + + ··· + , 4 2 n 251

as desired. ?F? 95.1. Prove that for any positive real numbers x, y and any positive integers m, n,

(m−1)(n−1) xm+n + y m+n +(m + n − 1) (xm y n + xn y m ) ≥ mn xm+n−1 y + xy m+n−1 . (Austrian-Polish Competition 1995) First solution: The inequality is equivalent to
(m − 1)(n − 1) xm+n + y m+n − xm y n − xn y m ≥
≥ mn xm+n−1 y + xy m+n−1 − xm y n − xn y m , or (m − 1)(n − 1)(xm − y m )(xn − y n ) ≥ mnxy(xm−1 − y m−1 )(xn−1 − y n−1 ). Due to symmetry, we may assume that x ≥ y > 0. Then, we find that the above inequality follows from multiplying the following inequalities √ (m − 1)(xm − y m ) ≥ m xy(xm−1 − y m−1 ), and

√ (n − 1)(xn − y n ) ≥ n xy(xn−1 − y n−1 ).

So, it suffices to prove that these two inequalities holds. Let us first prove the former inequality. Denote x = t2 y where t ≥ 1, then we can rewrite this inequality as f (t) = (m − 1)(t2m − 1) − mt(t2m−2 − 1) ≥ 0. We have   f 0 (t) = m (2m − 2)t2m−1 − (2m − 1)t2m−2 + 1 ≥ 0, since by the AM-GM Inequality, it follows that (2m − 2)t2m−1 + 1 = (2m − 2) · t2m−1 + 1 · 1 ≥ (2m − 1) t2m−1


2m−2 2m−1

1

· 1 2m−1

= (2m − 1)t2m−2 . This means that f (t) is increasing for all t ≥ 1 and thus, it is clear that f (t) ≥ f (1) = 0 for any t ≥ 1. This proves the first inequality. Similarly, we can prove that the second inequality holds, and it ends our proof. Note that the equality holds if and only if x = y or m = 1 or n = 1. Second solution: If x = y, then it is trivial. Conversely, in case x 6= y, we transform the inequality as follows mn(x − y)(xm+n−1 − y m+n−1 ) ≥ (m + n − 1)(xm − y m )(xn − y n ), 252

or equivalently, xm+n−1 − y m+n−1 xm − y m xn − y n ≥ · . (m + n − 1)(x − y) m(x − y) n(x − y) (we have assumed that x > y). The last relation can also be written Zx (x − y)

m+n−2

t y

Zx dt ≥

m−1

t

Zx dt ·

y

tn−1 dt,

y

and this follows from the Chebyshev’s Inequality for integrals. ?F? 95.2. Let a, b, c, d be positive real numbers. Prove that a+c b+d c+a d+b + + + ≥ 4. a+b b+c c+d d+a (Baltic Way 1995) Solution: According to the AM-GM Inequality, we have a+c b+d c+a d+b + + + = a+b b+c c+d d+a (a + c) (a + b + c + d) (b + d) (a + b + c + d) = + (a + b) (c + d) (b + c) (d + a)   4 (a + c) 4 (b + d) ≥ (a + b + c + d) + [(a + b) + (c + d)]2 [(b + c) + (d + a)]2 a+b+c+d =4· = 4, a+b+c+d as desired. Note that the equality holds if and only if a = c and b = d. ?F? 95.3. Let x, y, z be positive real numbers. Prove that xx y y z z ≥ (xyz)

x+y+z 3

. (Canada 1995)

Solution: The original inequality can be written as 1 x ln x + y ln y + z ln z ≥ (x + y + z)(ln x + ln y + ln z). 3 Due to symmetry, we may assume that x ≥ y ≥ z. Since the function f (x) = ln x is increasing for any positive reals x, this assumption also implies that ln x ≥ ln y ≥ ln z. Therefore, we can see that the above inequality is a direct consequence of the Chebyshev’s Inequality. Note that the equality holds if and ony if x = y = z. ?F? 253

95.4. If a, b, c are positive real numbers such that abc = 1, then 1 1 1 3 + 3 + 3 ≥ . + c) b (c + a) c (a + b) 2

a3 (b

(IMO 1995) Solution: Applying the AM-GM Inequality, we have s b+c 1 b+c 1 + ≥2 = . 3 3 a (b + c) 4bc 4a bc (b + c) a Adding this to the two analogous inequalities, we get   1 1 1 1 1 1 1 b+c c+a a+b ≥ + + . + 3 + 3 + + + 3 a (b + c) b (c + a) c (a + b) 4 bc ca ab a b c Therefore 1 1 1 1 + 3 + 3 ≥ 3 a (b + c) b (c + a) c (a + b) 2



1 1 1 + + a b c



3 ≥ 2

r 3

1 3 = . abc 2

The proof is completed. Note that the equality holds if and only if a = b = c = 1. ?F? 95.5. Suppose that x1 , x2 , . . . , xn are real numbers satisfying |xi − xi+1 | < 1 and xi ≥ 1 for i = 1, 2, . . . , n (xn+1 = x1 ). Prove the following inequality x1 x2 xn + + ··· + < 2n − 1. x2 x3 x1 (India 1995) x1 x2 xn First solution: Since · ··· = 1, there exists an index k such that x2 x3 x1 xk < 1. On the other hand, since |xi − xi+1 | < 1 and xi ≥ 1 for all i, we find xk+1 xi that xi < xi+1 + 1 ≤ 2xi+1 , and thus < 2 for all i. From these arguments, xi+1 we get x1 x2 xn X xi xk + + ··· + = + < 2(n − 1) + 1 = 2n − 1, x2 x3 x1 xi+1 xk+1 i6=k

as desired. Second solution: Similar to the preceding solution, one may prove that xi < 2xi+1 for all i. Now, we rewrite the inequality as       x1 x2 xn 2− + 2− + ··· + 2 − > 1, x2 x3 x1 or

n X 2xi+1 − xi i=1

xi+1 254

> 1.

Since 2xi+1 − xi > 0, the Cauchy Schwarz Inequality implies n X n X 2xi+1 − xi i=1

xi+1

≥

n X

xi

i=1 n X

x2i

i=1 n X

2

i=1

+

n X

i=1 n X

x2i −

i=1 n X

> xi xi+1

n X

2

xi xi+1

x2i

xi

x2i +

i=1 n X

2

n X

−

i=1 n X

xi xi+1

xi ·

i=1 n X

x2i −

i=1

i=1

!2

i=1

=

xi+1 (2xi+1 − xi )

i=1 n X

>

!2

i=1

xi 2

xi xi · 2

= 1.

This proves our inequality. ?F? 95.6. (a) Find the maximum value of the expression x2 y−y 2 x when 0 ≤ x ≤ 1 and 0 ≤ y ≤ 1. (b) Find the maximum value of the expression x2 y +y 2 z +z 2 x−y 2 x−z 2 y −x2 z when 0 ≤ x ≤ 1, 0 ≤ y ≤ 1 and 0 ≤ z ≤ 1. (United Kingdom 1995) Solution: (a) By the AM-GM Inequality, we have 2



2

x y − y x = xy(x − y) ≤ xy(1 − y) ≤ y(1 − y) ≤

y+1−y 2

2

1 = , 4

1 and the equality holds when x = 1 and y = . Therefore, the maximum value 2 1 of the desired expression is . 4 (b) Due to the cyclicity, we may assume that x = max {x, y, z} . Then, applying the AM-GM Inequality, we find that x2 y + y 2 z + z 2 x − y 2 x − z 2 y − x2 z = (x − y)(x − z)(y − z) ≤ (x − y)(x − z)y ≤ xy(x − y)   1 y + x − y 2 x3 = ≤ . ≤x· 2 4 4 1 The equality holds for example when x = 1, y = and z = 0. Therefore, the 2 1 maximum value of the desired expression is . 4 ?F? 95.7. Let a, b, c be real numbers satisfying a < b < c, a + b + c = 6 and ab + bc + ca = 9. Prove that 0 < a < 1 < b < 3 < c < 4. 255

(United Kingdom 1995) Solution: From the given condition, we have a < 2 < c. Now, let us prove the following chain of identities 3(a − 1)(a − 3) − (a − b)(a − c) = 3(b − 1)(b − 3) − (b − c)(b − a) = 3(c − 1)(c − 3) − (c − a)(c − b) = 0. Indeed, we have 0 = 9 − ab − bc − ca = 9 + a2 − 2a(b + c) − (a − b)(a − c) = 9 + a2 − 2a(6 − a) − (a − b)(a − c) = 3(a − 1)(a − 3) − (a − b)(a − c). Since a < b < c, it is clear that (a − b)(a − c) > 0,

(b − c)(b − a) < 0,

(c − a)(c − b) > 0.

And hence, the above identities imply that (a − 1)(a − 3) > 0, (b − 1)(b − 3) < 0 and (c − 1)(c − 3) > 0, from which we deduce that (with noting a < 2 < c) a < 1 < b < 3 < c. This yields that (1 − a)(1 − b) < 0 and (3 − b)(3 − c) < 0. Therefore, according to the identities 0 = 9 − ab − bc − ca = 10 − (c + 1)(a + b) − (1 − a)(1 − b) = 10 − (c + 1)(6 − c) − (1 − a)(1 − b) = (c − 1)(c − 4) − (1 − a)(1 − b), and 0 = 9 − ab − bc − ca = 18 − (a + 3)(b + c) − (3 − b)(3 − c) = 18 − (a + 3)(6 − a) − (3 − b)(3 − c) = a(a − 3) − (3 − b)(3 − c), we infer that (c − 1)(c − 4) < 0 and a(a − 3) < 0, from which we obtain c < 4 and a > 0. Thus, from these arguments, we conclude that 0 < a < 1 < b < 3 < c < 4, as desired. ?F? 95.8. Find the greatest constant k such that for any positive integer n which is not a square, √
√
1 + n sin π n > k. (Vietnam (IMO training camp) 1995) π Solution: We will prove that is the best constant. For any positive integer 2 i, we must have    p  p k < 1 + i2 + 1 sin π i2 + 1 . 256

 √  Because sin π i2 + 1 = sin √

π , we deduce that +1+i  p  k π √ ≥ sin π i2 + 1 > √ , i2 + 1 + i i2 + 1 + 1 π π is from which it follows that k ≤ . Now, let us prove that the constant 2 2 good. Clearly, the inequality can be rewritten as √
π sin π n > √ . 2 ( n + 1) i2

We have two cases 1 Case 1. {n} ≤ . Of course, 2 √ √ √ n ≥ n− n−1= √ and because sin x ≥ x − √
n ≥ sin √ sin π

or 6

√

n+

√

n−1

,

x3 , we find that 6

n+

π √

π 1 ≥√ √ − n−1 n−1+ n 6

Let us prove that the last quantity is at least 2+

n+

1 √



π √ √ n−1+ n

3 .

π √ . This comes down to 2 (1 + n)

√ n− n−1 π2 √ >
2 , √ √ 1+ n 3 n+ n−1

√

n−1


2

+3

√

n+

√


√
n − 1 > π2 1 + n ,

and it is clear. √ 1 1 Case 2. {n} > . Let x = 1 − { n} < and let n = k 2 + p, 1 ≤ p ≤ 2k. 2 2 1 Because {n} > , we have p ≥ k + 1. Then it is easy to see that x ≥ 2 1 √ and so, it suffices to prove that k + 1 + k 2 + 2k x π . √ sin ≥  √ 2 k + 1 + k + 2k 2 1 + k2 + k Using again the inequality sin x ≥ x −

x3 , we infer that 6

√ √ 2 k 2 + k − k 2 + 2k − k + 1 π2 √ >  2 . √ 1 + k2 + k 3 1 + k + k 2 + 2k √ √ But from the Cauchy Schwarz Inequality, we have 2 k 2 + k− k 2 + 2k−k ≥ 0.  2  √ √ π2  And because the inequality 1 + k + k 2 + 2k > 1 + k 2 + k holds, 3 this case is also solved. 257

?F? 96.1. Let m and n be positive integers such that n ≤ m. Prove that 2n · n! ≤

(m + n)! ≤ (m2 + m)n . (m − n)! (APMO 1996)

Solution: The quantity in the middle is (m + n)(m + n − 1) · · · (m − n + 1). If we pair of terms of the form (m + x) and (m + 1 − x), we get products which do not exceed m(m + 1), since the function f (x) = (m + x)(m + 1 − x) is a 1 concave parabola with maximum at x = . From this, the right inequality 2 follows. For the left, we need only show (m + x)(m + 1 − x) ≥ 2x for x ≤ n, this rearranges to (m − x)(m + 1 + x) ≥ 0, which holds because m ≥ n ≥ x. ?F? 96.2. Let a, b, c be the lengths of the sides of a triangle. Prove that √ √ √ √ √ √ a + b − c + b + c − a + c + a − b ≤ a + b + c, and determine when equality occurs. (APMO 1996) Solution: By the triangle inequality, b + c − a and c + a − b are positive. For any positive x, y, we have √ √
2 √ x+ y 2(x + y) ≥ x + y + 2 xy = by the AM-GM Inequality, with equality for x = y. Substituting x = a + b − c, y = b + c − a, we get √ √ √ a + b − c + b + c − a ≤ 2 a, which added to the two analogous inequalities yields the desired result. The equality holds for a + b − c = b + c − a = c + a − b, i.e. a = b = c. ?F? 96.3. The real numbers x, y, z, t satisfy the equalities x + y + z + t = 0 and x2 + y 2 + z 2 + t2 = 1. Prove that −1 ≤ xy + yz + zt + tx ≤ 0. (Austrian-Polish Competition 1996) Solution: The inner expression is (x + z)(y + t) = −(x + z)2 , so the second inequality is obvious. As for the first, note that 1 1 = (x2 + z 2 ) + (y 2 + t2 ) ≥ [(x + z)2 + (y + t)2 ] ≥ |(x + z)(y + t)| 2 by two applications of the power mean inequality. 258

?F? 96.4. Let a, b, c be positive real numbers, such that a + b + c = that ab + bc + ca ≥ 9(a + b + c).

√

abc. Prove

(Belarus 1996) Solution: Applying the well-known inequality (x + y + z)2 ≥ 3(xy + yz + zx) with x = ab, y = bc, z = ca, we get (ab + bc + ca)2 ≥ 3abc(a + b + c) = 3(a + b + c)3 , and thus ab + bc + ca ≥ (a + b + c)

p 3(a + b + c).

On the other hand, the AM-GM Inequality implies s  √ a+b+c 3 a + b + c = abc ≤ , 3 p from which we deduce that 3(a + b + c) ≥ 9. Therefore p ab + bc + ca ≥ (a + b + c) 3(a + b + c) ≥ 9(a + b + c), as claimed. Note that the equality holds if and only if a = b = c = ?F?

√

3.

96.5. Suppose n ∈ N, x0 = 0, xi > 0 for i = 1, 2, . . . , n, and x1 +x2 +· · ·+xn = 1. Prove that 1≤

n X i=1

√

xi π √ < . 2 1 + x0 + · · · + xi−1 · xi + · · · + xn (China 1996)

Solution: The left inequality follows from the fact that p √ 1 1 + x0 + · · · + xi−1 · xi + · · · + xn ≤ (1 + x0 + · · · + xn ) = 1, 2 so that the middle quantity is at least x1 +· · ·+xn = 1. For the right inequality, let θi = arcsin(x0 + · · · + xi ) for i = 0, . . . , n, so that p √ 1 + x0 + · · · + xi−1 · xi + · · · + xn = cos θi−1 , and the desired inequality is n X sin θi − sin θi−1 i=1

cos θi−1

<

π . 2

Now note that sin θi − sin θi−1 = 2 cos

θi + θi−1 θi − θi−1 sin < (θi − θi−1 ) cos θi−1 , 2 2 259

using the facts that θi−1 < θi and that sin x < x for x > 0, so that n X sin θi − sin θi−1

cos θi−1

i=1

<

n X π (θi − θi−1 ) = θn − θ0 < , 2 i=1

as claimed. ?F? 96.6. (a) Find the minimum value of xx for x a positive real number. (b) If x and y are positive real numbers, show that xy + y x > 1. (France 1996) First solution: (a) Since xx = ex log x and ex is an increasing function of x, it suffices to determine the minimum of x log x. This is easily done by setting its 1 1 derivative 1+log x to zero, yielding x = . The second derivative is positive e x for x > 0, so the function is everywhere convex, and the unique extremum is 1 indeed a global minimum. Hence xx has minimum value e− e . (b) If x ≥ 1, then xy ≥ 1 for y > 0, so we may assume 0 < x, y < 1. Without loss of generality, assume x ≤ y, now note that the function f (x) = xy + y x has derivative f 0 (x) = xy log x + y x−1 . Since y x ≥ xx ≥ xy for x ≤ y and 1 ≥ − log x, we see that f 0 (x) > 0 for 0 ≤ x ≤ y and so the minimum of x f occurs with x = 0, in which case f (x) = 1, since x > 0, we have strict inequality. Second solution: We present another way to prove (b). Firstly, we will a for any a, b ∈ (0, 1). Indeed, from the Bernoull’s prove that ab ≥ a + b − ab Inequality, it follows that a1−b = [1+(a−1)]1−b ≤ 1+(a−1)(1−b) = a+b−ab, and thus the conclusion follows. Now, if x or y is at least 1, we are done. Otherwise, let 0 < x, y < 1. In this case, we apply the above observation and find that xy + y x ≥

x y x y + > + = 1. x + y − xy x + y − xy x+y x+y ?F?

96.7. Let a, b, c be positive real numbers such that abc = 1. Prove that a5

bc ca ab + 5 + 5 ≤ 1. 5 5 + b + ab b + c + bc c + a5 + ca (IMO Shortlist 1996)

Solution: For any positive numbers x, y, we have (x3 − y 2 )(x2 − y 2 ) ≥ 0, and hence x5 + y 5 ≥ x2 y 2 (x + y). According to this inequality, we have ab a2 b2 c a2 b2 c c = ≤ = . 5 5 5 5 2 2 2 2 2 2 a + b + ab a +b +a b c a b (a + b) + a b c a+b+c 260

Adding this to the two analogous inequalities, we get the desired result. Note that the equality holds if and ony if a = b = c = 1. ?F? 96.8. Let a and b be positive real numbers with a + b = 1. Prove that a2 b2 1 + ≥ . a+1 b+1 3 (Hungary 1996) First solution: Using the condition a + b = 1, we can reduce the given inequality to homogeneous one 1 a2 b2 ≤ + , 3 (a + b)[a + (a + b)] (a + b)[b + (a + b)] or a2 b + ab2 ≤ a3 + b3 , which follows from (a3 + b3 ) − (a2 b + ab2 ) = (a − b)2 (a + b) ≥ 0. 1 The equality holds if and only if a = b = . 2 Second solution: By the Cauchy Schwarz Inequality, we have that a2 b2 (a + b)2 1 + ≥ = , a+1 b+1 a+1+b+1 3 as desired. ?F? 96.9. Prove the following inequality for positive real numbers x, y, z,   1 1 1 9 (xy + yz + zx) + + ≥ . 2 2 2 (y + z) (z + x) (x + y) 4 (Iran 1996) First solution: Without loss of generality, we may assume that x ≥ y ≥ z. Denote P (x, y, z) =

1 1 1 9 + + − . 2 2 2 (y + z) (z + x) (x + y) 4(xy + yz + zx)

We have to show that P (x, y, z) ≥ 0. Note that P (t, t, z) =

z(z − t)2 ≥0 2t2 (2z + t)(z + t)2

for any positive reals t ≥ z > 0. Therefore, a key to solve this problem is to find a suitable number t ≥ z such that P (x, y, z) ≥ P (t, t, z). 261

We will prove that the number t =

x+y satisfies this property, i.e. 2

1 2 9 9 1 + − ≥ − . 2 2 2 2 (x + z) (x + z) (t + z) 4(xy + yz + zx) 4(t + 2zt) With noting that 2 1 1 2 2 − + − x+z y+z (x + z)(y + z) (t + z)2 (x − y)2 (x − y)2 = + , (x + z)2 (y + z)2 2(x + z)(y + z)(t + z)2

1 1 2 + − = (x + z)2 (x + z)2 (t + z)2



and 9 9 9(x − y)2 − = , 4(xy + yz + zx) 4(t2 + 2zt) 16(t2 + 2tz)(xy + yz + zx) we can rewrite the above inequality as (x − y)2 9(x − y)2 (x − y)2 + ≥ , (x + z)2 (y + z)2 2(x + z)(y + z)(t + z)2 16(t2 + 2tz)(xy + yz + zx) or equivalently, 1 (x +

z)2 (y

+

z)2

+

1 9 ≥ . 2 2(x + z)(y + z)(t + z) 16(xy + yz + zx)(t2 + 2tz)

We have 4(xy + yz + zx) − 3(x + z)(y + z) = xy + yz + zx − 3z 2 ≥ 0, then 4 (x + z)(y + z) ≤ (xy + yz + zx), 3 and hence, it follows that 1 (x +

z)2 (y

+

z)2

≥

9 9 ≥ . 2 16(xy + yz + zx) 16(xy + yz + zx)(t2 + 2zt)

Therefore, the last inequality is valid and our proof is completed. Note that the equality holds if and only if x = y = z. Second solution: Due to symmetry, we can assume that x ≥ y ≥ z > 0. xy + yz + zx yz x Now, since = + , the original inequality can be 2 (y + z) y+z (y + z)2 written as x y z xy yz zx 9 + + + + + ≥ . y + z z + x x + y (x + y)2 (y + z)2 (z + x)2 4   x y 1 1 Since + = (x + y + z) + − 1, this inequality is y+z z+x x+z y+z equivalent to   1 xy 1 z yz zx 17 (x + y + z) + + + + + ≥ , 2 2 2 x+z y+z x + y (x + y) (y + z) (z + x) 4 262

or  (x + y + z)

 4(x + y + z) 1 1 4 z + + − + + x + z y + z x + y + 2z x + y + 2z x+y xy yz zx 17 + + + ≥ . 2 2 2 (x + y) (y + z) (z + x) 4

From this, it is not hard for us to write the inequality in the form M (x − y)2 + N z ≥ 0, where 1 x+y+z − (x + z)(y + z)(x + y + 2z) 4(x + y)2 x+y+z 1 ≥ − (x + y)(y + x)(x + y + 2z) 4(x + y)2 3x + 3y + 2z ≥ 0, = 4(x + y)2 (x + y + 2z)

M=

and 1 x y 4 + + − 2 2 x + y (x + z) (y + z) x + y + 2z   1 4 1 z z 1 + − + − − = 2 x + z y + z x + y + 2z x + y (x + z) (y + z)2 (x − y)2 1 z z = + − − . 2 (x + z)(y + z)(x + y + 2z) x + y (x + z) (y + z)2

N=

Since

z y (y − z)(x2 − yz) − = − ≤ 0, (x + z)2 (x + y)2 (x + y)2 (x + z)2

we deduce that (x − y)2 1 y z + − − 2 (x + y)(y + y)(x + y + 2y) x + y (x + y) 4yz 2 2 3 (x − y) (x − y) (x − y) = − = ≥ 0. 2 2y(x + y)(x + 3y) 4y(x + y) 4y(x + y)2 (x + 3y)

N≥

Therefore, in the last inequality, each element of it is nonnegative, and hence, the result follows. ?F? 96.10. Let n ≥ 2 be a fixed natural number and let a1 , a2 , . . . , an be positive numbers whose sum is 1. Prove that for any positive numbers x1 , x2 , . . . , xn whose sum is 1, n X n − 2 X ai x2i 2 xi xj ≤ + , n−1 1 − ai i=1

1≤i 3 and a1 , a2 , . . . , an be real numbers such that a1 +a2 +· · ·+an ≥ n and a21 + a22 + · · · + a2n ≥ n2 . Prove that max {a1 , a2 , . . . , an } ≥ 2. (USA 1999) First solution: The most natural idea is to suppose that ai < 2 for all i and n n X X to substitute xi = 2 − ai > 0. Then we have (2 − xi ) ≥ n or xi ≤ n, i=1

and also n2 ≤

n X

a2i =

n X

i=1

(2 − xi )2 = 4n − 4

i=1

i=1

n X i=1

Now, using the fact that xi > 0, we obtain

n X i=1

x2i

<

xi +

n X

x2i .

i=1 n X

!2 xi

, which com-

i=1

bined with the above inequality yields n2 < 4n − 4

n X i=1

xi +

n X

!2 xi

i=1

294

≤ 4n + (n − 4)

n X i=1

xi .

(we have used the fact that

n X

xi ≤ n). Thus, we have (n − 4)

i=1

0, which is clearly impossible since n ≥ 4 and

n X

n X

! xi − n

>

i=1

xi ≤ n. So, our assumption

i=1

was wrong and consequently max{a1 , a2 , . . . , an } ≥ 2. Second solution: Assume that ai < 2 for all i = 1, 2, . . . , n, then we have a1 + a2 + · · · + ai − 2(i − 1) < 2

∀i = 1, 2, . . . , n.

Now, notice that for any m, n < 2, we have (m + n − 2)2 + 22 − m2 − n2 = 2(2 − m)(2 − n) > 0. Using this inequality in combination with the above observation, we have (a1 + a2 − 2)2 + 22 > a21 + a22 , [a1 + a2 + a3 − 2(3 − 1)]2 + 22 > (a1 + a2 − 2)2 + a23 , ····················· [a1 + a2 + . . . + an − 2(n − 1)]2 + 22 > [a1 + a2 + . . . + an−1 − 2(n − 2)]2 + a2n . Summing up these inequalities, we get [a1 + a2 + · · · + an − 2(n − 1)]2 + 4(n − 1) > a21 + a22 + · · · + a2n . Since ai < 2, n ≥ 4 and a1 + a2 + . . . + an ≥ n, we have n − 2 ≥ 2 > a1 + a2 + · · · + an − 2(n − 1) ≥ 2 − n. From here, we deduce that [a1 + a2 + · · · + an − 2(n − 1)]2 ≤ (n − 2)2 , and hence a21 + a22 + · · · + a2n < (n − 2)2 + 4(n − 1) = n2 , which contradicts with the given hypothesis. So, our assumption was wrong, or in the other words, we must have max{a1 , a2 , . . . , an } ≥ 2. ?F?  π 99.9. Let a0 , a1 , . . . , an be numbers from the interval 0, such that 2    π π π tan a0 − + tan a1 − + · · · + tan an − ≥ n − 1. 4 4 4 Prove that tan a0 tan a1 · · · tan an ≥ nn+1 . (USA 1999) Solution: Let tan ai = xi for all i = 0, 1, . . . , n. It follows from the hypothesis that for each i, xi > 0, and xn − 1 x0 − 1 x1 − 1 + + ··· + ≥ n − 1, x0 + 1 x1 + 1 xn + 1 295

or equivalently, 1≥

1 1 1 + + ··· + . x0 + 1 x1 + 1 xn + 1

Now, for each i (0 ≤ i ≤ n), the AM-GM Inequality gives us X 1 1 n xi =1− ≥ ≥ sY , xi + 1 xi + 1 xj + 1 j6=i (xj + 1) n j6=i

and hence, it follows that v u (x + 1)n uYi xi ≥ n u . n t (xj + 1) j6=i

Setting i = 0, 1, . . . , n, we get the n + 1 similar inequalities and multiplying them up, we deduce that x0 x1 · · · xn ≥ nn+1 ,

tan a0 tan a1 · · · tan an ≥ nn+1 ,

or equivalently,

as desired. The equality holds if and only if a0 = a1 = · · · = an = arctan n. ?F? 99.10. Let x, y, z > 1. Prove that 2 +2yz

xx

yy

2 +2zx

zz

2 +2xy

≥ (xyz)xy+yz+zx . (USA (Shortlist) 1999)

Solution: The original inequality is equivalent to


x2 + 2yz ln x + y 2 + 2zx ln y + z 2 + 2xy ln z ≥ (xy + yz + zx) ln (xyz) , or (x − y) (x − z) ln x + (y − x) (y − z) ln y + (z − x) (z − y) ln z ≥ 0. Without loss of generality, we may assume that x ≥ y ≥ z. By this assumption, we have (x−y)(x−z) ≥ 0, (z−x)(z−y) ≥ 0. Therefore, with noting x, y, z > 1, we infer that (x − y) (x − z) ln x ≥ (x − y)(x − z) ln y,

and

(z − x)(z − y) ln z ≥ 0.

From this, we can see that the left hand side of the above inequality is not less than (x − y)(x − z) ln y + (y − z)(y − x) ln y = (x − y)2 ln y ≥ 0. Our proof is completed. Note that the equality holds if and only if x = y = z. ?F? 296

00.1. Let a, b be real numbers and a 6= 0. Prove that a2 + b2 +

1 b √ + ≥ 3. a2 a (Austria 2000)

Solution: Applying the AM-GM Inequality, we get 1 b a +b + 2 + = a a 2

2

r   √ 3 3 3 1 2 2 2 + a + 2 ≥ a + 2 ≥ 2 a2 · 2 = 3, b+ 2a 4a 4a 4a

1 3 as desired. The equality holds if and only if b = − and a2 = 2 , i.e. when 4a r r ! r r ! 2a 14 4 4 3 4 3 1 4 4 (a, b) equals , or − . ,− , 4 2 3 4 2 3 ?F? 00.2. For all real numbers a, b, c ≥ 0 such that a + b + c = 1, prove that 2 ≤ (1 − a2 )2 + (1 − b2 )2 + (1 − c2 )2 ≤ (1 + a)(1 + b)(1 + c). (Austrian-Polish Competition 2000) Solution: Let us first prove the left inequality. Without loss of generality, 1 assume that a = max {a, b, c} . By this assumption, we have a ≥ . Now, with 3 noting that b2 + c2 ≤ (b + c)2 = (1 − a)2 ≤ 1, we apply the Cauchy Schwarz Inequality and get  2 2 − (1 − a)2 (1 + 2a − a2 )2 (2 − b2 − c2 )2 (1 − b ) + (1 − c ) ≥ ≥ = . 2 2 2 2 2

2 2

Therefore, it suffices to prove that (1 − a2 )2 +

(1 + 2a − a2 )2 ≥ 2. 2

After some simple computations, we see that (1 − a2 )2 +

(1 + 2a − a2 )2 (1 − a)2 (1 + a)(3a − 1) −2= , 2 2

1 . This ends the proof for the left 3 inequality. Note that the equality holds if and only if (a, b, c) equals (1, 0, 0), or (0, 1, 0), or (0, 0, 1). which is clearly nonnegative since a ≥

For the right inequality, to prove it, let us notice that for every nonnegative numbers x, 9(1 − x)(1 + x)2 + 15x2 − 10x − 9 = −x(1 − 3x)2 ≤ 0. 297

This implies that (1 − x)(1 + x)2 ≤

9 + 10x − 15x2 . 9

Using this inequality, we deduce that X

(1 − a2 )2 =

X 1X (1 − a)(9 + 10a − 15a2 ), (1 − a)(1 − a)(1 + a)2 ≤ 9

and hence, it suffices to prove that X (1 − a)(9 + 10a − 15a2 ) ≤ 9(1 + a)(1 + b)(1 + c). Now, expanding with noting that a + b + c = 1, we can rewrite this inequality as X X X 15 a3 − 25 a2 + 10 ≤ 9 ab + 9abc, or 15

X

a3 − 25

X

a2

 X  X 3 X  X  a + 10 a ≤9 a ab + 9abc.

The last one can be simplified to X 4 ab(a + b) ≥ 24abc, which is obviously true by the AM-GM Inequality. Therefore, the right hand side inequality is proved. It is easy to see that the equality holds   if and only 1 1 1 if (a, b, c) is one of the triples (1, 0, 0), (0, 1, 0), (0, 0, 1), , , . 3 3 3 ?F? 00.3. Let a, b, c, x, y, z be positive real numbers. Prove that a3 b3 c3 (a + b + c)3 + + ≥ . x y z 3 (x + y + z) (Belarus 2000) First solution: From the Cauchy Schwarz Inequality, we have a3 x

+

b3 y

+

c3 z

 √ √ √ 2 a a+b b+c c ≥

x+y+z

.

Using this inequality, it suffices to prove that  √ √ √ 2 3 a a + b b + c c ≥ (a + b + c)3 . Because of the homogeneity of this inequality, we may homogenize a+b+c = 3, and hence, it becomes √ √ √ a a + b b + c c ≥ 3. 298

Now, applying the Bernoulli’s Inequality, we have √ 3 3 a a = [1 + (a − 1)] 2 ≥ 1 + (a − 1). 2 Adding this to the two analogous inequalities, we conclude that √ √ √ 3 a a + b b + c c ≥ 3 + (a + b + c − 3) = 3, 2 as desired. Second solution: By the Holder’s Inequality, 

a3 b3 c3 + + x y z

1/3

(1 + 1 + 1)1/3 (x + y + z)1/3 ≥ a + b + c.

Cubing both sides and then dividing both sides by 3(x+y+z) gives the desired result. ?F? 00.4. Suppose that the real numbers a1 , a2 , . . . , a100 satisfy (i) a1 ≥ a2 ≥ · · · ≥ a100 ≥ 0; (ii) a1 + a2 ≤ 100; (iii) a3 + a4 + · · · + a100 ≤ 100. Determine the maximum possible value of a21 + a22 + · · · + a2100 , and find all possible sequences a1 , a2 , . . . , a100 which achieve this maximum. (Canada 2000) Solution: We have a21 + a22 − a23 − (a1 + a2 − a3 )2 = −2(a1 − a3 )(a2 − a3 ) ≤ 0, and hence a21 + a22 ≤ a23 + (a1 + a2 − a3 )2 ≤ a23 + (100 − a3 )2 . Note that the equality in this inequality holds iff a2 = a3 and a1 + a2 = 100. Now, since a3 ≥ a4 ≥ · · · ≥ a100 , we have a24 + · · · + a2100 ≤ a3 a4 + · · · + a3 a100 = a3 (a4 + · · · + a100 ) ≤ a3 (100 − a3 ), with equality iff a24 = a3 a4 , . . . , a2100 = a3 a100 and a3 (a3 +a4 +· · ·+a100 −100) = 0. In addition, since a1 ≥ a2 ≥ a3 and a1 + a2 ≤ 100, we deduce that a3 ≤ 50, and hence a3 (a3 − 50) ≤ 0 with equality iff a3 (a3 − 50) = 0. From these inequalities, we infer that a21 + a22 + · · · + a2100 ≤ a23 + (100 − a3 )2 + a23 + a3 (100 − a3 ) = 10000 + 2a3 (a3 − 50) ≤ 10000. The equality holds if and only if the equality must hold in each inequality found above - that is, we must have: (a) a1 + a2 = 100, a2 = a3 , 299

(b) a24 = a3 a4 , . . . , a2100 = a3 a100 , a3 (a3 + a4 + · · · + a100 − 100) = 0, (c) a3 (a3 − 50) = 0. Solving the system of equations (a), (b) and (c) (with noting at the given hypothesis), we can find that the equality holds only when the sequence a1 , a2 , . . . , a100 equals 100, 0, 0, . . . , 0,

or

50, 50, 50, 50, 0, 0, . . . , 0. ?F?

00.5. Show that

s 3



2 (a + b)

1 1 + a b

r

 ≥

3

a + b

r 3

b a

for all positive real numbers a and b, and determine when the equality occurs. (Czech-Slovak 2000) √ First solution: Multiplying both sides of the desired inequality by 3 ab gives the equivalent inequality q √ √ 3 3 3 a2 + b2 ≤ 2 (a + b)2 . √ a = x and 3 b = y, we see that it suffices to prove that q
3
2 3 2 2 x + y ≤ 2 (x3 + y 3 )2 , or equivalently, x2 + y 2 ≤ 2 x3 + y 3 .

Setting

√ 3

Now, by applying the Cauchy Schwarz Inequality and the Chebyshev’s Inequality, we get x2 + y 2


3


√

√



√ p 2 x3 + y · y 3 ≤ x2 + y 2 (x + y) x3 + y 3


2 ≤ 2 x3 + y 3 x3 + y 3 = 2 x3 + y 3 , = x2 + y 2

x·

as desired. It is easy to see that the equality holds if and only if a = b. Second solution: By the Power Mean Inequality, we have r 3

   

a + b 2

r 3  r r 2 a b 3 b +    a a  b  ≤  , 2   

a b with equality if and only if = , or equivalently a = b. It is easy to see that b a the desired result follows from this inequality and the identity r

a + b

r !2   b 1 1 = (a + b) + . a a b ?F? 300

00.6. Let a, b, c be positive real numbers such that abc = 1. Prove that 18 1 + ab2 1 + bc2 1 + ca2 + + ≥ 3 . c3 a3 b3 a + b3 + c3 (Hong Kong 2000) Solution: Multiplying both sides of the desired inequality by a3 + b3 + c3 > 0 gives the equivalent inequality    
1
ab2 bc2 ca2 1 1 3 3 3 3 3 3 a +b +c + a +b +c + + + 3 + 3 ≥ 18. a3 b3 c3 c3 a b Now, applying the AM-GM Inequality, we have r   √
1 1 1 1 3 3 3 3 3 3 3 3 a +b +c + 3 + 3 ≥3 a b c ·3 = 9, 3 3 a b c a b3 c3 and
a3 + b3 + c3



ab2 bc2 ca2 + 3 + 3 c3 a b



√ 3

≥ 3 a3 b3 c3 · 3

r 3

a3 b3 c3 = 9. a3 b3 c3

Adding up these two inequalities, we get the result. It is easy to see that the equality holds if and only if a = b = c = 1. ?F? 00.7. Let x, y, and z denote positive real numbers, each less than 4. Prove 1 1 1 1 1 1 , + , and + is greater that at least one of the numbers + x 4−y y 4−z z 4−x than or equal to 1. (Hungary 2000) Solution: From the Cauchy Schwarz Inequality, we have 1 1 4 + ≥ = 1. x 4−x x + (4 − x) 1 1 1 1 1 1 + , + , and + were all less x 4−y y 4−z z 4−x than 1, their sum S would be less than 3. However,       1 1 1 1 1 1 S= + + + + + ≥ 3. x 4−x y 4−y z 4−z

Now, if the three numbers

This contradiction proves the requested result. ?F? 00.8. Let a, b, c be positive real numbers such that abc = 1. Prove the inequality     1 1 1 a+ −1 b+ −1 c + − 1 ≤ 1. b c a (IMO 2000) 301

First solution: Without loss of generality, we can suppose that b = min{a, b, c}, 1 then it is clear that b ≤ 1. Now, replacing c = , the inequality becomes ab     1 1 1 a + − 1 (b + ab − 1) + − 1 ≤ 1, b ab a or (ab + 1 − b)(ab + b − 1)(b + 1 − ab) ≤ ab2 . Because (ab + b − 1) + (b + 1 − ab) = 2b > 0, we can see that max{ab + b − 1, b + 1 − ab} > 0. On the other hand, it is obvious that ab + 1 − b > 0, therefore if min{ab + b − 1, b + 1 − ab} ≤ 0, the left hand side of the above inequality is not greater than 0 while the right hand side is, so the inequality is trivial. Otherwise, one can see that 0 < (ab + 1 − b)(ab + b − 1) = a2 b2 − (b − 1)2 ≤ a2 b2 , 0 < (ab + b − 1)(b + 1 − ab) = b2 − (ab − 1)2 ≤ b2 , 0 < (b + 1 − ab)(ab + 1 − b) = 1 − b2 (a − 1)2 ≤ 1. Multiplying these three inequalities and taking the square root, we can get the desired result. The equality holds if and only if a = b = c = 1. Second solution: Using the condition abc = 1, it is straightforward to verify the equalities     1 1 1 2= a−1+ +c b−1+ , a b c     1 1 1 2= b−1+ +a c−1+ , b c a     1 1 1 2= c−1+ +b a−1+ . c a c 1 In particular, they show that at most one of the numbers u = a − 1 + , b 1 1 v = b − 1 + , w = c − 1 + is negative. If there is such a number, we have c a     1 1 1 a−1+ b−1+ c−1+ = uvw < 0 < 1. b c a And if u, v, w ≥ 0, the AM-GM Inequality yields r 1 c uv, 2 = u + cv ≥ 2 a a

r 1 a 2 = v + aw ≥ 2 vw, b b

1 2 = w + aw ≥ 2 c

r

b wu. c

a b c a b c Thus, uv ≤ , vw ≤ , wu ≤ , so (uvw)2 ≤ · · = 1. Since u, v, w ≥ 0, c a b c a b this completes the proof. 302

x y Third solution: From the given hypothesis, we may substitute a = , b = , y z z c = , where x, y, z > 0. By this substitution, our inequality becomes x   x z y x z y −1+ −1+ −1+ ≤ 1, y y z z x x or xyz ≥ (y + z − x) (z + x − y) (x + y − z) . By expanding, we find that the last inequality is equivalent to x3 + y 3 + z 3 + 3xyz ≥ xy(x + y) + yz(y + z) + zx(z + x), which is just the Schur’s Inequality (in the special case third degree). ?F? 00.9. Let x, y ≥ 0 with x + y = 2. Prove that x2 y 2 (x2 + y 2 ) ≤ 2. (Ireland 2000) First solution: After homogenizing it, we need to prove  2

x+y 2

6

≥ x2 y 2 (x2 + y 2 ),

or

(x + y)6 ≥ 32x2 y 2 (x2 + y 2 ).

(Now, forget the constraint x + y = 2!) In case xy = 0, it clearly holds. We now assume that xy 6= 0. Because of the homogeneity of the inequality, this means that we may normalize to xy = 1. Then, it becomes     1 6 1 2 x+ ≥ 32 x + 2 , x x

or

p3 ≥ 32(p − 2),

  1 2 where p = x + ≥ 4. Our job is now to minimize F (p) = p3 − 32(p − 2) x r 32 0 2 on [4, ∞). Since F (p) = 3p − 32 ≥ 0, where p ≥ , F is (monotone) 3 increasing on [4, ∞). So, F (p) ≥ F (4) = 0 for all p ≥ 4. Second solution: As in the first solution, we prove that (x + y)6 ≥ 32(x2 + y 2 )(xy)2 for all x, y ≥ 0. In case x = y = 0, it’s clear. Now, if x2 +y 2 > 0, then x2 + y 2 we may normalize to x2 +y 2 = 2. Setting p = xy, we have 0 ≤ p ≤ =1 2 and (x + y)2 = x2 + y 2 + 2xy = 2 + 2p. It now becomes (2 + 2p)3 ≥ 64p2 ,

or

p3 − 5p2 + 3p + 1 ≥ 0.

3 2 0 We  F (p) =  want to minimize F (p) = p − 5p + 3p + 1 on [0, 1]. We compute 1 1 3 p− (p − 3). We find that F is monotone increasing on 0, and 3 3

303



 1 monotone decreasing on , 1 . Since F (0) = 1 and F (1) = 0, we conclude 3 that F (p) ≥ F (1) = 0 for all p ∈ [0, 1]. Third solution: We show that (x + y)6 ≥ 32(x2 + y 2 )(xy)2 where x ≥ y ≥ 0. We make the substitution u = x + y and v = x − y. Then, we have u ≥ v ≥ 0. It becomes  2  2 2 u + v2 u − v2 u6 ≥ 32 , or u6 ≥ (u2 + v 2 )(u2 − v 2 )2 . 2 4 Note that u4 ≥ u4 − v 4 ≥ 0 and that u2 ≥ u2 − v 2 ≥ 0. So, u6 ≥ (u4 − v 4 )(u2 − v 2 ) = (u2 + v 2 )(u2 − v 2 )2 .

Fourth solution: According to the AM-GM Inequality, we find that   x+y 2 = 1, xy ≤ 2 and xy(x2 + y 2 ) =

1 1 · 2xy · (x2 + y 2 ) ≤ 2 2



2xy + x2 + y 2 2

2 = 2.

Therefore, by combining these two results, we conclude that x2 y 2 (x2 + y 2 ) = xy · xy(x2 + y 2 ) ≤ 1 · 2 = 2, as desired. ?F? 00.10. The real numbers a, b, c, x, y, z satisfy a ≥ b ≥ c > 0 and x ≥ y ≥ z > 0. Prove that a2 x2 b2 y 2 c2 z 2 3 + + ≥ . (by + cz)(bz + cy) (cz + ax)(cx + az) (ax + by)(ay + bx) 4 (Korea 2000) First solution: Denote the left hand side of the given inequality by S. Because a ≥ b ≥ c and x ≥ y ≥ z, by the Rearrangement Inequality, we have bz + cy ≤ by + cz so (by + cz)(bz + cy) ≤ (by + cz)2 ≤ 2[(by)2 + (cz)2 ]. Setting m = (ax)2 , n = (by)2 , p = (cz)2 , we obtain a2 x2 a2 x2 m ≥ = . (by + cz)(bz + cy) 2[(by)2 + (cz)2 ] 2(n + p) Adding this to the two analogous inequalities, we find that   1 m n p S≥ + + . 2 n+p p+m m+n 304

On the other hand, according to the Nesbitt’s Inequality, it is easy to see that m n p 3 + + ≥ , n+p p+m m+n 2 3 and therefore S ≥ , as desired. 4 Second solution: According to the Cauchy p Schwarz Inequality, we have p by + cz ≤ (b2 + c2 )(y 2 + z 2 ) and bz + cy ≤ (b2 + c2 )(z 2 + y 2 ). Multiplying these inequalities, we get (by + cz)(bz + cy) ≤ (b2 + c2 )(y 2 + z 2 ), and hence x2 a2 x2 a2 · . ≥ 2 (by + cz)(bz + cy) b + c2 y 2 + z 2 Adding this to the two analogous inequalities, we find that a2 x2 b2 y 2 c2 z 2 + + ≥ (by + cz)(bz + cy) (cz + ax)(cx + az) (ax + by)(ay + bx) x2 b2 y2 c2 z2 a2 · + · + · . ≥ 2 b + c2 y 2 + z 2 c2 + a2 z 2 + x2 a2 + b2 x2 + y 2 On the other hand, from the given hypothesis, it is easy to verify that a2 b2 c2 ≥ ≥ , b2 + c2 c2 + a2 a2 + b2

x2 y2 z2 ≥ ≥ . y2 + z2 z 2 + x2 x2 + y 2

Therefore, by the Chebyshev’s Inequality, we have a2 x2 b2 y2 c2 z2 · + · + · ≥ b2 + c2 y 2 + z 2 c2 + a2 z 2 + x2 a2 + b2 x2 + y 2    1 a2 b2 c2 x2 y2 z2 3 ≥ + 2 + 2 + 2 + 2 ≥ , 2 2 2 2 2 2 2 2 3 b +c c +a a +b y +z z +x x +y 4 where the last inequality holds according to the Nesbitt’s Inequality

m + n+p

p 3 n + ≥ . From this and the above estimation, we can get the p+m m+n 2 result. ?F? 00.11. Let x, y, z be real numbers. Prove that √ x2 + y 2 + z 2 ≥ 2 (xy + yz) . (Macedonia 2000) Solution: By the AM-GM Inequality, we find that r r    2  y2 y y2 y2 2 2 2 2 2 2 2 x +y +z = x + + +z ≥2 x · +2 ·z 2 2 2 2 √ √ = 2 (|xy| + |yz|) ≥ 2(xy + yz), 305

y as desired. It is easy to see that the equality holds if and only if x = z = . 2 ?F? 00.12. Let a, b, x, y, z be positive real numbers. Prove that x y z 3 + + ≥ . ay + bz az + bx ax + by a+b (MOSP 2000) Solution: Using the Cauchy Schwarz Inequality in combination with the wellknown inequality (x + y + z)2 ≥ 3(xy + yz + zx), we deduce that x y z (x + y + z)2 + + ≥ ay + bz az + bx ax + by x(ay + bz) + y(az + bx) + z(ax + by) 1 (x + y + z)2 3 = · ≥ , a + b xy + yz + zx a+b as desired. It is easy to see that the equality holds if and only if x = y = z = 1. ?F? 00.13. Let ABC be an acute triangle. Prove that 

cos A cos B

2

 +

cos B cos C

2

 +

cos C cos A

2 + 8 cos A cos B cos C ≥ 4. (MOSP 2000)

Solution: Since 4−8 cos A cos B cos C = 4(cos2 A+cos2 B +cos2 C), it suffices to prove that 

cos A cos B

2

 +

cos B cos C

2

 +

cos C cos A

2

≥ 4(cos2 A + cos2 B + cos2 C).

Now, applying the AM-GM Inequality in combination with the well-known 1 inequality cos A cos B cos C ≤ , we have 8 r       cos A 2 cos A 2 cos B 2 cos4 A 3 + + ≥3 2 cos B cos B cos C cos B cos2 C 3 cos2 A = √ 3 cos2 A cos2 b cos2 C 3 cos2 A ≥ r = 12 cos2 A. 1 3 82 Adding this inequality with its analogous forms and dividing both sides of the resulting inequality by 3, we obtain desired result. It is easy to see that π equality holds if and only if A = B = C = . 3 ?F? 306

00.14. Let a, b, c be positive real numbers such that min {a, b} ≥ c. Prove that p p √ c(a − c) + c(b − c) ≤ ab. (MOSP 2000) Solution: By the Cauchy Schwarz Inequality, we have i2 hp p c(a − c) + c(b − c) ≤ [c + (b − c)] [(a − c) + c] = ab. Taking square root of each side of this inequality yields the desired result. ?F? 00.15. Let a1 , a2 , . . . , an be nonnegative real numbers. Prove that √ a1 + a2 + · · · + an − n a1 a2 · · · an ≥ n n√ o √ √ √ √ √ 1 ≥ min ( a1 − a2 )2 , ( a2 − a3 )2 , . . . , ( an − a1 )2 . 2 (MOSP 2000) Solution: Denote an+1 = a1 , because n n√ √ 2 √ √ 2 √ √ 2o X √ √ n min ( a1 − a2 ) , ( a2 − a3 ) , . . . , ( an − a1 ) ≤ ( ai − ai+1 )2 , i=1

one can see that the original inequality is deduced from the following inequality n

√ n

1X √ √ a1 + a2 + · · · + an − n a1 a2 · · · an ≥ ( ai − ai+1 )2 . 2 i=1

This is equivalent to √ √ √ a1 +a2 +· · ·+an −n n a1 a2 · · · an ≥ a1 +a2 +. . .+an −( a1 a2 + · · · + an a1 ) , or

√

a1 a2 + · · · +

√

√ an a1 ≥ n n a1 a2 · · · an .

The last one is clearly true by the AM-GM Inequality, so our proof is completed. Note that the equality holds if and only if a1 = a2 = · · · = an . ?F? 00.16. Let (an ) be an infinite sequence of positive numbers such that a1 a4 a 2 + + ··· + k ≤ 1 1 2 k for all k. Prove that

a1 a2 an + + ··· + < 2, 1 2 n

for all n. (MOSP 2000) 307

Solution: One can see that we can put n = k 2 +p where k ≥ 1 and 0 ≤ p ≤ 2k. Then ak2 +p a1 a2 an a1 a2 + + ··· + = + + ... + 2 1 2 n 1 2 k +p ak2 +p a 2 a1 a2 ≤ + + ··· + 2 + · · · + 2k +2k 1 2 k +p k + 2k   k X ai2 a2 = + · · · + 2i +2i . i2 i + 2i i=1

Since (an ) is an decreasing sequence, it is clear that   ai2 +2i a i2 1 1 1 + ··· + 2 a i2 . ≤ 2+ 2 + ··· + 2 i2 i + 2i i i +1 i + 2i On the other hand, we have     1 1 1 1 1 1 + ··· + 2 = 2 + ··· + 2 + 2 + ... + 2 i2 i + 2i i i +i−1 i +i i + 2i i i+1 2 < 2+ 2 = . i i +i i Therefore, for any i = 1, 2, . . . , k, we have a2 ai2 2a 2 + · · · + 2i +2i ≤ i , 2 i i + 2i i and from this, it follows that k X a1 a2 an a i2 + + ··· + . Accordingly, this 2 3 assumption allows us to proceed the following estimation 1 1 1 ab + bc + ca ab + bc + ca 1 + + = + + b+c c+a a+b  a+b a + c b + c    ab 1 ac + = b+c+ + b+c a+b a+c ab ac a(1 + bc) ≥2+ + =2+ a+b a+c a(a + b + c) + bc a(1 + bc) a(1 + bc) ≥2+ ≥2+ 2a + bc 2a + 2abc 1 5 =2+ = . 2 2 Therefore, in any cases, we always have 1 1 5 1 + + ≥ . b+c c+a a+b 2 The equality holds if and only if (a, b, c) is a permutation of (1, 1, 0). ?F? 00.18. Let a1 , a2 , . . . , an be positive real numbers such that 1 1 1 + + ··· + ≤ 1. a1 a2 an Prove that for any positive integer k, (ak1 − 1)(ak2 − 1) · · · (akn − 1) ≥ (nk − 1)n . (MOSP 2000) 1 , and then ai b1 + b2 + · · · + bn ≤ 1. By this substitution, we need to prove that      1 1 1 −1 − 1 · · · k − 1 ≥ (nk − 1)n . bn bk1 bk2

Solution: In order to prove the given inequality, we put bi =

Firstly, we will show that      1 1 1 −1 − 1 ... − 1 ≥ (n − 1)n . b1 b2 bn Indeed, by the AM-GM Inequality, we have sY

X bj b1 + b2 + · · · + bn 1 −1≥ −1= bi bi 309

j6=i

bi

(n − 1) n−1 ≥

j6=i

bi

bj .

Setting i = 1, 2, . . . , n, we get the n similar inequalities and multiplying them up, we can get the above inequality. Now, since k is an positive integer, the Holder’s Inequality implies !  Y  Y n  n n  Y 1 1 1 1 −1 = −1 · 1 + + · · · + k−1 k bi bi b bi i i=1 i=1 i=1 ! n Y 1 1 ≥ (n − 1)n 1 + + · · · + k−1 bi bi i=1 #n " 1 1 + ··· + √ ≥ (n − 1)n 1 + √
k−1 . n n b1 b2 . . . bn b1 b2 . . . bn On the other hand, from the condition b1 +b2 +· · ·+bn ≤ 1, we can easily deduce √ 1 that n b1 b2 · · · bn ≤ . Using this in combination with the above inequality, n we get  n  Y 1 − 1 ≥ (n − 1)n (1 + n + · · · + nk−1 )n = (nk − 1)n , k b i i=1 as desired. Note that the inequality becomes equality if and only if a1 = a2 = · · · = an = n. ?F? 00.19. Let a, b, c be positive real numbers. Prove that  2 √ 3 a + b + c + abc 1 1 1 1 + + + √ ≥ . 3 a + b b + c c + a 2 abc (a + b)(b + c)(c + a) (MOSP 2000) First solution: Applying the Cauchy Schwarz Inequality, we have √ a2

b2

c2

3

2 abc

1 1 1 1 + + + √ = 2 + + + 3 a + b b + c c + a 2 abc a (b + c) b2 (c + a) c2 (a + b) 2abc  2 √ a + b + c + 3 abc ≥ 2 . a (b + c) + b2 (c + a) + c2 (a + b) + 2abc On the other hand, it is easy to verify that a2 (b+c)+b2 (c+a)+c2 (a+b)+2abc = (a + b)(b + c)(c + a). Therefore, the above inequality implies  2 √ 3 a + b + c + abc 1 1 1 1 + + + √ ≥ . a + b b + c c + a 2 3 abc (a + b)(b + c)(c + a) The equality holds if and only if a = b = c = 1. Second solution: The original inequality is equivalent to 2 X √ (a + b)(b + c)(c + a)  3 √ (a + b)(a + c) + ≥ a + b + c + abc , 2 3 abc 310

or √ √ (a + b)(b + c)(c + a) 3 3 √ ≥ 2 abc(a + b + c) + a2 b2 c2 . 3 2 abc √ 3 Since ab + bc + ca ≥ 3 a2 b2 c2 , the above inequality follows from ab + bc + ca +

√ √ (a + b)(b + c)(c + a) 3 3 √ + 2 a2 b2 c2 ≥ 2 abc(a + b + c). 3 2 abc √ Multiplying both sides for 2 3 abc > 0, we can rewrite this inequality as √ 3 (a + b)(b + c)(c + a) + 4abc ≥ 4 a2 b2 c2 (a + b + c). Without loss of generality, we can assume that a ≥ b ≥ c. Now, we write the inequality in form h  i  √ √ √ 3 3 3 (b + c) (a + b)(a + c) − 4 a2 b2 c2 ≥ 4 a2 b2 c2 a − abc . Because (a + b)(a + c) = a2 + ab + ac + bc, it is equivalent to     √ √ √ 3 3 3 (b + c) a2 + ab + bc + ca − 4 a2 b2 c2 ≥ 4 a2 b2 c2 a − abc . √ 3 Using again the estimation ab + bc + ca ≥ 3 a2 b2 c2 , it suffices to prove that     √ √ √ 3 3 3 (b + c) a2 − a2 b2 c2 ≥ 4 a2 b2 c2 a − abc , or equivalently,

  √ √ 3 3 (b + c) a + abc ≥ 4 a2 b2 c2 .

Of course, this is obvious since by the AM-GM Inequality, we have q   √ √ √ √ 3 3 3 (b + c) a + abc ≥ 2 bc · 2 a abc = 4 a2 b2 c2 . ?F? 00.20. For any integer n ≥ 2, consider n−1 positive real numbers a1 , a2 , . . . , an−1 having sum 1, and n real numbers b1 , b2 , . . . , bn . Prove that b21 +

b22 b2 b2 + 3 + . . . + n ≥ 2b1 (b2 + b3 + . . . + bn ). a1 a2 an−1 (Romania 2000)

Solution: By the Cauchy Schwarz Inequality, we have b22 b2 b2 (b2 + b3 + · · · + bn )2 + 3 + ··· + n ≥ = (b2 + b3 + · · · + bn )2 . a1 a2 an−1 a1 + a2 + · · · + an−1 From this inequality, we deduce that b21 +

b22 b2 b2 + 3 + · · · + n ≥ b21 + (b2 + b3 + · · · + bn )2 a1 a2 an−1 ≥ 2 |b1 (b2 + b3 + · · · + bn )| ≥ 2b1 (b2 + b3 + · · · + bn ), 311

as desired. ?F? 00.21. Positive real numbers x, y, z satisfy xyz = 1. Prove that the following inequality holds x2 + y 2 + z 2 + x + y + z ≥ 2 (xy + yz + zx) . (Russia 2000) Solution: From the AM-GM Inequality and the given hypothesis, we have √ x + y + z ≥ 3 3 xyz = 3. Therefore, it suffices to prove that x2 + y 2 + z 2 + 3 ≥ 2 (xy + yz + zx) . Multiplying both sides of this inequality by x + y + z > 0, we can rewrite it as (x2 + y 2 + z 2 )(x + y + z) + 3(x + y + z) ≥ 2(x + y + z)(xy + yz + zx). X  X  X  X  X X x = x3 + xy(x + y) and x xy = x2 Because X X xy(x + y) + 3xyz = xy(x + y) + 3, the latter inequality is equivalent to x3 + y 3 + z 3 + 3 (x + y + z) ≥ xy (x + y) + yz (y + z) + zx (z + x) + 6. Now, we use the AM-GM Inequality and the Schur’s Inequality (applied for third degree), and get   x3 + y 3 + z 3 + 3 (x + y + z) = x3 + y 3 + z 3 + (x + y + z) + 2 (x + y + z)
√ √ ≥ x3 + y 3 + z 3 + 3 3 xyz + 2 · 3 3 xyz
√ = x3 + y 3 + z 3 + 3xyz + 2 · 3 3 xyz ≥ xy (x + y) + yz (y + z) + zx (z + x) + 6. Therefore, the last inequality is valid and so, our proof is completed. Note that the equality holds if and only if x = y = z = 1. ?F? 00.22. Let x1 , x2 , . . . , xn be real numbers (n ≥ 2), satisfying the conditions −1 < x1 < x2 < · · · < xn < 1 and 13 13 x13 1 + x2 + · · · + xn = x1 + x2 + · · · + xn .

Prove that 13 13 x13 1 y1 + x2 y2 + · · · + xn yn < x1 y1 + x2 y2 + · · · + xn yn

for any real numbers y1 < y2 < · · · < yn . (Russia 2000) 312

Solution: Using the Abel’s summation formula, we have n X

x i yi −

i=1

n X

x13 i yi =

i=1

n X

yi (xi − x13 i )

i=1

= y1

n X

(xi − x13 i )+

i=1

=

n X

(yi − yi−1 )

i=2

n X

(xj − x13 j )

j=i

n X

n X (yi − yi−1 ) (xj − x13 j ).

i=2

j=i

Therefore, in order to prove the desired inequality, it suffices to prove that each term of the above sum is positive, and since yi − yi−1 > 0 for any i = 2, . . . , n, it suffices to prove that n X (xj − x13 j )>0 j=i

for any i = 2, . . . , n. Indeed, if xi > 0 then we have 1 > xn > · · · > xi > 0, from which it follows that xj > x13 j for any j = i, . . . , n, and therefore n X (xj −x13 j ) > 0. Alternatively, if xi ≤ 0 then we have −1 < x1 < . . . < xi ≤ 0, j=i

from which it follows that xj < x13 j for any j = 1, . . . , i − 1, so it is clear that i−1 X (xj − x13 j ) < 0, j=1

and since

n X

i−1 X (xj − x13 ) = − (xj − x13 j j ), we conclude that

j=i

j=1 n X (xj − x13 j ) > 0. j=i

This completes our proof. ?F? 00.23. Show that for all n ∈ N and x ∈ R, sinn 2x + (sinn x − cosn x)2 ≤ 1. (Russia 2000) Solution: For n = 0 and n = 1, the inequality becomes equality. In case n ≥ 2, denote a = sin x, b = cos x, then we have a2 + b2 = 1. The left hand side of the desired inequality equals (2ab)n + (an − bn )2 = a2n + b2n + (2n − 2)an bn , while the right hand side equals 2

2 n

1 = (a + b ) = a

2n

2n

+b

+

n−1 X j=1

313

 n 2(n−j) 2j a b . j

n−1 X

 n 2(n−j) 2j It thus suffices to prove that a b ≥ (2n − 2)an bn . We can do so j j=1   n−1 n−1 X n X  n 2(n−j) 2j by viewing a b as a sum of = 2n − 2 terms of the form j j j=1

j=1

a2(n−j) b2j , and then applying the AM-GM Inequality to these terms. ?F? 00.24. Let n ≥ 3 be a positive integer. Prove that for all positive real numbers a1 , a2 , . . . , an , we have a1 + a2 a2 + a3 an + a1 a1 + a2 + a3 an + a1 + a2 √ √ · ··· ≤ ··· . 2 2 2 2 2 2 2 (Saint Petersburg 2000) Solution: We take the indices of the ai modulo n. Observe that 4(ai−1 + ai + ai+1 )2 = [(2ai−1 + ai ) + (ai + 2ai+1 )]2 ≥ 4(2ai−1 + ai )(ai + 2ai+1 ) by the AM-GM Inequality. Equivalently, (ai−1 + ai + ai+1 )2 ≥ (2ai−1 + ai )(ai + 2ai+1 ). In addition to this inequality, note that (2ai−1 + ai )(ai + 2ai−1 ) ≥ 2(ai−1 + ai )2 since ai is positive number for each i. Multiplying these two inequalities together for i = 1, 2, . . . , n, and then taking the square root of each side of the resulting inequality, gives the desired result. ?F? 00.25. Let n ≥ 3 be an integer. Prove that for positive numbers x1 ≤ x2 ≤ · · · ≤ xn , xn x1 x1 x2 xn−1 xn + + ··· + ≥ x1 + x2 + · · · + xn . x2 x3 x1 (Saint Petersburg 2000) First solution: Suppose that 0 ≤ x ≤ y and 0 < a ≤ 1. We have 1 ≥ a and y ≥ x ≥ ax, implying that (1 − a)(y − ax) ≥ 0 or ax + ay ≤ a2 x + y. Dividing both sides of this final inequality by a, we find that y x + y ≤ ax + . a   xn−1 x1 We may set (x, y, a) = xn , xn · , in (1) to find that x2 x2 xn +

xn x1 xn−1 xn xn−1 xn ≤ + . x2 x2 x1 314

(1)

(2)

  xi+1 xi+1 , Furthermore, for i = 1, 2, . . . , n−2, we may set (x, y, a) = x1 , xn−1 · x2 xi+2 in (1) to find that xi + xn−1 ·

xi+1 xi+1 xi+1 xi+2 xi+1 xi+2 ≤ xi · + xn−1 · · = xi · + xn−1 · . x2 xi+2 x2 xi+1 xi+2 xi

Summing this inequality for i = 1, 2, . . . , n − 2 yields   x1 x2 xn−2 xn−1 xn−1 xn (x1 + · · · + xn−2 ) + xn−1 ≤ + ··· + + . x3 xn x2 Summing this last inequality with (2) yields the desired inequality. Second solution: We will prove the inequality by induction on n. For n = 3, it becomes x1 x2 x3 x1 x2 x3 + + ≥ x1 + x2 + x3 , x3 x2 x1 which is true since by the AM-GM Inequality, we have  X   x1 x2 x3 x1 x2 x3 x1 x2 x3 x1 + + = + ≥ 2(x1 + x2 + x3 ). 2 x3 x2 x1 x3 x2 Now, suppose that the equality holds for n ≥ 3 and let us prove it for n + 1, that is to prove x1 x2 xn−2 xn−1 xn−1 xn xn xn+1 xn+1 x1 +· · ·+ + + + ≥ x1 +x2 +· · ·+xn +xn+1 . x3 xn xn+1 x1 x2 Since x1 ≤ x2 ≤ · · · ≤ xn , we can apply the inductive hypothesis to get x1 x2 xn−2 xn−1 xn−1 xn xn x1 + ··· + + + ≥ x1 + x2 + · · · + xn . x3 xn x1 x2 Therefore, in order to prove that the inequality holds for n + 1, it suffices to prove that xn−1 xn xn xn+1 xn+1 x1 xn−1 xn xn x1 + + − xn+1 ≥ + . xn+1 x1 x2 x1 x2 We have xn−1 xn xn xn+1 xn+1 x1 xn−1 xn xn x1 + + − − − xn+1 = xn+1 x1 x2 x1 x2 xn (xn+1 − xn−1 ) x1 (xn+1 − xn ) xn−1 xn − x2n+1 = + + x1 x2 xn+1 xn (xn+1 − xn + xn − xn−1 ) x1 (xn+1 − xn ) xn−1 xn − x2n + x2n − x2n+1 = + + x1 x2 xn+1     xn x1 xn xn xn = (xn+1 − xn ) + − − 1 + (xn − xn−1 ) − . x1 x2 xn+1 x1 xn+1 Because xn+1 ≥ xn ≥ xn−1 , we find that xn x1 + ≥2 x1 x2

r

xn xn ≥1≥ , and x1 xn+1

xn xn ≥2≥ + 1. x2 xn+1 315

Therefore, the last quantity is obviously nonnegative and it implies that xn−1 xn xn xn+1 xn+1 x1 xn−1 xn xn x1 + + − xn+1 ≥ + , xn+1 x1 x2 x1 x2 as desired. Our proof is completed. ?F?
3 00.26. Let a, b, c, d be positive real numbers such that c2 + d2 = a2 + b2 . Prove that a3 b3 + ≥ 1. c d (Singapore 2000) Solution: From the given hypothesis and the Cauchy Schwarz Inequality, we have  r ! r !2   2    3 3 3 b a b3  √
2 √ 2 a + (ac + bd) =  + ac + bd c d c d !2 r r
2 a3 √ b3 √ = a2 + b2 · ac + · bd ≥ c d p = (a2 + b2 ) (c2 + d2 ) ≥ ac + bd. This implies that a3 b3 + ≥ 1, c d as desired. ?F? 00.27. Given that x, y, z are positive real numbers satisfying xyz = 32, find the minimum value of x2 + 4xy + 4y 2 + 2z 2 . (United Kingdom 2000) Solution: By the AM-GM Inequality, we have that x2 + 4xy + 4y 2 + 2z 2 = (x − 2y)2 + 4xy + 4xy + 2z 2 ≥ 4xy + 4xy + 2z 2 p p ≥ 3 3 4xy · 4xy · 2z 2 = 3 3 32(xyz)2 = 96. The equality holds when x = 2y and 4xy = 2z 2 , i.e. when x = 4, y = 2, z = 4. ?F? 00.28. Prove that for any nonnegative real numbers a, b, c, the following inequality holds   √ 2 √ √ √ 2 √ √
2 a+b+c √ 3 − abc ≤ max a− b , b− c , c− a . 3 (USA 2000) 316

First solution: We prove the stronger inequality √ √ 2 √ √ √ 2 √ √
2 3 a− b + b− c + c− a . a + b + c − 3 abc ≤

(1)

The conclusion is immediate if abc = 0, so we assume that a, b, c > 0. By multiplying a, b, c by a suitable factor, we may reduce to the case abc = 1. Without loss of generality, assume that a and b are both greater than or equal to 1, or both less than or equal to 1. The desired inequality now becomes √ √ √ a + b + c − 2 ab − 2 bc − 2 ca + 3 ≥ 0. √ 1 √ 1 1 , bc = √ , and ca = √ , we find that it is equivalent ab a b √ 1 2 2 P =a+b+ − 2 ab − √ − √ + 3 ≥ 0. ab a b

By replacing c = to

We have √ √ 2 1 P = a− b + − ab  √ √ 2 1 = a− b + √ a  √ √ 2 1 a− b + √ = a

2 2 √ − √ +3 a b 2 2  1 1 1 1 −1 + √ −1 + − − +1 ab a b b 2    2  1 1 1 −1 − 1 ≥ 0. −1 + √ −1 + a b b

Therefore the inequality (1) is proved and from it, the result follows immediately. Note that the equality holds if and only if a = b = c. Second solution: We again prove the stronger inequality (1), which can be written X [a − 2(ab)1/2 + (abc)1/3 ] ≥ 0. sym

But this inequality follows from adding the two inequalities X [a − 2a2/3 b1/3 + (abc)1/3 ] ≥ 0, sym

and X

(a2/3 b1/3 + a1/3 b2/3 − 2a1/2 b1/2 ) ≥ 0.

sym

The first of these is the Schur’s Inequality with x = a1/3 , y = b1/3 , z = c1/3 , while the second follows from the AM-GM Inequality. Third solution: Without loss of generality, assume that b is between a and c. The desired inequality reads √ √
3 a + b + c − 3 abc ≤ 3 a + c − 2 ac . As a function of b, the right side minus the left side is concave (its second 2(ac)1/3 derivative is − ), so its minimum value in the range [a, c] occurs at 3b5/3 317

one of the endpoints. Thus, without loss of generality, we may assume a = b. Moreover, we may rescale the variables to get a = b = 1. Now the claim reads 2c + 3c1/3 + 1 ≥ c1/2 . 6 This is an instance of weighted AM-GM Inequality. ?F? 01.1. Let a, b, c ≥ 0 such that a + b + c ≥ abc. Prove that the following inequality holds √ a2 + b2 + c2 ≥ 3abc. (Balkan 2001) Solution: From the AM-GM Inequality, we (a + b + c)3 ≥ 27abc. Multiplying both sides of this inequality by a + b + c > 0 and using the given hypothesis, we deduce that (a + b + c)4 = (a + b + c)3 (a + b + c) ≥ 27a2 b2 c2 . Taking the square root of each side of this inequality, we get √ (a + b + c)2 ≥ 3 3abc. On the other hand, it is clear that 3(a2 + b2 + c2 ) ≥ (a + b + c)2 from the Cauchy Schwarz Inequality. Combining this and the above inequality, we√get the result. It is easy to that the equality holds if and only if a = b = c = 3. ?F? 01.2. Let x1 , x2 , x3 be real numbers in [−1, 1], and let y1 , y2 , y3 be real numbers in [0, 1). Find the maximum possible value of the expression 1 − x1 1 − x2 1 − x3 · · . 1 − x2 y3 1 − x3 y1 1 − x1 y2 (Belarus 2001) Solution: Since 1−x1 y2 = (1−x1 )y2 + (1−y2 ) > 0, 1 +x1 ≥ 0 and 1 −y2 > 0, we have 1 − x1 2 (1 + x1 )(1 − y2 ) − =− ≤ 0. 1 − x1 y2 1 + y2 (1 + y2 )(1 − x1 y2 ) From this, we deduce that 1 − x1 2 ≤ ≤ 2. 1 − x 1 y2 1 + y2 Similarly, we have

1 − x2 ≤ 2, 1 − x2 y3

1 − x3 ≤ 2. 1 − x 3 y1

318

Now, note that from the given hypothesis, three expressions

1 − x1 1 − x2 , , 1 − x1 y2 1 − x2 y3

1 − x3 must be nonnegative. From this note and the three inequalities 1 − x 3 y1 above, we get 1 − x1 1 − x2 1 − x3 1 − x1 1 − x2 1 − x3 · · = · · ≤ 8. 1 − x2 y3 1 − x3 y1 1 − x1 y2 1 − x1 y2 1 − x2 y3 1 − x3 y1

On the other hand, it is easy to see that the equality can be attained (for example, take x1 = x2 = x3 = −1, y1 = y2 = y3 = 0), therefore the searched maximum is 8. ?F? 01.3. Let x and y be any two real numbers. Prove that 3(x + y + 1)2 + 1 ≥ 3xy. Under what conditions does equality hold? (Colombia 2001) x+y , then we have xy ≤ t2 (according to the 2 AM-GM Inequality). And hence, it suffices to prove that First solution: Denote t =

3(2t + 1)2 + 1 ≥ 3t2 . We have 3(2t+1)2 +1−3t2 = (3t+2)2 which is clearly nonnegative. Therefore, the above inequality holds and our proof is completed. Note that the equality 2 holds if and only if x = y = − . 3 Second solution: For any real numbers X and Y , we have   Y 2 3Y 2 2 2 X + Y + XY = X + + ≥ 0, 2 4 with equality if and only if X = Y = 0. Let x and y be 2 2 Letting X = x + and Y = y + , we obtain 3 3  2     2 2 2 2 x+ + y+ + x+ y+ 3 3 3

any two real numbers.

2 3

Expanding and multiplying by 3 gives 3x2 + 3y 2 + 3xy + 6x + 6y + 4 ≥ 0. This may be written as 3(x + y + 1)2 − 3xy + 1 ≥ 0, from which we arrive at the desired inequality. ?F? 319

 ≥ 0.

01.4. Let n (n ≥ 2) be an integer and let a1 , a2 , . . . , an be positive real numbers. Prove the inequality (a31 + 1)(a32 + 1) · · · (a3n + 1) ≥ (a21 a2 + 1) · · · (a2n a1 + 1). (Czech-Slovak-Polish 2001) First solution: We try to apply the Cauchy Schwarz Inequality for each factor of the product in the right hand side. It is natural to write (1+a21 a2 )2 ≤ (1 + a31 )(1 + a1 a22 ), since we need 1 + a31 , which appears in the left hand side. Similarly, we can write (1 + a22 a3 )2 ≤ (1 + a32 )(1 + a2 a23 ),

...,

(1 + a2n a1 )2 ≤ (1 + a3n )(1 + an a21 ).

Multiplying we obtain [(a21 a2 +1) · · · (a2n a1 +1)]2 ≤ [(a31 +1) · · · (a3n +1)][(1+a1 a22 ) · · · (1+an a21 )]. (∗) Now, we use again the same argument to find that [(1+a1 a22 ) · · · (1+an a21 )]2 ≤ [(a31 +1) · · · (a3n +1)][(a21 a2 +1) · · · (a2n a1 +1)]. (∗∗) Thus, if (a21 a2 + 1) · · · (a2n a1 + 1) ≥ (1 + a1 a22 ) · · · (1 + an a21 ), then (∗) will give the answer, otherwise (∗∗) will. It is easy to see that the equality holds if and only if a1 = a2 = · · · = an . Second solution: From the Holder’s Inequality, (a31 + 1)1/3 (a31 + 1)1/3 (a32 + 1)1/3 ≥ a21 a2 + 1, we get (a31 + 1)2 (a32 + 1) ≥ (a21 a2 + 1)3 . In the same manner, we can establish the folllowing inequalities a31 + 1


2



3 a32 + 1 ≥ a21 a2 + 1 ,

...,

a3n + 1


2



3 a31 + 1 ≥ a2n a1 + 1 .

Multiplying them up and taking the cube root of each side of the resulting inequality, we get the desired result. ?F? 01.5. Prove that for any positive real numbers a, b, c, the following inequality holds a b c √ +√ +√ ≥ 1. 2 2 2 a + 8bc b + 8ca c + 8ab (IMO 2001) First solution: Applying the AM-GM Inequality, we have √

a2

a 2a(a + b + c) 2a(a + b + c) √ = ≥ . 2 (a + b + c)2 + a2 + 8bc + 8bc 2(a + b + c) a + 8bc 320

By establishing the similar inequalities for the two other expressions, we get X X a a √ ≥ 2(a + b + c) . 2 + a2 + 8bc 2 (a + b + c) a + 8bc On the other hand, the Cauchy Schwarz Inequality implies that a

X

=

(a + b + c)2 + a2 + 8bc

X

≥X =

a2 a(a + b + c)2 + a3 + 8abc (a + b + c)2 [a(a + b + c)2 + a3 + 8abc]

(a + b + c)2 . (a + b + c)3 + a3 + b3 + c3 + 24abc

It coud be said that X

√

a 2(a + b + c)3 ≥ . (a + b + c)3 + a3 + b3 + c3 + 24abc a2 + 8bc

Therefore, in order to prove the original inequality, it suffices to prove that (a + b + c)3 ≥ a3 + b3 + c3 + 24abc. Of course, this is trivial since by the AM-GM Inequality, we have (a + b + c)3 = a3 + b3 + c3 + 3(a + b)(b + c)(c + a) ≥ a3 + b3 + c3 + 24abc. The equality occurs iff a = b = c. Second solution: First we shall prove that 4

√

a a3 ≥ 4 4 4 , a2 + 8bc a3 + b3 + c3

or equivalently, that 

4

4

4

a3 + b3 + c3

2

2

≥ a 3 (a2 + 8bc).

The AM-GM inequality yields 

4

4

4

a3 + b3 + c3

2

 4  4  8 4 4 4 4 − a3 = b3 + c3 a3 + a3 + b3 + c3 2

2

2

1

1

2

≥ 2b 3 c 3 · 8a 3 b 3 c 3 = 8a 3 bc. Thus 

4

4

4

a3 + b3 + c3

2

8

2

2

≥ a 3 + 8a 3 bc = a 3 (a2 + 8bc),

so our statement is proved. Similarly, we have 4

b b3 √ ≥ 4 4 4 , 2 b + 8ca a3 + b3 + c3

4

and 321

c c3 √ ≥ 4 4 4 . 2 c + 8ab a3 + b3 + c3

Adding these three inequalities yields √

a2

a b c +√ +√ ≥ 1. 2 2 + 8bc b + 8ca c + 8ab

Third solution: To remove the square roots, we make the following substitution a b c x= √ , y=√ , z=√ . 2 2 2 a + 8bc b + 8ca c + 8ab Clearly, x, y, z ∈ (0, 1). Our aim is to show that x + y + z ≥ 1. We notice that a2 x2 , = 8bc 1 − x2 Hence 1 = 512



b2 y2 , = 8ac 1 − y2

x2 1 − x2



y2 1 − y2

c2 z2 . = 8ab 1 − z2



z2 1 − z2

 .

Thus, we need to show that x+y +z ≥ 1 where 0 < x, y, z < 1 and (1−x2 )(1− y 2 )(1 − z 2 ) = 512(xyz)2 . We use contradiction method. Assume that there exist x, y, z satisfying both conditions 0 < x, y, z < 1, (1−x2 )(1−y 2 )(1−z 2 ) = 512(xyz)2 and also x+y +z < 1. We will prove that this is impossible. Indeed, from 1 > x + y + z, it follows that (1 − x2 )(1 − y 2 )(1 − z 2 ) > > [(x + y + z)2 − x2 ][(x + y + z)2 − y 2 ][(x + y + z)2 − z 2 ] = (x + x + y + z)(y + z)(x + y + y + z)(z + x)(x + y + z + z)(x + y) 1

1

1

1

1

1

≥ 4(x2 yz) 4 · 2(yz) 2 · 4(y 2 zx) 4 · 2(zx) 2 · 4(z 2 xy) 4 · 2(xy) 2 = 512(xyz)2 . This is a contradiction. ?F?

01.6. Let a, b, c be positive real numbers such that abc = 1. Prove that ab+c bc+a ca+b ≤ 1. (India 2001) Solution: The desired inequality is equivalent to (b + c) ln a + (c + a) ln b + (a + b) ln c ≤ 0. Without loss of generality, we may assume that a ≥ b ≥ c. By this assumption, we have b + c ≤ c + a ≤ a + b, and ln a ≥ ln b ≥ ln c. Therefore, by the Chebyshev’s Inequality, we get X

1 [(b + c) + (c + a) + (a + b)] (ln a + ln b + ln c) 3 2 = (a + b + c) ln(abc) = 0, 3

(b + c) ln a ≤

322

as claimed. It is easy to see that the equality holds if and only if a = b = c = 1. ?F? 01.7. Let x, y, z be positive real numbers such that xyz ≥ xy + yz + zx. Prove that xyz ≥ 3 (x + y + z) . (India 2001) Solution: Applying the well-known inequality (a + b + c)2 ≥ 3(ab + bc + ca) for the triple (a, b, c) = (xy, yz, zx), we deduce that (xy + yz + zx)2 ≥ 3xyz(x + y + z). On the other hand, from the given hypothesis, we have x2 y 2 z 2 ≥ (xy + yz + zx)2 . It follows that x2 y 2 z 2 ≥ 3xyz(x + y + z). Dividing both sides of the last inequality by xyz > 0, we get the desired result. Note that the equality holds if and only if x = y = z = 3. ?F? 01.8. Prove that for any real numbers a, b, c the following inequality holds (b + c − a)2 (c + a − b)2 (a + b − c)2 ≥ (b2 + c2 − a2 )(c2 + a2 − b2 )(a2 + b2 − c2 ). (Japan 2001) First solution: The inequality is equivalent to 3(a2 + b2 + c2 )(b + c − a)2 (c + a − b)2 (a + b − c)2 ≥ ≥ 3(a2 + b2 + c2 )(b2 + c2 − a2 )(c2 + a2 − b2 )(a2 + b2 − c2 ). By the Cauchy Schwarz Inequality, we have 3(a2 + b2 + c2 ) ≥ (a + b + c)2 , and thus it is suffice to prove that (a + b + c)2 (b + c − a)2 (c + a − b)2 (a + b − c)2 ≥ ≥ 3(a2 + b2 + c2 )(b2 + c2 − a2 )(c2 + a2 − b2 )(a2 + b2 − c2 ). Since  X X 2 (a + b + c)2 (b + c − a)2 (c + a − b)2 (a + b − c)2 = 2 a2 b2 − a4 and  X X  3(a2 + b2 + c2 )(b2 + c2 − a2 )(c2 + a2 − b2 )(a2 + b2 − c2 ) = 3 2 a4 b4 − a8 , the last inequality is equivalent to  X  X X 2 X  2 xy − x2 ≥ 3 2 x2 y 2 − x4 , 323

where x = a2 , y = b2 , z = c2 . This last one rewrites as X 4 x2 (x − y)(x − z) ≥ 0, which is a particular case of the Schur’s Inequality. Note that the equality holds iff a = b = c, or c = 0 and a = b, or b = 0 and c = a, or a = 0 and b = c.

Second solution: It is clear that we only need to consider the case a2 , b2 , c2 are the sidelengths of a triangle (because in the conversed case, the given inequality is trivial). Then, we will prove that [a2 − (b − c)2 ]2 ≥ (a2 + b2 − c2 )(a2 + c2 − b2 ). Indeed, it is equivalent to a4 − 2a2 (b − c)2 + (b − c)4 ≥ a4 − (b − c)2 (b + c)2 ≥ 0, or (b − c)2 (b2 + c2 − a2 ) ≥ 0. Of course, this is true and so the claim is proved. Proceeding by the same way, we can establish the similar inequalities as follows [b2 − (c − a)2 ]2 ≥ (b2 + c2 − a2 )(b2 + a2 − c2 ), [c2 − (a − b)2 ]2 ≥ (c2 + a2 − b2 )(c2 + b2 − a2 ). Multiplying these three inequalities, we get [a2 −(b−c)2 ]2 [b2 −(c−a)2 ]2 [c2 −(a−b)2 ]2 ≥ (a2 +b2 −c2 )2 (b2 +c2 −a2 )2 (c2 +a2 −b2 )2 , which is equivalent to (a + b − c)4 (b + c − a)4 (c + a − b)4 ≥≥ (a2 + b2 − c2 )2 (b2 + c2 − a2 )2 (c2 + a2 − b2 )2 . Now, taking square root for both sides, we can get the result. ?F? 01.9. Let a, b, c be the sidelengths of an acute-angled triangle. Prove that (a + b + c)(a2 + b2 + c2 )(a3 + b3 + c3 ) ≥ 4(a6 + b6 + c6 ). (Japan 2001) Solution: By the Cauchy Schwarz Inequality, we have (a + b + c)(a3 + b3 + c3 ) ≥ (a2 + b2 + c2 )2 , and therefore it is enough to prove that (a2 + b2 + c2 )3 ≥ 4(a6 + b6 + c6 ). 324

Set x = a2 , y = b2 , z = c2 . Since a, b, c are the sidelengths of an acute-angled triangle, (it is well-known and easy to prove that) the numbers x, y, z are theirselves the sidelengths of a triangle. The inequality to prove now becomes (x + y + z)3 ≥ 4(x3 + y 3 + z 3 ). According to the triangle Inequality, we proceed as follows (x + y + z)3 = x3 + y 3 + z 3 + 3x2 (y + z) + 3y 2 (z + x) + 3z 2 (x + y) + 6xyz ≥ x3 + y 3 + z 3 + 3x2 (y + z) + 3y 2 (z + x) + 3z 2 (x + y) ≥ x3 + y 3 + z 3 + 3x2 · x + 3y 2 · y + 3z 2 · z = 4(x3 + y 3 + z 3 ). Note that the equality holds if and only if the triangle is degenerated, having sidelengths of the type (0, t, t), where t is a positive real number. ?F? 01.10. Prove that for any real numbers x1 , x2 , . . . , xn , y1 , y2 , . . . , yn such that x21 + x22 + · · · + x2n = y12 + y22 + · · · + yn2 = 1, ! n X x k yk . (x1 y2 − x2 y1 )2 ≤ 2 1 − k=1

(Korea 2001) First solution: We clearly have the inequality X

2

(x1 y2 − x2 y1 ) ≤

n X

2

(xi yj − xj yi ) =

i=1

1≤i 0, p p (a2 b + b2 c + c2 a) (ab2 + bc2 + ca2 ) ≥ abc+ 3 (a3 + abc) (b3 + abc) (c3 + abc). (Korea 2001) Solution: Dividing by abc, it becomes s s    2  2  a b c c a b a2 b c + + + + ≥ abc + 3 +1 +1 +1 . c a b a b c bc ca ab a b c After the substitution x = , y = , z = , we obtain the constraint xyz = 1. b c a It takes the form s  x y  z p (x + y + z) (xy + yz + zx) ≥ 1 + 3 +1 +1 +1 . z x y 326

From the constraint xyz = 1, we find two identities  y  z x +1 +1 + 1 = (z + x)(x + y)(y + z), z x y and (x + y + z) (xy + yz + zx) = (x+y)(y+z)(z+x)+xyz = (x+y)(y+z)(z+x)+1. p p 3 Letting p = 3 (x + y)(y + z)(z + x), the inequality now q becomes p √+ 1 ≥ √ √ 1 + p. Applying the AM-GM Inequality, we have p ≥ 3 2 xy · 2 yz · 2 zx = 2. It follows that (p3 + 1) − (1 + p)2 = p(p + 1)(p − 2) ≥ 0. The proof is completed. Note that the equality holds if and only if a = b = c. ?F? 01.13. Prove that if a, b, c > 0 have product 1, then (a + b)(b + c)(c + a) ≥ 4(a + b + c − 1). (MOSP 2001) First solution: Using the identity (a + b)(b + c)(c + a) = (a + b + c)(ab + bc + ca) − 1, we reduce the problem to the following one ab + bc + ca +

3 ≥ 4. a+b+c

Now, we can apply the AM-GM Inequality in the following form s 3 ab + bc + ca 3 (ab + bc + ca)3 ab + bc + ca + =3· + ≥44 . a+b+c 3 a+b+c 9(a + b + c) And so, it is enough to prove that (ab + bc + ca)3 ≥ 9(a + b + c). But this is easy, because we clearly have ab + bc + ca ≥ 3 and (ab + bc + ca)2 ≥ 3abc(a + b + c) = 3(a + b + c). It is easy to see that the equality holds if and only if a = b = c = 1. 8 Second solution: We will use the fact that (a + b)(b + c)(c + a) ≥ (a + b + 9 c)(ab + bc + ca). So, it is enough to prove that 2 1 (ab + bc + ca) + ≥ 1. 9 a+b+c Using the AM-GM Inequality, we can write s 2 1 (ab + bc + ca)2 (ab + bc + ca) + ≥33 ≥ 1, 9 a+b+c 81(a + b + c) 327

because (ab + bc + ca)2 ≥ 3abc(a + b + c) = 3(a + b + c). Third solution: By homogenizing, we can write the inequality as √ 3 (a + b)(b + c)(c + a) + 4abc ≥ 4 a2 b2 c2 (a + b + c). Without loss of generality, we can assume that a ≥ b ≥ c. Now, we write the inequality in form i   h √ √ √ 3 3 3 (b + c) (a + b)(a + c) − 4 a2 b2 c2 ≥ 4 a2 b2 c2 a − abc . Because (a + b)(a + c) = a2 + ab + ac + bc, it is equivalent to     √ √ √ 3 3 3 (b + c) a2 + ab + bc + ca − 4 a2 b2 c2 ≥ 4 a2 b2 c2 a − abc . √ 3 Using again the estimation ab + bc + ca ≥ 3 a2 b2 c2 , it suffices to prove that     √ √ √ 3 3 3 (b + c) a2 − a2 b2 c2 ≥ 4 a2 b2 c2 a − abc , or equivalently,

  √ √ 3 3 (b + c) a + abc ≥ 4 a2 b2 c2 .

Of course, this is obvious since by the AM-GM Inequality, we have q   √ √ √ √ 3 3 3 (b + c) a + abc ≥ 2 bc · 2 a abc = 4 a2 b2 c2 . ?F? 01.14. Show that the inequality n X i=1

  X n n ixi ≤ + xii 2 i=1

holds for every integer n ≥ 2 and all real numbers x1 , x2 , . . . , xN ≥ 0. (Poland 2001) Solution: From the Bernoulli’s Inequality, for each nonnegative real number x and for each positive integer k, we have xk = [1 + (x − 1)]k ≥ 1 + k(x − 1), and from this, we deduce that kx ≤ xk + k − 1, with equality, for k ≥ 2, if and only if x = 1. Using this for each xi and summing leads to   X n n n X X n i ixi ≤ [xi + (i − 1)] = + xii , 2 i=1

i=1

i=1

as desired. Note that the equality holds if and only if x2 = · · · = xn = 1 (and x1 ≥ 0 is arbitrary). ?F? 01.15. Let a and b be positive real numbers in the interval (0, 1]. Prove that √

1 1 2 +√ ≤√ . 2 2 1 + ab 1+a 1+b 328

(Russia 2001) Solution: Because a and b are positive real numbers, there are angles x and y, with 0◦ < x, y < 90◦ , such that tan x = a and tan y = b. The desired inequality is clearly true when a = b. Hence we assume that a 6= b, or equivalently, x 6= y. 1 Then 1 + a2 = sec2 x and √ = cos x. Note that 1 + a2 1 + ab =

cos x cos y + sin x sin y cos(x − y) = cos x cos y cos x cos y

by the addition and subtraction formulas. The desired inequality reduces to r cos x cos y cos x + cos y ≤ 2 . cos(x − y) Squaring both sides, we can rewrite it as cos2 x + cos2 y + 2 cos x cos y ≤

4 cos x cos y . cos(x − y)

Because 0◦ < |x − y| < 90◦ , it follows that 0 < cos(x − y) < 1. Hence 2 cos x cos y . It suffices to show that 2 cos x cos y ≤ cos(x − y) cos(x − y)(cos2 x + cos2 y) ≤ 2 cos x cos y, or cos(x − y)(cos 2x + cos 2y + 2) ≤ 4 cos x cos y by the double-angle formulas. By the sum-to-product formulas, the last inequality is equivalent to cos(x − y)[2 cos(x − y) cos(x + y) + 2] ≤ 2[cos(x − y) + cos(x + y)], or cos 2(x − y) cos(x + y) ≤ cos(x + y), which is clearly true, because for 0 < a, b ≤ 1, we have 0◦ < x, y ≤ 45◦ , and so 0◦ < x + y ≤ 90◦ and cos(x + y) > 0. This completes our proof. ?F? 01.16. Let a, b, c, x, y, z be positive real numbers such that x + y + z = 1. Prove that p ax + by + cz + 2 (xy + yz + zx)(ab + bc + ca) ≤ a + b + c. (Ukraine 2001) First solution: We will use Inequality twice. First, we p √ the Cauchy Schwarz can write ax + by + cz ≤ a2 + b2 + c2 · x2 + y 2 + z 2 and then we apply again the Cauchy Schwarz Inequality to obtain p ax + by + cz + 2 (xy + yz + zx)(ab + bc + ca) ≤ s X qX qX q X 2 2 ≤ a · x + 2 xy · 2 ab cyc

≤

qX

a2 + 2

X

ab ·

329

qX

x2 + 2

X

xy =

X

a.

Second solution: The inequality being homogeneous in a, b, c, so we can assume that a + b + c = 1. We apply this time the AM-GM Inequality and we find that the left hand side of the original inequality is not greater than ax + by + cz + xy + yz + zx + ab + bc + ca. Consequently, 1 − x2 − y 2 − z 2 1 − a2 − b2 − c2 + 2 2 ≤ 1 − ax − by − cz,

xy + yz + zx + ab + bc + ca =

the last one being equivalent to (x − a)2 + (x − b)2 + (x − c)2 ≥ 0. Third solution: Applying the AM-GM Inequality, we have p 2 (ab + bc + ca)(xy + yz + zx) = s   yz(b + c) bc(y + z) x(b + c) + a(y + z) + =2 b+c y+z ≤ a(y + z) +

bc(y + z) yz(b + c) + x(b + c) + . b+c y+z

Therefore, if we denote with P the left hand side of the original inequality, then we find that P ≤ ax + by + cz + a(y + z) +

bc(y + z) yz(b + c) + x(b + c) + . b+c y+z

And so, it suffices to prove that ax + by + cz + a(y + z) +

bc(y + z) yz(b + c) + x(b + c) + ≤ (a + b + c)(x + y + z), b+c y+z

or bz + cy ≥

bc(y + z) yz(b + c) + . b+c y+z

The last one is true since by the Cauchy Schwarz Inequality, we have bz + cy −

bc(y + z) b2 z + c2 y (b2 z + c2 y)(y + z) = = b+c b+c (b + c)(y + z) √ √
2 b zy + c yz yz(b + c) ≥ = . (b + c)(y + z) y+z ?F?

01.17. Let a, b, c be positive real numbers such that a + b + c ≥ abc. Prove that at least two of the inequalities 2 3 6 + + ≥ 6, a b c

2 3 6 + + ≥ 6, b c a 330

2 3 6 + + ≥6 c a b

are true. (USA 2001) 1 1 1 Solution: Denote x = , y = , z = , then we have xy + yz + zx ≥ 1, and a b c we are required to prove that: There are at least two of the inequalities 2x + 3y + 6z ≥ 6,

2y + 3z + 6x ≥ 6,

2z + 3x + 6y ≥ 6

are true. We have that (2x + 3y + 6z) + (2y + 3z + 6x) = 8x + 5y + 9z, and (8x + 5y + 9z)2 − 144(xy + yz + zx) = 64x2 − 64xy + 25y 2 − 54yz + 81z 2 = 16(2x − y)2 + 9(y − 3z)2 ≥ 0, therefore √ (2x + 3y + 6z) + (2y + 3z + 6x) ≥ 12 xy + yz + zx ≥ 12, and it follows that max {2x + 3y + 6z, 2y + 3z + 6x} ≥ 6. Without loss of generality, we may assume that max {2x + 3y + 6z, 2y + 3z + 6x} = 2x + 3y + 6z, then from the above inequality, we have 2x + 3y + 6z ≥ 6. On the other hand, proceeding by the same way as above, we can also find that (2y + 3z + 6x) + (2z + 3x + 6y) ≥ 12, from which it follows max {2y + 3z + 6x, 2z + 3x + 6y} ≥ 6. These arguments show that there are at least 2 inequalities among the given inequalities holds. ?F? 01.18. Prove that for any nonnegative real numbers a, b, c such that a2 + b2 + c2 + abc = 4, we have 0 ≤ ab + bc + ca − abc ≤ 2. (USA 2001) First solution: From the condition, at least one of a, b and c does not exceed 1, say a ≤ 1. Then ab + bc + ca − abc = a(b + c) + bc(1 − a) ≥ 0. To obtain the equality, we have a(b+c) = bc(1−a) = 0. If a = 1, then b+c = 0 or b = c = 0, which contradicts the given condition a2 + b2 + c2 + abc = 4. 331

Hence 1 − a 6= 0 and only one of b and c is 0. Without loss of generality, say b = 0. Therefore b + c > 0 and a = 0. Plugging a = b = 0 back into the given condition gives c = 2. By permutation, the lower bound holds if and only if (a, b, c) is one of the triples (2, 0, 0), (0, 2, 0), and (0, 0, 2). Now we prove the upper bound. Let us note that at least two of the three numbers a, b, and c are both greater than or equal to 1 or less than or equal to 1. Without loss of generality, we assume that the numbers with this property are b and c. Then we have (1 − b)(1 − c) ≥ 0.

(1)

The given equality a2 + b2 + c2 + abc = 4 and the inequality b2 + c2 ≥ 2bc imply a2 + 2bc + abc ≤ 4, or bc(2 + a) ≤ 4 − a2 . Dividing both sides of the last inequality by 2 + a yields bc ≤ 2 − a.

(2)

Combining (1) and (2) gives ab + bc + ca − abc ≤ a(b + c) + (2 − a) − abc = 2 − a(1 + bc − b − c) = 2 − a(1 − b)(1 − c) ≤ 2, as desired. The last equality holds if and only if b = c and a(1 − b)(1 − c) = 0. Hence, the equality for√the√upper if and √ holds √ only √ if
(a, b, c) is one of
√bound
the triples (1, 1, 1), 0, 2, 2 , 2, 0, 2 , and 2, 2, 0 . Second solution: We won’t prove again the lower part, since this is an easy problem. Let us concentrate on the upper bound. Let a ≥ b ≥ c and let a = x + y, b = x − y (x ≥ y ≥ 0). The hypothesis becomes x2 (2 + c) + y 2 (2 − c) = 4 − c2 , and we have to prove that (x2 − y 2 )(1 − c) ≤ 2(1 − xc). Since 2+c 2 y2 = 2 + c − x , the problem asks to prove the inequality 2−c 4x2 − (4 − c2 ) (1 − c) ≤ 2(1 − xc). 2−c 2+c 2 Of course, we have c ≤ 1 and 0 ≤ y 2 = 2 + c − x , therefore x2 ≤ 2 − c 2 − c  √  √ and hence x ≤ 2 − c. Now, consider the function f : 0, 2 − c → R, f (x) = 2(1 − cx) −

4x2 − (4 − c2 ) (1 − c). 2−c

8x(1 − c) We have f 0 (x) = −2c − ≤ 0 and thus f is decreasing and f (x) ≥ 2−c

√ √ f 2 − c . So we have to prove that f 2 − c ≥ 0, or equivalently √
2 1 − c 2 − c ≥ (2 − c)(1 − c). 332

By some simple calculations, we find that it is equivalent to c 1 − 0, clearly true. Thus, the problem is solved.

√


2 2−c ≥

Third solution: We give another way to prove the upper part ab + bc + ca − abc ≤ 2. Letting a = 2p, b = 2q, c = 2r, we get p2 + q 2 + r2 + 2pqr = 1. Therefore, h π i we may put a = 2 cos A, b = 2 cos B, c = 2 cos C for some A, B, C ∈ 0, with 2 A + B + C = π. We are required to prove 1 cos A cos B + cos B cos C + cos C cos A − 2 cos A cos B cos C ≤ . 2 π Assume that A ≥ or 1 − 2 cos A ≥ 0. Note that 3 cos A cos B + cos B cos C + cos C cos A − 2 cos A cos B cos C = = cos A(cos B + cos C) + cos B cos C(1 − 2 cos A). We apply the Jensen’s Inequality to deduce cos B + cos C ≤ 32 − cos A. Note that 2 cos B cos C = cos(B − C) + cos(B + C) ≤ 1 − cos A. These imply that cos A(cos B + cos C) + cos B cos C(1 − 2 cos A) ≤     3 1 − cos A ≤ cos A − cos A + (1 − 2 cos A). 2 2     3 1 − cos A However, it’s easy to verify that cos A − cos A + (1−2 cos A) = 2 2 1 . 2 ?F? 01.19. Let x, y, z be positive real numbers satisfying  √ √ 1 1 (i) √ ≤ z ≤ min x 2, y 3 ; 2 √ 2√ (ii) x + ≥ 6; √ √z 3 √ (iii) y 3 + z 10 ≥ 2 5. 1 2 3 Find the maximum of P (x, y, z) = 2 + 2 + 2 . x y z (Vietnam 2001) √ √ 1 2 First solution: From the first condition, we have 2 ≥ ≥ , and it z x 2 1 z 1 follows that 2 ≤ 2, 2 ≤ . The second condition leads us to x2 + 3z 2 ≥ 3, z x 2 2 2z 2 2 and from it, we deduce that + 2 ≥ 2 . Thus 3 x x   1 1 2 1 1 2 2z 2 1 z2 + = 2 + 2 − 2 ≤ + 2 + 2 1− 2 x2 z 2 x z x 3 x z x   2 2 2 2z z 8 ≤ + 2 +2 1− 2 = . 3 x x 3 333

Similarly, using the first and third conditions, we get 1 1 13 + 2 ≤ . 2 y z 5 From these two inequalities, we obtain   8 26 1 1 1 118 1 + 2 ≤ + = . P = 2 + 2 +2 2 x z y z 3 3 15 118 if and only if In addition, from above proofs, it is easy to see that P = 15 r r 3 5 1 x= ,y = , z = √ . Obviously, these values of x, y, z satisfy (i), (ii) 2 3 2 118 . and (iii). So max P = 15 Second solution: From the given conditions, we have x, y > z and r    √ √ 10 √ x≥ 3 2−z , y ≥ 2−z . 3 To find the maximum of P, we need to consider two cases: √ √ √
1 6 Case 1. 3 2 − z ≥ z. In this case, we find that √ ≤ z ≤ √ , and 2 3+1 3 1 2 3 1 2 + 2 + 2+ 2 ≤ √
2 + 10 √ 2
2 x y z z 3 2−z 2−z 3 3 14 = √
2 + 2 = f (z). z 15 2 − z Now, we find that f 00 (z) =

28 18 + √
4 > 0, therefore f (z) is convex, 4 z 5 2−z

√ 1 6 and since √ ≤ z ≤ √ , it follows that 2 3+1 (  ( √ !) √ )  1 6 118 58 + 29 3 118 f (z) ≤ max f √ ,f √ = max , = . 15 15 15 2 3+1 Case 2.

√

3

√

√




2 − z ≤ z. In this case, we find that z ≥ √

6 . Therefore, 3+1

by using the inequalities x > z and y > z, we get 1 2 3 6 + 2+ 2 ≤ 2 ≤ 2 x y z z

√ 6 118 √ !2 = 4 + 2 3 < 15 . 6 √ 3+1

Actually, we have proved in both cases the following inequality P =

1 2 3 118 + 2+ 2 ≤ . 2 x y z 15 334

r

r

5 1 , z = √ , we can easily see that the conditions 3 2 118 118 . Therefore max P = . (i), (ii) and (iii) are still satisfied and also P = 15 15 ?F?

Moreover, let x =

3 ,y = 2

01.20. Let x, y, z be positive real numbers such that 2 (i) ≤ z ≤ min {x, y} ; 5 4 (ii) xz ≥ ; 15 1 (iii) yz ≥ . 5 Determine the maximum possible value of P (x, y, z) =

1 2 3 + + . x y z (Vietnam 2001)

4 1 First solution: From the given condition, we have x ≥ ,y ≥ . To find 15z 5z the maximum value of P, we will consider two cases 4 2 2 The first case is when ≥ z. In this case, we have ≤ z ≤ √ . Now, 15z 5 15 observe that 15z 3 55 3 1 2 3 + + ≤ + 2 · 5z + = z + = f (z). x y z 4 z 4 z It is easy to check that f (z) is convex, therefore ( √ )      10 5 2 2 = max 13, f (z) ≤ max f ,f √ = 13, 5 3 15 4 2 2 = ,y = and hence, we deduce that P ≤ 13 with equality iff z = , x = 5 15z 3 1 1 = . 5z 2 4 2 ≤ z. In this case, we have z ≥ √ , and hence The second case is when 15z 15 √ 1 2 3 6 + + ≤ ≤ 3 15 < 13. x y z z 1 2 2 So, in both cases, we always have P ≤ 13 with equality iff x = , y = , z = . 3 2 5 Therefore, the searched maximum of P is 13. Second solution: Applying the AM-GM Inequality in combination with the given hypothesis, we deduce that p √ 3x + 5z ≥ 2 15xz ≥ 4, 4y + 5z ≥ 4 5yz ≥ 4. 335

From this, it follows that 2 5z 3 ≤ + , x 2x 2

2 5z ≤ + 2. y 2y

Using these two inequalities with noting that 1 −

z 2 z ≥ 0, 1 − ≥ 0 and z ≥ , x y 5

we have 1 1 2 1 z  5z 3 1 z + = + 1− ≤ + + 1− x z x z x 2x 2 z x 3 5 z 5z + + 1− = 4, ≤ 2x 2 2 x and     1 1 z 5z z 2 1 1 1− ≤ 1− + = + +2+ y z y z y 2y z y   5 z 9 5z +2+ 1− = . ≤ 2y 2 y 2 Therefore, we conclude that     1 2 3 1 1 1 1 9 P = + + = + +2 + ≤ 4 + 2 · = 13, x y z x z y z 2 1 2 4 , yz = , z = , i.e. when with equality if and only if 3x = 4y = 5z, xz = 15 5 5 2 1 2 x = ,y = ,z = . 3 2 5 ?F? 1 2 3 01.21. Find the minimum value of the expression + + where a, b, c are a b c positive real numbers such that 21ab + 2bc + 8ca ≤ 12. (Vietnam 2001) 1 2 3 First solution: Let x = , y = , z = . Then it is easy to check that a b c the condition of the problem becomes 2xyz ≥ 2x + 4y + 7z. And we need to minimize x + y + z. But from the hypothesis, we find that z(2xy − 7) ≥ 2x + 4y 2x + 4y . Now, we transform the expression so that and hence 2xy > 7, z ≥ 2xy − 7 after one application of the AM-GM Inequality, the numerator 2xy − 7 should vanish r   2x + 14 2x + 4y 11 7 11 7 x x+y+z ≥ x+y+ = x+ + y − + ≥ x+ +2 1 + 2 . 2xy − 7 2x 2x 2xy − 7 2x x 7 3+ 7 x and so x + y + z ≥ But, it is immediate to prove that 2 1 + 2 ≥ x 2 3 9 15 5 +x+ ≥ . We have equality for x = 3, y = , z = 2. Therefore, in the 2 x 2 2 15 1 4 3 initial problem, the answer is , achieved for a = , b = , c = . 2 3 5 2 r

336

Second solution: We use the same substitution and reduce the problem to finding the minimum value of x + y + z when 2xyz ≥ 2x + 4y + 7z. Applying the weighted AM-GM Inequality, we find that  x+y+z ≥ 1

5x 2

2 5

1

(3y)

1 3

7



15z 4

1

4

15

.

1

7

And also 2x + 4y + 7z ≥ 10 5 · 12 3 · 5 15 · x 5 · y 3 · z 15 . This means that 225 (x + y + z)2 (2x + 4y + 7z) ≥ xyz. Because 2xyz ≥ 2x + 4y + 7z, we will 2 have 225 15 (x + y + z)2 ≥ , or x + y + z ≥ , 4 2 5 with equality for x = 3, y = , z = 2. 2 ?F? 02.1. Let x, y, z be positive real numbers such that 1 1 1 + + = 1. x y z Prove that √

x + yz +

√

y + zx +

√

z + xy ≥

√

xyz +

√

x+

√

y+

√

z.

(APMO 2002) 1 1 1 , b = , c = . Then from the given hypothesis, we have x y z a + b + c = 1, and the original inequality becomes r r r 1 1 1 1 1 1 1 1 1 1 +√ +√ +√ , + + + + + ≥√ a bc b ca c ab a c abc b Solution: Let a =

which is equivalent to √ √ √ √ √ √ a + bc + b + ca + c + ab ≥ ab + bc + ca + 1, or

Xp √ √ √ a(a + b + c) + bc ≥ ab + bc + ca + a + b + c.

By the Cauchy Schwarz Inequality, we have Xp Xp X √  X X√ a(a + b + c) + bc = (a + b)(a + c) ≥ a + bc = a+ ab. Therefore, the last inequality is true and thus, our problem is solved. Note that the equality holds if and only if x = y = z = 3. ?F? 02.2. Let a, b, c be positive real numbers. Prove that a3 b3 c3 a2 b2 c2 + + ≥ + + . b2 c2 a2 b c a 337

(Balkan (Shortslit) 2002) Solution: Using the AM-GM Inequality, we have a3 2a2 +a≥ , 2 b b

b3 2b2 +b≥ , 2 c c

c3 2c2 +c≥ . 2 a a

Adding up these three inequalities, we deduce that  2  c3 b2 c2 a3 b3 a + 2 + 2 ≥2 − (a + b + c). + + b2 c a b c a From this inequality, we see that it suffices to prove that a2 b2 c2 + + ≥ a + b + c, b c a which is true since the AM-GM Inequality yields a2 + b ≥ 2a, b

b2 + c ≥ 2b, c

c2 + a ≥ 2c. a

The equality occurs if and only if a = b = c. ?F? 02.3. If a, b, c are positive real numbers such that abc = 2, then √ √ √ a3 + b3 + c3 ≥ a b + c + b c + a + c a + b. (Balkan (Shortlist) 2002) Solution: Because 2(a3 + b3 + c3 ) − a2 (b + c) − b2 (c + a) − c2 (a + b) =       = a3 + b3 − ab(a + b) + b3 + c3 − bc(b + c) + c3 + a3 − ca(a + b) = (a − b)2 (a + b) + (b − c)2 (b + c) + (c − a)2 (c + a) ≥ 0, it suffices to prove that  √  √ √ a2 (b + c) + b2 (c + a) + c2 (a + b) ≥ 2 a b + c + b c + a + c a + b . On the other hand, the Cauchy Schwarz Inequality yields 2 √ √ 1 √ a b+c+b c+a+c a+b , a2 (b + c) + b2 (c + a) + c2 (a + b) ≥ 3 so it is enough to prove that √ √ √ a b + c + b c + a + c a + b ≥ 6, which is true since from the AM-GM Inequality, we have q p √ √ √ 3 a b + c + b c + a + c a + b ≥ 3 abc (a + b)(b + c)(c + a) q √ 3 ≥ 3 abc 8abc = 6. 338

The equality holds if and only if a = b = c = ?F?

√ 3

2.

02.4. Let a, b, c be real numbers such that a2 +b2 +c2 = 1. Prove the inequality a2 b2 c2 3 + + ≥ . 1 + 2bc 1 + 2ca 1 + 2ab 5 (Bosnia and Herzegovina 2002) Solution: In view of the inequality 2xy ≤ x2 + y 2 and the observation that 1 + 2bc = a2 + (b + c)2 > 0, etc., we see that a2 b2 c2 a2 b2 c2 + + ≥ + + 2 2 2 2 1 + 2bc 1 + 2ca 1 + 2ab 1+b +c 1+c +a 1 + a2 + b2 b2 c2 a2 + + = 2 − a2  2 − b2 2 − c2  1 1 1 = −3 + 2 + + . 2 − a2 2 − b2 2 − c2 On the other hand, according to the Cauchy Schwarz Inequality, one can find that 1 1 1 9 9 + + ≥ = , 2 − a2 2 − b2 2 − c2 6 − a2 − b2 − c2 5 from which it follows a2 b2 c2 9 3 + + ≥ −3 + 2 · = , 1 + 2bc 1 + 2ca 1 + 2ab 5 5 1 as desired. Note that the equality holds if and only if a = b = c = ± √ . 3 ?F? 02.5. Show that for any positive real numbers a, b, c, we have a3 b3 c3 + + ≥ a + b + c. bc ca ab (Canada 2002) First solution: From the AM-GM Inequality, we get r 3 a3 3 a +b+c≥3 · b · c = 3a. bc bc Adding this to the two analogous inequalities, we get the desired result. Note that the equality holds if and only if a = b = c. Second solution: Multiplying both sides of the desired inequality by abc > 0, we can rewrite it as a4 + b4 + c4 ≥ abc (a + b + c) . 339

Now, we apply the well-known inequality x2 + y 2 + z 2 ≥ xy + yz + zx with (x, y, z) = (a2 , b2 , c2 ) and (x, y, z) = (ab, bc, ca) to get a4 + b4 + c4 ≥ a2 b2 + b2 c2 + c2 a2 ,

and a2 b2 + b2 c2 + c2 a2 ≥ abc(a + b + c).

Combining these two inequalities, we get the desired result. ?F? 02.6. Assume (P1 , P2 , . . . , Pn ) (n ≥ 2) is an arbitrary permutation of (1, 2, . . . , n). Prove that 1 1 1 n−1 + + ... + > . P1 + P2 P2 + P3 Pn−1 + Pn n+2 (China 2002) Solution: Denote with A the left hand side of the desired inequality. By the Cauchy Schwarz Inequality, we find that (n − 1)2 (P1 + P2 ) + (P2 + P3 ) + · · · + (Pn−1 + Pn ) (n − 1)2 (n − 1)2 = = 2(P1 + P2 + · · · + Pn ) − P1 − Pn 2(1 + 2 + · · · + n) − P1 − Pn 2 2 (n − 1)2 (n − 1) (n − 1) = ≥ = n(n + 1) − P1 − Pn n(n + 1) − 1 − 2 (n − 1)(n + 2) − 1 2 (n − 1) n−1 > = , (n − 1)(n + 2) n+2

A≥

as desired. ?F? 02.7. Let x, y be positive real numbers such that x + y = 2. Prove that x3 y 3 (x3 + y 3 ) ≤ 2. (India 2002) Solution: According to the AM-GM Inequality and the given hypothesis, we find that x3 y 3 (x3 + y 3 ) = 2x3 y 3 (x2 − xy + y 2 ) = 2 · xy · xy · xy · (x2 − xy + y 2 )  4 xy + xy + xy + x2 − xy + y 2 ≤2 4  4 2 (x + y) =2 = 2, 4 as desired. The equality holds if and only if x = y = 1. ?F? 340

02.8. For any positive real numbers a, b, c, show that the following inequality holds a b c c+a a+b b+c + + ≥ + + . b c a c+b a+c b+a (India 2002) First solution: Rewrite the inequality in the form       a a+c b a+b c b+c − +1 + − +1 + − + 1 ≥ 3, b b+c c a+c a b+a or

b2 + ca c2 + ab a2 + bc + + ≥ 3. b(b + c) c(c + a) a(a + b)

From this, using the AM-GM Inequality, we find that this inequality is deduced from (a2 + bc)(b2 + ca)(c2 + ab) ≥ abc(a + b)(b + c)(c + a). Since (a2 + bc)(b2 + ca) − ab(a + c)(b + c) = c(a + b)(a − b)2 , we get (a2 + bc)(b2 + ca) ≥ ab(a + c)(b + c), and similarly, (b2 + ca)(c2 + ab) ≥ bc(a + b)(c + a) and (a2 + bc)(c2 + ab) ≥ ac(a + b)(b + c). Thus (a2 + bc)2 (b2 + ca)2 (c2 + ab)2 ≥ a2 b2 c2 (a + b)2 (b + c)2 (c + a)2 , and so (a2 + bc)(b2 + ca)(c2 + ab) ≥ abc(a + b)(b + c)(c + a). It is easy to see that the equality holds if and only if a = b = c. Second solution: Let us take x =

b c a , y = , z = . Observe that b c a

a+c 1 + xy 1−x = =x+ . b+c 1+y 1+y Using similar relations, the problem reduces to proving that if xyz = 1, then x−1 y−1 z−1 + + ≥ 0. y+1 z+1 x+1 By expanding, we find that it is equivalent to (x2 − 1)(z + 1) + (y 2 − 1)(x + 1) + (z 2 − 1)(y + 1) ≥ 0, or (xy 2 + yz 2 + zx2 ) + (x2 + y 2 + z 2 ) ≥ (x + y + z) + 3. But this inequality is very easy. Indeed, using the AM-GM Inequality, we have xy 2 + yz 2 + zx2 ≥ 3 and so it remains to prove that x2 + y 2 + z 2 ≥ x + y + z, which follows from the inequalities x2 + y 2 + z 2 ≥

(x + y + z)2 ≥ x + y + z. 3 341

?F? 02.9. Let x1 , x2 , . . . , xn be positive real numbers. Prove that √ xn x1 x2 + ··· + < n. + 2 2 2 2 2 2 1 + x1 1 + x1 + x2 1 + x1 + x2 + · · · + xn (India 2002) Solution: Applying the Cauchy Schwarz Inequality, we can see that the left hand side of the original inequality is not greater than s √ x2n x22 x21 + ··· + . + n 2 2 2 2 2 2 (1 + x1 ) (1 + x1 + x2 ) (1 + x1 + x22 + · · · + x2n )2 Now, we have x21 x21 1 ≤ =1− , 2 (1 + x1 )2 1 + x21 1 + x21 and for all 2 ≤ i ≤ n, we can assert that x2i x2i ≤ (1 + x21 + · · · + x2i )2 (1 + x21 + · · · + x2i−1 )(1 + x21 + · · · + x2i ) 1 1 − . = 2 2 2 1 + x1 + · · · + xi−1 1 + x1 + · · · + x2i Adding the preceding expressions, we obtain n X i=1

x2i 1 ≤1− < 1. 2 2 2 2 (1 + x1 + · · · + xi ) 1 + x1 + · · · + x2n

From this, it follows that the left hand side of the original inequality is less √ than n. This is what we want to prove. ?F? 02.10. Let a, b, c, d be the positive real numbers such that 1 1 1 1 + + + = 1. 4 4 4 1+a 1+b 1+c 1 + d4 Prove that abcd ≥ 3. (Latvia 2002) First solution: We need to prove the inequality a4 b4 c4 d4 ≥ 81. After making the substitution A=

1 , 1 + a4

B=

1 , 1 + b4

C=

1 , 1 + c4

D=

1 , 1 + d4

a4 =

1−A , A

b4 =

1−B , B

c4 =

1−C , C

d4 =

1−D . D

we obtain

342

The constraint becomes A + B + C + D = 1 and the inequality can be written as 1−A 1−B 1−C 1−D · · · ≥ 81, A B C D or B+C +D C +D+A D+A+B A+B+C · · · ≥ 81, A B C D or (B + C + D)(C + D + A)(D + A + B)(A + B + C) ≥ 81ABCD. However, this is an immediate consequence of the AM-GM Inequality (B + C + D)(C + D + A)(D + A + B)(A + B + C) ≥ 1

1

1

1

≥ 3 (BCD) 3 · 3 (CDA) 3 · 3 (DAB) 3 · 3 (ABC) 3 . √ Note that the equality holds if and only if a = b = c = d = 4 3. Second solution: We can write a2 = tan A, b2 = tan B, c2 = tan C, d2 = π . Then, the algebraic identity becomes the tan D, where A, B, C, D ∈ 0, 2 following trigonometric identity cos2 A + cos2 B + cos2 C + cos2 D = 1. Applying the AM-GM Inequality, we obtain 2

sin2 A = 1 − cos2 A = cos2 B + cos2 C + cos2 D ≥ 3 (cos B cos C cos D) 3 . Similarly, we obtain 2

sin2 B ≥ 3 (cos C cos D cos A) 3 , 2

sin2 C ≥ 3 (cos D cos A cos B) 3 , 2

sin2 D ≥ 3 (cos A cos B cos C) 3 . Multiplying these four inequalities, we get the result. ?F? 02.11. Prove that for any positive real numbers a, b, c, we have b c a + + ≤ 1. 2a + b 2b + c 2c + a (Moldova 2002) Solution: The original inequality is equivalent to       2a 2b 2c 1− + 1− + 1− ≥ 1, 2a + b 2b + c 2c + a or

b c a + + ≥ 1, 2a + b 2b + c 2c + a 343

which is true since by Cauchy Schwarz Inequality, we have b c a (b + c + a)2 + + ≥ = 1. 2a + b 2b + c 2c + a b (2a + b) + c (2b + c) + a (2c + a) It is easy to see that the equality holds if and only if a = b = c. ?F? 02.12. Positive numbers α, β, x1 , x2 , . . . , xn (n ≥ 1) satisfy the condition x1 + x2 + · · · + xn = 1. Prove that x31 x32 x3n 1 + + ··· + ≥ . αx1 + βx2 αx2 + βx3 αxn + βx1 n(α + β) (Moldova 2002) First solution: By the AM-GM Inequality, we have αxi + βxi+1 1 3xi x3i + + 2 ≥ , αxi + βxi+1 n(α + β)2 n (α + β) n(α + β) for all i = 1, 2, . . . , n. Therefore n X i=1

Since

n X αxi + βxi+1 i=1

n

n

i=1

i=1

X αxi + βxi+1 X 3xi x3i 1 + + ≥ . αxi + βxi+1 n(α + β)2 n(α + β) n(α + β)

n(α + β)2

have that

n X i=1

which yields

n X 1 3xi 3 = and = , we thus n(α + β) n(α + β) n(α + β) i=1

2 3 x3i + ≥ , αxi + βxi+1 n(α + β) n(α + β) n X i=1

1 x3i ≥ . αxi + βxi+1 n(α + β)

The equality holds if and only if xi =

1 for all i = 1, 2, . . . , n. n

Second solution: By the Holder’s Inequality, we have that !" n # !3 n n X X X x3i n (αxi + βxi+1 ) ≥ xi = 1, αxi + βxi+1 i=1

i=1

and since

n X

i=1

(αxi + βxi+1 ) = α + β, we conclude that

i=1 n X i=1

x3i 1 ≥ . αxi + βxi+1 n(α + β) ?F? 344

02.13. Let a, b, c be positive real numbers. Prove that 

2a b+c

2



3

+

2b c+a

2



3

+

2c a+b

2 3

≥ 3. (MOSP 2002)

Solution: By the AM-GM Inequality, we have that 

2a b+c

2 3

2a 6a 3a = p ≥ = . 3 2a + (b + c) + (b + c) a + b+c 2a · (b + c) · (b + c)

Similarly, we have 

2b c+a

2

3

3b ≥ , a+b+c



2c a+b

2 3

≥

3c . a+b+c

Adding up these three inequalities, we get the result. It is easy to see that the equality holds if and only if a = b = c. ?F? 02.14. If a, b, c ∈ (0, 1), prove that p √ abc + (1 − a)(1 − b)(1 − c) < 1. (Romania 2002) √ √ √ √ Firstpsolution: Observe thatp x < 3 x for x ∈ (0, 1). Thus abc < 3 abc and (1 − a)(1 − b)(1 − c) < 3 (1 − a)(1 − b)(1 − c). On the other hand, the AM-GM Inequality implies √ a+b+c 3 abc ≤ , 3

and

p (1 − a) + (1 − b) + (1 − c) 3 (1 − a)(1 − b)(1 − c) ≤ . 3

Therefore √ abc +

p a + b + c (1 − a) + (1 − b) + (1 − c) (1 − a)(1 − b)(1 − c) < + = 1, 3 3

as desired. Second solution: We have p √ √ √ √ √ abc + (1 − a)(1 − b)(1 − c) < b · c + 1 − b · 1 − c < 1, by the Cauchy Schwarz Inequality.  π Third solution: Let a = sin2 x, b = sin2 y, c = sin2 z, where x, y, z ∈ 0, . 2 The inequality becomes sin x · sin y · sin z + cos x · cos y · cos z < 1, 345

and it follows from the inequalities sin x · sin y · sin z + cos x · cos y · cos z < sin x · sin y + cos x · cos y < cos(x − y) ≤ 1. ?F? 02.15. Given positive real numbers a, b, c and x, y, z, for which a+x = b+y = c + z = 1. Prove that   1 1 1 (abc + xyz) ≥ 3. + + ay bz cx (Russia 2002) First solution: Our inequality is equivalent to   1 1 1 [abc + (1 − a)(1 − b)(1 − c)] + + ≥ 3. a(1 − b) b(1 − c) c(1 − a) Since a + x = b + y = c + z = 1 and x, y, z are positive numbers, we have 0 < a, b, c < 1. And thus, there exist positive numbers m, n, p such that 1 1 1 a= ,b = ,c = . By this substitution, we have m+1 n+1 p+1 abc + (1 − a)(1 − b)(1 − c) =

mnp + 1 , (m + 1)(n + 1)(p + 1)

and 1 1 1 (m + 1)(n + 1) (n + 1)(p + 1) (p + 1)(m + 1) + + = + + . a(1 − b) b(1 − c) c(1 − a) n p m Therefore, the above inequality can be written as mnp + 1 mnp + 1 mnp + 1 + + ≥ 3. m(n + 1) n(p + 1) p(m + 1) By the AM-GM Inequality, we have 1 + mnp 1 + mnp 1 + mnp + + +3= m(1 + n) n(1 + p) p(1 + m) 1 + m + mn + mnp 1 + n + np + mnp 1 + p + pm + mnp = + + m(1 + n) n(1 + p) p(1 + m) (1 + m) + mn(1 + p) (1 + n) + np(1 + m) (1 + p) + pm(1 + n) = + + m(1 + n) n(1 + p) p(1 + m)     1+m 1+n 1+p n(1 + p) p(1 + m) m(1 + n) = + + + + + m(1 + n) n(1 + p) p(1 + m) 1+n 1+p 1+m 3 √ ≥ √ + 3 3 mnp ≥ 6. 3 mnp From this, we deduce that mnp + 1 mnp + 1 mnp + 1 + + ≥ 3, m(n + 1) n(p + 1) p(m + 1) 346

as desired. It is easy to see that the equality holds if and only if m = n = p = 1, 1 i.e. if and only if a = b = c = x = y = z = . 2 Second solution: Proceeding similar as above, we see that it suffices to prove that the inequality   1 1 1 ≥3 (1 + abc) + + a(1 + b) b(1 + c) c(1 + a) holds for any positive real numbers a, b, c. Note that in general we have abc 6= 1, v w u so we cannot apply the substitutions a = , b = , c = with positive u v w v w u reals u, v, w. However, we can substitute a = k · , b = k · , c = k · , u v w where k, u, v, w are positive real numbers. Then abc = k 3 , and our inequality simplifies to  
v w u 3 + + 1+k ≥ 3. k (v + kw) k (w + ku) k (u + kv) This can be rewritten as 1 + k3 · k



u v w + + v + kw w + ku u + kv

 ≥ 3.

Now, using the Cauchy Schwarz Inequality and the well-known (u + v + w)2 ≥ 3(uv + vw + wu), we have u v w (u + v + v)2 3 + + ≥ ≥ . v + kw w + ku u + kv (k + 1)(uv + vw + wu) k+1 So, it is sufficient to prove that

k3 + 1 3 · ≥ 3, which is true. k k+1

Third solution: We will give another proof for the inequality   1 1 1 + + ≥ 3. (1 + abc) a(1 + b) b(1 + c) c(1 + a) We have X 1 + abc a(1 + b)

−3=

X 1 − a − ab + abc

=

X (1 − ab) − a(1 − bc)

a(1 + b) a(1 + b) X 1 − ab X 1 − bc X 1 − ab X 1 − ab = − = − a(1 + b) 1+b a(1 + b) 1+a   X X 1 1 (1 − ab)2 = (1 − ab) − = . a(1 + b) 1 + a a(1 + a)(1 + b)

The last quantity is obviously nonnegative, so we must have   1 1 1 (1 + abc) + + ≥ 3, a(1 + b) b(1 + c) c(1 + a) as desired. 347

Fourth solution: We present another approach to the inequality   1 1 1 ≥ 3. (1 + abc) + + a(1 + b) b(1 + c) c(1 + a) Multiplying each side of this inequality by rewrite it as

X (1 + a)(1 + c) a

or

X1

≥

(1 + a)(1 + b)(1 + c) > 0, we can 1 + abc

3(1 + a)(1 + b)(1 + c) , 1 + abc

Xa

3(1 + a)(1 + b)(1 + c) +3≥ . b 1 + abc X X Because (1 + a)(1 + b)(1 + c) = a+ ab + abc + 1, it is equivalent to X X X X1 Xa 3 a+3 ab + ≥ , a+ a b 1 + abc or X X1 X Xa X X abc a+ + a2 c + ≥2 a+2 ab. a b But this follows from the inequalities +

a

a2 bc +

X

a+

b ≥ 2ab, c

and a2 c +

1 ≥ 2a, c

b2 ca +

c ≥ 2bc, a

c2 ab +

b2 a +

1 ≥ 2b, a

c2 b +

a ≥ 2ca, b

1 ≥ 2c. b

Fifth solution: In the same manner with all above solutions, we shall prove that   1 1 1 (abc + 1) + + ≥3 a(1 + b) b(1 + c) c(1 + a) for positive real numbers a, b, c. Squaring both sides and using the well-known inequality (x + y + z)2 ≥ 3(xy + yz + zx), we see that it suffices to prove X 1 (abc + 1)2 · 3 ≥ 9, ab(1 + b)(1 + c) or equivalently, X (abc + 1)2 ·

a+

X

ab

abc(1 + a)(1 + b)(1 + c)

≥ 3.

Now, using the AM-GM Inequality, we have X X a+ ab 1 1 = ≥ 1 + abc 1 + abc (1 + a)(1 + b)(1 + c) √ X X +1 +1 √ 3 3 3 abc + 3 a2 b2 c2 a+ ab √ √ √ 3 3 3 abc + 3 a2 b2 c2 3 3 abc = = √ √ 3 2 . 3 3 abc + 1 abc + 1 348

So, it is enough to prove that √ 3 3 abc (abc + 1) · √ 2 ≥ 3, abc 3 abc + 1 2

which is not hard to prove. ?F? 02.16. Let x, y, z be positive real numbers with sum 3. Prove that √

x+

√

y+

√ z ≥ xy + yz + zx. (Russia 2002)

First solution: By applying the Holder’s Inequality, we have
√
2 2 √ √ x+ y+ z x + y 2 + z 2 ≥ (x + y + z)3 = 27, and hence, it suffices to prove that
x2 + y 2 + z 2 (xy + yz + zx)2 ≤ 27. This is true since by the AM-GM Inequality, we have 2

2

x +y +z

2




x2 + y 2 + z 2 + 2(xy + yz + zx) (xy + yz + zx) ≤ 3  3 2 (x + y + z) = = 27. 3 2



3

Note that the equality holds if and only if x = y = z = 1. Second solution: Rewrite the inequality in the form √ √ √ x2 + 2 x + y 2 + 2 y + z 2 + 2 z ≥ x2 + y 2 + z 2 + 2(xy + yz + zx). Since x2 + y 2 + z 2 + 2(xy + yz + zx) = (x + y + z)2 = 9, it is equivalent to √ √ √ x2 + 2 x + y 2 + 2 y + z 2 + 2 z ≥ 9. Now, from the AM-GM Inequality, we have 2

√

2

x +2 x=x +

√

x+

√

q √ √ 3 x ≥ 3 x2 · x · x = 3x.

Therefore √ √ √ x2 + 2 x + y 2 + 2 y + z 2 + 2 z ≥ 3(x + y + z) = 9, as desired. ?F? 349

02.17. Let a1 , a2 , . . . , an and b1 , b2 , . . . , bn be real numbers between 1001 and 2002 inclusive. Suppose a21 + · · · + a2n = b21 + · · · + b2n . Prove that n X a3 i

i=1

bi

n

17 X 2 ai . 10

≤

i=1

Determine when equality holds. (Singapore 2002) Solution: For each i, we have (2ai − bi )(2bi − ai ) ≥ 0, that is

2002 1001 ai 1 ≤ = ≤ = 2, and therefore 2 2002 bi 1001

5ai bi ≥ 2a2i + 2b2i . Multiplying this inequality by

(1)

ai > 0, we get bi

5a2i ≥ 2

a3i + 2ai bi . bi

(2)

4 From (1), we have 2ai bi ≥ (a2i + b2i ). Using this inequality in (2), we obtain 5 5a2i ≥ 2

a3i 4 + (a2i + b2i ), bi 5

which may be rewritten as a3i 21 2 ≤ a2i − b2i . bi 10 5

(3)

Note that the equality in (3) holds if and only if bi = 2ai or ai = 2bi , that is, if and ony if (ai , bi ) = (1001, 2002) or (ai , bi ) = (2002, 1001). Now, summing n n X X over i in (3) and recalling b2i = a2i , we get i=1 n X a3

i=1 n

n

n

i=1

i=1

i=1

21 X 2 2 X 2 17 X 2 ≤ ai − bi = ai , bi 10 5 10 i

i=1

as desired. The equality holds if and only if, for each i, either (ai , bi ) = n n X X 2 (1001, 2002) or (ai , bi ) = (2002, 1001). The condition bi = a2i can be i=1

i=1

n . 2 Thus, the equality holds if and only if n is even and (ai , bi ) = (1001, 2002) for half of the subscripts i while (ai , bi ) = (2002, 1001) for the other half. ?F?   1 02.18. Let a, b, c, d be real numbers contained in the interval 0, . Prove 2 that rewritten as

10012 p

+ (n −

p)20022

=

20022 p

+ (n −

p)10012 ,

which is p =

a4 + b4 + c4 + d4 (1 − a)4 + (1 − b)4 + (1 − c)4 + (1 − d)4 ≥ . abcd (1 − a)(1 − b)(1 − c)(1 − d) 350

(Taiwan 2002) First solution: Without loss of generality, we may assume that a ≥ b ≥ c ≥ d. Then, by noting at the identity x4 + y 4 + z 4 + t4 = (x2 − y 2 )2 + (z 2 − t2 )2 + 2(x2 y 2 + z 2 t2 ), we can see that the original inequality follows from adding the following three inequalities (a2 − b2 )2 (a − b)2 (2 − a − b)2 ≥ , abcd (1 − a)(1 − b)(1 − c)(1 − d) (c2 − d2 )2 (c − d)2 (2 − c − d)2 ≥ , abcd (1 − a)(1 − b)(1 − c)(1 − d) and

2(a2 b2 + c2 d2 ) 2[(1 − a)2 (1 − b)2 + (1 − c)2 (1 − d)2 ] ≥ . abcd (1 − a)(1 − b)(1 − c)(1 − d)

Hence, in order to prove it, it suffices to show that these three inequalities holds. Indeed, the first one is equivalent to (1 − c)(1 − d) cd and since



1 −1 a



1 −1 b



 ≥

2 2 −1 , a+b

(1 − c)(1 − d) ≥ 1, it suffices to prove that cd    2  1 2 1 −1 −1 ≥ −1 . a b a+b

This is true, because by the AM-GM Inequality, we have 

1 −1 a



1 −1 b



1−a−b 4(1 − a − b) = +1≥ +1= ab (a + b)2



2 −1 a+b

2 .

Since the proof for (c − d)2 (2 − c − d)2 (c2 − d2 )2 ≥ abcd (1 − a)(1 − b)(1 − c)(1 − d) is similar to the previous one, we are left to show that a2 b2 + c2 d2 (1 − a)2 (1 − b)2 + (1 − c)2 (1 − d)2 ≥ , abcd (1 − a)(1 − b)(1 − c)(1 − d) or equivalently,  f

ab cd



 ≥f

(1 − c)(1 − d) (1 − a)(1 − b)

 ,

1 where f is the real-valued function such that f (x) = x + . Now, since f (x) x is monotonically increasing for all x ≥ 1, and moreover since ab ≥ 1, cd

(1 − c)(1 − d) ≥ 1, (1 − a)(1 − b) 351

we only need to show that (1 − c)(1 − d) ab ≥ , cd (1 − a)(1 − b)

or

a(1 − a) b(1 − b) · ≥ 1. c(1 − c) d(1 − d)

But this is obviously true, because (a − c)(1 − a − c) a(1 − a) −1= ≥ 0, c(1 − c) c(1 − c) and

b(1 − b) (b − d)(1 − b − d) −1= ≥ 0. d(1 − d) d(1 − d)

Note that the equality holds iff a = b = c = d.   1 Second solution: First, we claim that for any x, y, z, t ∈ 0, , the in2 equality holds (1 − x)4 + (1 − y)4 (x2 − y 2 )2 [zt(1 − x)(1 − y) − xy(1 − z)(1 − t)] + ≤ (1 − x)(1 − y) x2 y 2 zt ≤  Indeed, since z, t ∈

1 0, 2

(1 − x)(1 − y)(x4 + y 4 ) . x2 y 2

 , we have

zt(1 − x)(1 − y) − xy(1 − z)(1 − t) (1 − z)(1 − t) = (1 − x)(1 − y) − xy · zt zt < (1 − x)(1 − y) − xy = 1 − x − y, and thus, it suffices to prove that (1 − x)(1 − y)(x4 + y 4 ) (1 − x)4 + (1 − y)4 (x2 − y 2 )2 (1 − x − y) − ≥ , x2 y 2 (1 − x)(1 − y) x2 y 2 which is equivalent to (note that mn(u2 + v 2 ) − uv(m2 + n2 ) = (um − vn)(un − vm)) (x − y)2 (1 − x − y)(x + y − x2 − y 2 )(x + y − 2xy) (x2 − y 2 )2 (1 − x − y) ≥ . x2 y 2 (1 − x)(1 − y) x2 y 2 This can be simplified into (x + y − x2 − y 2 )(x + y − 2xy) ≥ (x + y)2 (1 − x)(1 − y). Applying the Cauchy Schwarz Inequality, we have (x + y − x2 − y 2 )(x + y − 2xy) = [x(1 − x) + y(1 − y)][x(1 − y) + y(1 − x)] h p i2 p ≥ x (1 − x)(1 − y) + y (1 − y)(1 − x) = (x + y)2 (1 − x)(1 − y). 352

This proves our claim. Now, turning back to the main problem. Because the inequality is symmetric, we may assume without loss of generality that d ≥ c ≥ b ≥ a. Then, applying the above claim by replacing (x, y, z, t) with (a, b, c, d) and (c, d, a, b), we get (1 − a)4 + (1 − b)4 (1 − a)(1 − b)(a4 + b4 ) M (a2 − b2 )2 ≤ − , (1 − a)(1 − b)(1 − c)(1 − d) a2 b2 (1 − c)(1 − d) a2 b2 cd(1 − c)(1 − d) (1 − c)4 + (1 − d)4 (1 − c)(1 − d)(c4 + d4 ) M (c2 − d2 )2 ≤ + , (1 − a)(1 − b)(1 − c)(1 − d) c2 d2 (1 − a)(1 − b) abc2 d2 (1 − a)(1 − b) where M = cd(1 − a)(1 − b) − ab(1 − c)(1 − d) ≥ 0. From these two inequalities, we can see that the original inequality is deduced from cd(1 − a)(1 − b)(a4 + b4 ) ab(1 − c)(1 − d)(c4 + d4 ) + + ab(1 − c)(1 − d) cd(1 − a)(1 − b) +

M (c2 − d2 )2 M (a2 − b2 )2 − ≤ a4 + b4 + c4 + d4 . cd(1 − a)(1 − b) ab(1 − c)(1 − d)

Since a4 + b4 −

cd(1 − a)(1 − b)(a4 + b4 ) M (a4 + b4 ) =− , ab(1 − c)(1 − d) ab(1 − c)(1 − d)

and c4 + d4 −

M (c4 + d4 ) ab(1 − c)(1 − d)(c4 + d4 ) = , cd(1 − a)(1 − b) cd(1 − a)(1 − b)

one can see that the last inequality is equivalent to M (a4 + b4 ) M (a2 − b2 )2 M (c4 + d4 ) M (c2 − d2 )2 − ≤ − , ab(1 − c)(1 − d) ab(1 − c)(1 − d) cd(1 − a)(1 − b) cd(1 − a)(1 − b) that is 2abM 2cdM ≤ . (1 − c)(1 − d) (1 − a)(1 − b) Of couse, this inequality is true because of M ≥ 0 and the fact x(1 − x) ≥ 1 y(1 − y) for any > x ≥ y > 0. The proof is now completed. 2 ?F? 02.19. Let x, y, z be positive real numbers such that x2 + y 2 + z 2 = 1. Prove that 1 x2 yz + y 2 zx + z 2 xy ≤ . 3 (United Kingdom 2002) First solution: From the Cauchy Schwarz Inequality and the hypothesis, we have p √ x2 yz + y 2 zx + z 2 xy = xyz(x + y + z) ≤ xyz 3(x2 + y 2 + z 2 ) = 3xyz. 353

p 1 x2 + y 2 + z 2 On the other hand, = ≥ 3 x2 y 2 z 2 , using the AM-GM Inequal3 3 1 ity, hence xyz ≤ √ . The desired inequality follows immediately by combin3 3 1 ing the two results. Note that the equality holds if and only if x = y = z = √ . 3 Second solution: From the identity (a+b+c)2 = (a2 +b2 +c2 )+2(ab+bc+ca), and the well-known inequality a2 + b2 + c2 ≥ ab + bc + ca, it follows that (a + b + c)2 ≥ 3(ab + bc + ca). Setting a = xy, b = yz, c = zx in this inequality yields 3(x2 yz + y 2 zx + z 2 xy)2 ≤ (xy + yz + zx)2 ≤ (x2 + y 2 + z 2 ) = 1, from which we get 1 x2 yz + y 2 zx + z 2 xy ≤ , 3 as claimed. ?F? 02.20. Let a, b, c be real numbers satisfying a2 + b2 + c2 = 9. Prove the inequality 2(a + b + c) − abc ≤ 10. (Vietnam 2002) First solution: If there is a negative number among a, b, c (for example b < 0). Then, from the AM-GM Inequality, we have 2(a + b + c) − abc ≤ 2(a + b + c) − b ·

a2 + c2 2

a2 + 4 c2 + 4 a2 + c2 + + 2b − b · 2 2 2 2 2 9−b 9−b =4+ + 2b − b · 2 2 2 (b − 3)(b + 1) = 10 + ≤ 10. 2 Let us consider now the case a, b, c ≥√0. Without loss of generality, we may assume that a ≥ b ≥ c, then 3 ≥ a ≥ 3. There are two cases to consider If ab ≥ 1, we have ≤

2(a + b + c) − abc ≤ 2(a + b + c) − c = 2a + 2b + c p ≤ (22 + 22 + 12 )(a2 + b2 + c2 ) = 9 < 10. If ab < 1, then it follows that a2 b2 + a2 c2 ≤ 2a2 b2 < 2, or b2 + c2 < yields

2 . a2

2(a + b + c) − abc ≤ 2(a + b + c) ≤ 2a + b2 + c2 + 2 2 2 26 < 2a + 2 + 2 < 2 · 3 + √
2 + 2 = < 10. a 3 3 354

This

Therefore, in any cases, we always have 2(a + b + c) − abc ≤ 10. The equality holds if and only if (a, b, c) is a permutation of (2, 2, −1). Second solution: Without loss of generality, we may assume that a2 ≤ b2 ≤ c2 . And since a2 + b2 + c2 = 9, this assumption gives us c2 ≥ 3 and 2ab ≤ 6. From this, applying the Cauchy Schwarz Inequality, we have [2(a + b + c) − abc]2 = [2(a + b) + (2 − ab)c]2 ≤ [4 + (2 − ab)2 ][(a + b)2 + c2 ] = (8 − 4ab + a2 b2 )(9 + 2ab) = 2a3 b3 + a2 b2 − 20ab + 72 = (ab + 2)2 (2ab − 7) + 100 ≤ 100. According to this inequality, it follows that 2(a + b + c) − abc ≤ |2(a + b + c) − abc| ≤ 10, as desired. Third solution: Because max{a, b, c} ≤ 3 and |abc| ≤ 10, it is enough to consider only the cases when a, b, c ≥ 0 or exactly one of the three numbers is negative. First, we will suppose that a, b, c are nonnegative. If abc ≥ 1, then we are done because p 2(a + b + c) − abc ≤ 2 3(a2 + b2 + c2 ) − 1 < 10. Otherwise, we may assume that a < 1. In this case, we have h i p p 2(a + b + c) − abc ≤ 2 a + 2(b2 + c2 ) = 2a + 2 18 − 2a2 ≤ 10. Now, assume that not all three numbers are nonnegative and let c < 0. Thus, the problem reduces to proving that for any nonnegative x, y, z whose sum of squares is 9, we have 4(x + y − z) + 2xyz ≤ 20. But we can write this inequality as (x − 2)2 + (y − 2)2 + (z − 1)2 ≥ 2xyz − 6z − 2. Because 2xyz − 6z − 2 ≤ z(x2 + y 2 ) − 6z − 2 = −z 3 + 3z − 2 = −(z − 1)2 (z + 2) ≤ 0, the inequality follows. √ Fourth solution: Of course, we have |a|, |b|, |c| ≤ 3 and |a+b+c|, |abc| ≤ 3 3. Also, we may assume of course that a, b, c are non-zero and √ that a ≤ b ≤ c. If c < 0, then we have 2(a + b + c) − abc < −abc ≤ 3 3 < 10. Also, if a ≤ b < 0 < c, then we have 2(a + b + c) < 2c ≤ 6 < 10 + abc because abc > 0. If a < 0 < b ≤ c, using the Cauchy Schwarz Inequality, we find that 2b + 2c − a ≤ 9. Thus 2(a + b + c) = (2b + 2c − a) + 3a ≤ 9 + 3a and it remains to prove that 3a − 1 ≤ abc. But a < 0 and 2bc ≤ 9 − a2 , so that it remains to 9a − a3 ≥ 3a − 1 or (a + 1)2 (a − 2) ≤ 0, which follows. So, we just show that 2 have to threat the case 0 < a ≤ b ≤ c. In this case we have 2b + 2c + a ≤ 9 and hence 2(a + b + c) ≤ 9 + a. So, we need to prove that a ≤ 1 + abc. This is clear if a < 1 and if a ≥ 1, we have b, c ≥ 1 and the inequality is again. Thus, the problem is solved. 355

?F? 03.1. If a, b, c > −1, then 1 + a2 1 + b2 1 + c2 + + ≥ 2. 2 2 1+b+c 1+c+a 1 + a + b2 (Laurentiu Panaitopol, Balkan 2003) Solution: We have 1 + b + c2 ≥ 1 + b > 0 and 1 + b + c2 ≤ hence 1 + a2 2(1 + a2 ) ≥ . 2 1+b+c 1 + b2 + 2(1 + c2 )

1 + b2 + 1 + c2 , 2

Setting x = 1 + a2 , y = 1 + b2 , z = 1 + c2 , it suffices to show that x y z + + ≥ 1. y + 2z z + 2x x + 2y Using the Cauchy Schwarz Inequality, we have y z (x + y + z)2 x + + ≥ y + 2z z + 2x x + 2y x(y + 2z) + y(z + 2x) + z(x + 2y) (x + y + z)2 = ≥ 1. 3(xy + yz + zx) Note that the equality holds if and only if a = b = c = 1. ?F? 03.2. Prove that if a, b and c are positive real numbers with sum 3, then b c 3 a + + ≥ . b2 + 1 c2 + 1 a2 + 1 2 (Bulgaria 2003) Solution: By the AM-GM Inequality, we have a ab2 ab2 ab = a − ≥ a − =a− . 2 2 b +1 b +1 2b 2 Adding this to the two analogous inequalities, we get a b c ab + bc + ca (a + b + c)2 3 + + ≥ a+b+c− ≥ a+b+c− = , b2 + 1 c2 + 1 a2 + 1 2 6 2 as desired. Note that the equality holds if and only if a = b = c = 1. ?F? 03.3. Let a, b, c, d be positive real numbers such that ab + cd = 1 and let x1 , x2 , x3 , x4 , y1 , y2 , y3 , y4 be real numbers such that x21 + y12 = x22 + y22 = x23 + y32 = x24 + y42 = 1. Prove that the following inequality holds  2  a + b2 c2 + d2 2 2 (ay1 + by2 + cy3 + dy4 ) + (ax4 + bx3 + cx2 + dx1 ) ≤ 2 + . ab cd 356

(China 2003) First solution: Fix a number k > 0. Using the Cauchy Schwarz Inequality, we have  
y12 y42 2 2 2 2 2 2 2 , + y2 + y3 + (ay1 + by2 + cy3 + dy4 ) ≤ a k + b + c + d k k k  
x24 x21 2 2 2 2 2 2 2 (ax4 + bx3 + cx2 + dx1 ) ≤ a k + b + c + d k , + x3 + x2 + k k and then, we deduce that (ay1 + by2 + cy3 + dy4 )2 + (ax4 + bx3 + cx2 + dx1 )2 ≤  2  x1 + y12 x24 + y42 2 2 2 2 2 2 2 2 ≤ (a k + b + c + d k) + x2 + y2 + x3 + y3 + k k     1 =2 + 1 k(a2 + d2 ) + (b2 + c2 ) k 2(b2 + c2 ) . = 2(a2 + b2 + c2 + d2 ) + 2k(a2 + d2 ) + k Now, we choose k > 0 such that k(a2 + d2 ) +

cd(a2 + b2 ) ab(c2 + d2 ) b2 + c2 = + , k ab cd

or f (k) = (a2 + d2 )k 2 −



 cd(a2 + b2 ) ab(c2 + d2 ) + k + b2 + c2 = 0. ab cd

We have 2 cd(a2 + b2 ) ab(c2 + d2 ) ∆f = + − 4(a2 + d2 )(b2 + c2 ) ab cd  2 2 2 2 a d (b + c2 ) + b2 c2 (a2 + d2 ) = − 4(a2 + d2 )(b2 + c2 ) a2 b2 c2 d2  2 2 2 2 a d (b + c2 ) − b2 c2 (a2 + d2 ) = ≥ 0. a2 b2 c2 d2 

Therefore, the number k > 0 satisfies the above equation is cd(a2 + b2 ) ab(c2 + d2 ) a2 d2 (b2 + c2 ) − b2 c2 (a2 + d2 ) + + ab cd abcd k= . 2(a2 + d2 ) From this value of k and the above arguments, we have that (ay1 + by2 + cy3 + dy4 )2 + (ax4 + bx3 + cx2 + dx1 )2 ≤ 2cd(a2 + b2 ) 2ab(c2 + d2 ) ≤ 2(a2 + b2 + c2 + d2 ) + + ab  2  2 cd2  a + b2 c2 + d2 a +b c2 + d2 = 2(ab + cd) + =2 + , ab cd ab cd 357

as desired. Second solution: Set u = ay1 + by2 , v = cy3 + dy4 , u1 = ax4 + bx3 and v1 = cx2 + dx1 . Then u2 ≤ (ay1 + by2 )2 + (ax1 − bx2 )2 = a2 + b2 + 2ab(y1 y2 − x1 x2 ), that is x1 x2 − y1 y2 ≤

a2 + b2 − u2 . 2ab

Similarly, we have v12 ≤ (cx2 + dx1 )2 + (dy1 − cy2 )2 = c2 + d2 − 2cd(y1 y2 − x1 x2 ), and then y1 y2 − x1 x2 ≤

c2 + d2 − v12 . 2cd

From this, we deduce that a2 + b2 − u2 c2 + d2 − v12 + ≥ (x1 x2 − y1 y2 ) + (y1 y2 − x1 x2 ) = 0, 2ab 2cd that is a2 + b2 c2 + d2 u2 v12 + ≤ + . ab cd ab cd Similarly, we also have a2 + b2 c2 + d2 v 2 u21 + ≤ + . cd ab ab cd Now, applying the Cauchy Schwarz Inequality, we get  2   2  u v2 u1 v12 2 2 (u + v) + (u1 + v1 ) ≤ (ab + cd) + + (ab + cd) + ab cd ab cd  2  u2 v12 v 2 u21 a + b2 c2 + d2 = + + + ≤2 + , ab cd cd ab ab cd as desired. Third solution: By the Cauchy Schwarz Inequality, we get   (ay1 + by2 )2 (cy3 + dy4 )2 2 (ay1 + by2 + cy3 + dy4 ) ≤ (ab + cd) + ab cd (ay1 + by2 )2 (cy3 + dy4 )2 = + ab cd a 2 b 2 c 2 d 2 = y1 + y2 + y3 + y4 + 2y1 y2 + 2y3 y4 , b a d c and   (ax4 + bx3 )2 (cx2 + dx1 )2 2 (ax4 + bx3 + cx2 + dx1 ) ≤ (ab + cd) + ab cd 2 2 (ax4 + bx3 ) (cx2 + dx1 ) = + ab cd a 2 b 2 c 2 d 2 = x4 + x3 + x2 + x1 + 2x1 x2 + 2x3 x4 . b a d c 358

Denote 2

2

P = (ay1 +by2 +cy3 +dy4 ) +(ax4 +bx3 +cx2 +dx1 ) −2



a2 + b2 c2 + d2 + ab cd

 .

Then from the above inequalities, we find that a 2 b 2 c 2 d 2 a b c y1 + y2 + y3 + y4 + 2y1 y2 + 2y3 y4 + x24 + x23 + x22 + b a d c b a d   d 2 a b c d + x1 + 2x1 x2 + 2x3 x4 − 2 + + + c b a d c         a 2 b 2 c 2 d 2 a 2 b 2 c 2 d 2 =− x + x − x + x − y + y − y + y + b 1 a 2 d 3 c 4 b 4 a 3 d 2 c 1 + 2x1 x2 + 2x3 x4 + 2y1 y2 + 2y3 y4

P ≤

≤ −2x1 x2 − 2x3 x4 − 2y4 y3 − 2y2 y1 + 2x1 x2 + 2x3 x4 + 2y1 y2 + 2y3 y4 = 0, as desired. ?F? 03.4. Let a1 , a2 , . . . , a2n be real numbers such that 2n−1 X

(ai − ai+1 )2 = 1.

i=1

Determine the maximum value of (an+1 + an+2 + · · · + a2n ) − (a1 + a2 + . . . + an ). (China 2003) Solution: For n = 1, we have (a1 − a2 )2 = 1, implying that a2 − a1 = ±1. Therefore, in this case, max {a2 − a1 } = 1. Suppose that n ≥ 2, let x1 = a1 , xi+1 = ai+1 − ai for i = 1, 2, . . . , 2n − 1. By this substitution, we have ai = x1 + · · · + xi for i = 1, 2, . . . , 2n and x22 + x23 + · · · + x22n = 1. Now, using the Cauchy Schwarz Inequality, we have (an+1 + an+2 + · · · + a2n ) − (a1 + a2 + · · · + an ) = = n(x1 + x2 + · · · + xn + xn+1 ) + (n − 1)xn+2 + · · · + x2n −

n X (n + 1 − i)xi i=1

= x2 + 2x3 + · · · + (n − 1)xn + nxn+1 + (n − 1)xn+2 + · · · + x2n q
≤ [12 + 22 + · · · + (n − 1)2 + n2 + (n − 1)2 + · · · + 12 ] x22 + x23 + · · · + x22n r p n(2n2 + 1) = 12 + 22 + · · · + (n − 1)2 + n2 + (n − 1)2 + · · · + 12 = . 3 The equality holds when √ 3i(i − 1) ai = p , 2 n(2n2 + 1)

an+i

√  2  3 2n − (n − i)(n − i + 1) p = 2 n(2n2 + 1) 359

for r all i = 1, 2, . . . , n. So, the maximum value of the desired expression is n(2n2 + 1) . 3 ?F?   3 , 5 . Prove that 03.5. Suppose x be a real number in the interval 2 √ √ √ √ 2 x + 1 + 2x − 3 + 15 − 3x < 2 19. (China 2003) Solution: From the Cauchy Schwarz Inequality and the given hypothesis, we have √ √ √ 2 x + 1 + 2x − 3 + 15 − 3x = √ √ √ √ = x + 1 + x + 1 + 2x − 3 + 15 − 3x p ≤ [(x + 1) + (x + 1) + (2x − 3) + (15 − 3x)] (12 + 12 + 12 + 12 ) √ √ = 2 x + 14 ≤ 2 19. √ √ √ We have equality if and only if x + 1 = 2x − 3 = 15 − 3x and x = 5, but this is impossible. Therefore, we conclude that √ √ √ √ 2 x + 1 + 2x − 3 + 15 − 3x < 2 19, as desired. ?F? 03.6. Let x1 , x2 , . . . , x5 be nonnegative real numbers such that 5 X i=1

Prove that

1 = 1. 1 + xi

5 X

xi 2 + 4 ≤ 1. x i=1 i (China 2003)

First solution: The original inequality is equivalent to 5  X i=1

or

5xi 1− 2 xi + 4



5 X (xi − 1)(xi − 4) i=1

x2i + 4

≥ 0,

≥ 0.

This last inequality can be written as 5 X x2i − 1 xi − 4 · ≥ 0. x2i + 4 xi + 1 i=1

360

Now, due to symmetry, we may assume without loss of generality that x1 ≥ x2 ≥ x3 ≥ x4 ≥ x5 . By this assumption, it is easy to check that x21 − 1 x2 − 1 x2 − 1 x2 − 1 x2 − 1 ≥ 22 ≥ 32 ≥ 42 ≥ 25 , 2 x2 + 4 x2 + 4 x3 + 4 x4 + 4 x5 + 4 and

x1 − 4 x2 − 4 x3 − 4 x4 − 4 x5 − 4 ≥ ≥ ≥ ≥ . x1 + 1 x2 + 1 x3 + 1 x4 + 1 x5 + 1 Therefore, by the Chebyshev’s Inequality, we have ! ! 5 5 5 X X xi − 4 x2i − 1 xi − 4 1 X x2i − 1 ≥ · 5 x +1 x2 + 4 xi + 1 x2 + 4 i=1 i i=1 i i=1 i ! ! 5 5 X X 1 x2i − 1 1− = 0. = xi + 1 x2i + 4 i=1

i=1

The equality holds if and only if x1 = x2 = x3 = x4 = x5 = 4. 1 1 − yi , then xi = > xi + 1 yi 1 − yi into the desired 0 and y1 + y2 + y3 + y4 + y5 = 1. Replacing xi by yi inequality, we see that it is equivalent to

Second solution: For each i (1 ≤ i ≤ 5), let yi =

5 X

yi − yi2 ≤ 1, 5yi2 − 2yi + 1 i=1

or

5 X 5yi − 5yi2 ≤ 5, 5yi2 − 2yi + 1 i=1

which is true because  X 5 5  5 X X 5yi − 5yi2 3yi + 1 3yi + 1 = −1 + = −5 2 2 2 5yi − 2yi + 1 5yi − 2yi + 1 5yi − 2yi + 1 i=1 i=1 i=1 5 X

5 5X 3yi + 1 = −5≤ (3yi + 1) − 5 = 5.  2 4 4 i=1 5 y − 1 i=1 + i 5 5

?F? 03.7. Find the greatest real number k such that, for any positive a, b, c with a2 > bc, (a2 − bc)2 > k(b2 − ca)(c2 − ab). (Japan 2003) Solution: We will prove that the greatest k is 4. First suppose that (a2 −bc)2 > k(b2 −ca)(c2 −ab) whenever a, b, c > 0 and a2 > bc. Let t ∈ (0, 1), Since 12 > t·t, we have (1 − t2 )2 > k(t2 − t)(t2 − t), from which we deduce that 

1+t t

2

361

> k.

It follows that  k ≤ lim

t→1

1+t t

2 = 4.

Now, we will show that (a2 − bc)2 > 4(b2 − ca)(c2 − ab) whenever a, b, c > 0 and a2 > bc. Assume on the contrary that (a2 − bc)2 ≤ 4(b2 − ca)(c2 − ab)

(1)

for some positive a, b, c such that a2 > bc, and define f (x) = (b2 − ca)x2 + (a2 − bc)x + (c2 − ab). From (1), either f (x) ≥ 0 for all real x or f (x) ≤ 0 for all real x. Actually, the former holds since f (x) = a2 + b2 + c2 − ab − bc − ca > 0 (note that a = b = c is excluded by a2 > bc, and so a2 + b2 + c2 > ab + bc + ca). It follows that b2 − ca is positive. Now, we write f (x) = (bx − c)2 − ag(x) − x(a2 − bc), where g(x) = cx2 − 2ax + b. Since a2 − bc > 0 and g c2

c b

=

c(c2 − ab) + b(b2 − ac) >0 b2 c

b2

(since − ab has the same sign as − ac by (1)), we have f b contradiction. This completes the proof. ?F?

< 0, a

03.8. Prove that in any acute triangle ABC, cot3 A + cot3 B + cot3 C + 6 cot A cot B cot C ≥ cot A + cot B + cot C. (MOSP 2003) Solution: Let x = cot A, y = cot B, z = cot C. Because xy + yz + zx = 1, it suffices to prove the homogeneous inequality x3 + y 3 + z 3 + 6xyz ≥ (x + y + z)(xy + yz + zx). But it is equivalent to x(x − y)(x − z) + y(y − z)(y − x) + z(z − x)(z − y) ≥ 0, which is the Schur’s Inequality (in the special case third degree). Note that π the equality holds if and only if A = B = C = . 3 ?F? 03.9. Let ai be positive real numbers, for all i = 1, 2, . . . , n, satisfying a1 + a2 + · · · + an =

1 1 1 + + ··· + . a1 a2 an

362

Prove that 1 1 1 + + ··· + ≤ 1. n − 1 + a1 n − 1 + a2 n − 1 + an (Vasile Cirtoaje, MOSP 2003) Solution: Notice that the inequality is equivalent to  n  n X X n−1 ai 1− ≥ 1, or ≥ 1. n − 1 + ai n − 1 + ai i=1

i=1

1 , then the condition from the hypothesis remains invariant, but in ai terms of bi ’s rewrites as

Set bi =

b1 + b2 + · · · + bn =

1 1 1 + + ··· + . b1 b2 bn

On the other hand, the inequality to prove becomes now n X i=1

1 ≥ 1. (n − 1)bi + 1

We proceed now by making use of the contradiction method. In this case, assume that there exist some numbers b1 , b2 , . . . , bn satisfying the above condition, but such that n X 1 < 1. (n − 1)bi + 1 i=1

Then there also exists some k < 1 such that n X i=1

1 = 1. (n − 1)kbi + 1

1 − xi 1 < 1, i.e. bi = , we have (n − 1)kbi + 1 (n − 1)kxi that x1 + x2 + · · · + xn = 1, and the given condition becomes

Furthermore, setting xi =

n X i=1

n

X (n − 1)kxi 1 − xi = , (n − 1)kxi 1 − xi i=1

and since k < 1, we also have that n n X X (n − 1)xi 1 − xi < . (n − 1)xi 1 − xi i=1

i=1

On the other hand, from the Cauchy Schwarz Inequality, we get   X   xj n n n  n X X X1 X X (n − 1)2  xi j6=i   2   X , = xi ≥ xi · X  = (n − 1) xi xj  xj  xj i=1

i=1

j6=i

i=1

i=1

j6=i

363

j6=i

and therefore X xj n n X X xi j6=i 2 X , ≥ (n−1) xi xj

or equivalently,

i=1

i=1

n n X X 1 − xi (n − 1)xi ≥ . (n − 1)xi 1 − xi i=1

i=1

j6=i

This obviously comes in contradiction with n n X X 1 − xi (n − 1)xi < , (n − 1)xi 1 − xi i=1

i=1

completing then our proof. Note that the equality holds if and only if a1 = a2 = · · · = an = 1. ?F? 03.10. Let a, b, c be nonnegative real numbers satisfying a2 + b2 + c2 = 1. Prove that √ a b c 1≤ + + ≤ 2. 1 + bc 1 + ca 1 + ab (Faruk Zejnulahi, MOSP 2003)   1 First solution: For all real numbers x contained in the interval 0, , we 2 2 1 x(1 − 2x) have 1 − x − = ≥ 0, and therefore 3 1+x 3(1 + x) 2 1 ≤ 1 − x. 1+x 3 1 Now, notice that max{ab, bc, ca} ≤ . In this case, we get 2       a b c 2 2 2 + + ≤ a 1 − bc + b 1 − ca + c 1 − ab 1 + bc 1 + ca 1 + ab 3 3 3 = a + b + c − 2abc = a(1 − 2bc) + b + c p ≤ [a2 + (b + c)2 ][(1 − 2bc)2 + 1] p = 2(2b2 c2 − 2bc + 1)(1 + 2bc) p √ = 2[1 − 2b2 c2 (1 − 2bc)] ≤ 2. For the right hand side of the inequality, note that and therefore

1 x2 −1+x = ≥ 0, 1+x 1+x

1 ≥ 1 − x, 1+x for all positive real numbers x. Hence b c a + + ≥ a(1 − bc) + b(1 − ca) + c(1 − ab) 1 + bc 1 + ca 1 + ab = a + b + c − 3abc p = a2 + b2 + c2 + 2(ab + bc + ca) − 3abc p = 1 + 2(ab + bc + ca) − 3abc, 364

and since √

1 + 2u − 1 =

√  2u 2u √ √ = ≥ 3−1 u 1 + 2u + 1 1+ 3

for any 0 ≤ u ≤ 1, we obtain p

√

 3 − 1 (ab + bc + ca) − 3abc √  ≥1+ 3 − 1 (ab + bc + ca)3/2 − 3abc √  √ 3/2 3 ≥1+ 3 − 1 3 a2 b2 c2 − 3abc  √  = 1 + 3 2 − 3 abc ≥ 1.

1 + 2(ab + bc + ca) − 3abc ≥ 1 +

This yields a b 1 + + ≥ 1, 1 + bc 1 + ca 1 + ab and the proof is completed. Note in the left inequality occurs  that the equality  1 1 iff (a, b, c) is a permutation of √ , √ , 0 , while the equality in the right 2 2 inequality occurs iff (a, b, c) is a permutation of (1, 0, 0). Second solution: First, we will show that (a + b + c)2 ≤ 2(1 + bc)2 , or equivalently, 2a(b + c) ≤ 1 + 2bc + 2b2 c2 . This can be rewritten as 2a(b + c) ≤ a2 + b2 + c2 + 2bc + 2b2 c2 , which is obviously true since a2 + b2 + c2 + 2bc − 2a(b + c) = (b + c − a)2 ≥ 0. Thus, we now have that √ √ √ √ a b c 2a 2b 2c + + ≤ + + = 2, 1 + bc 1 + ca 1 + ab a+b+c a+b+c a+b+c and since the AM-GM Inequality gives us a + abc ≤ a + it follows that

a(b2 + c2 ) a(1 − a2 ) (a − 1)2 (a + 2) =a+ =1− ≤1 2 2 2 a a[1 − (a + abc)] − a2 = ≥ 0, 1 + bc 1 + bc

and therefore a b c + + ≥ a2 + b2 + c2 = 1. 1 + bc 1 + ca 1 + ab ?F? 365

03.11. Let a, b, c be positive real numbers so that abc = 1. Prove that 1+

3 6 ≥ . a+b+c ab + bc + ca (Romania 2003)

1 1 1 and observe that xyz = 1. The Solution: We set x = , y = , z = a b c inequality is equivalent to 1+

3 6 ≥ . xy + yz + zx x+y+z

From (x + y + z)2 ≥ 3(xy + yz + zx), we get 1+

3 9 , ≥1+ xy + yz + zx (x + y + z)2

so it suffices to prove that 1+

9 6 ≥ . 2 (x + y + z) x+y+z

2 3 ≥ 0 and this ends the x+y+z proof. Note that the equality holds if and only if a = b = c = 1. ?F?

 The last inequality is equivalent to 1 −

03.12. Let a, b, c be positive real numbers. Prove that (2a + b + c)2 (2b + c + a)2 (2c + a + b)2 + + ≤ 8. 2a2 + (b + c)2 2b2 + (c + a)2 2c2 + (a + b)2 (USA 2003) First solution: Because the inequality is homogeneous, we can assume that a + b + c = 3. Then   (2a + b + c)2 a2 + 6a + 9 1 2(4a + 3) = 1+ = 2 2a2 + (b + c)2 3a − 6a + 9 3 2 + (a − 1)2   1 2(4a + 3) 4a + 4 ≤ 1+ = . 3 2 3 Thus

X (2a + b + c)2 1X ≤ (4a + 4) = 8, 2 2 2a + (b + c) 3

as desired. Note that the equality holds if and only if a = b = c. Second solution: Observe that 3−

(2a + b + c)2 2(b + c − a)2 = , 2a2 + (b + c)2 2a2 + (b + c)2 366

the desired inequality can be rewritten as (b + c − a)2 (c + a − b)2 (a + b − c)2 1 + + ≥ . 2 2 2 2 2 2 2a + (b + c) 2b + (c + a) 2c + (a + b) 2 By the Cauchy Schwarz Inequality, we find that X  i2 X 2 i hX (b + c − a)2 hX 2  2 2 = a2 . b 2a + (b + c) ≥ b(b + c − a) 2a2 + (b + c)2 Therefore, the above inequality is deduced from  X 2 X   b2 2a2 + (b + c)2 . 2 a2 ≥ By expanding, we see that it is equivalent to a4 + b4 + c4 + a2 b2 + b2 c2 + c2 a2 ≥ 2(a3 b + b3 c + c3 a), which is true since a4 + a2 b2 ≥ 2a3 b, b4 + b2 c2 ≥ 2b3 c, c4 + c2 a2 ≥ 2c3 a. b+c c+a a+b Third solution: Denote x = ,y = ,z = . We have to prove a b c that X (x − 1)2 X (x + 2)2 1 ≤ 8, or equivalently, ≥ . x2 + 2 x2 + 2 2 But, from the Cauchy Schwarz Inequality, we have X (x − 1)2 x2

+2

≥

(x + y + z − 3)2 . x2 + y 2 + z 2 + 6

It remains to prove that 2(x2 + y 2 + z 2 + 2xy + 2yz + 2zx − 6x − 6y − 6z + 9) ≥ x2 + y 2 + z 2 + 6, or (x + y + z)2 + 2(xy + yz + zx) − 12(x + y + z) + 12 ≥ 0. p Now xy + yz + zx ≥ 3 3 x2 y 2 z 2 ≥ 12 (because xyz ≥ 8), so we still have to prove that (x + y + z)2 + 24 − 12(x + y + z) + 12 ≥ 0, which is equivalent to (x + y + z − 6)2 ≥ 0, clearly true. ?F?  π , the following inequality holds 03.13. Prove that for any a, b, c ∈ 0, 2 X sin a · sin(a − b) · sin(a − c) ≥ 0. sin(b + c) (USA 2003) Solution: Let x = sin a, y = sin b, z = sin c. Then we have x, y, z > 0. It is easy to see that the following relations are true sin a · sin(a − b) · sin(a − c) · sin(a + b) · sin(a + c) = x(x2 − y 2 )(x2 − z 2 ). 367

Using similar relations for the other terms, we have to prove that x(x2 − y 2 )(x2 − z 2 ) + y(y 2 − z 2 )(y 2 − x2 ) + z(z 2 − x2 )(z 2 − y 2 ) ≥ 0. With the substitution x = √

u(u − v)(u − w) +

√

u, y =

√

√

v, z =

√

w, the inequality becomes

v(v − w)(v − u) +

√

w(w − u)(w − v) ≥ 0.

But this follows from the Schur’s Inequality and hence, our proof is completed. Note that the equality holds if and only if a = b = c. ?F? 04.1. Prove that (a2 + 2)(b2 + 2)(c2 + 2) ≥ 9(ab + bc + ca) for any positive real numbers a, b, c. (APMO 2004) First solution: We denote x = a2 + 2, y = b2 + 2, z = c2 + 2, and apply the AM-GM Inequality to obtain   1 a2 (b2 + 2) b2 (a2 + 2) ab ≤ + 2 a2 + 2 b2 + 2      1 2 2 2 2 = 1− 2 (b + 2) + 1 − 2 (a + 2) 2 a +2 b +2        1 2 2 1 2x 2y = 1− y+ 1− x = x+y− − . 2 x y 2 y x Proceeding the same way as this, we can get two other similar estimations for bc and ca. Adding up these three inequalities, one can see that ab + bc + ca ≤ x + y + z −

x y y z z x − − − − − . y x z y x z

And thus, we can reduce the original inequality into proving that   x y y z z x xyz ≥ 9 x + y + z − − − − − − , y x z y x z or equivalently, 9x 9x 9y 9y 9z 9z + + + + + + xyz ≥ 9x + 9y + 9z. y z z x x y The rest is simple as we have (by the AM-GM Inequality) r 9x 9x xyz 9x 9x xyz + + ≥33 · · = 9x. y z 3 y z 3 Note that the equality holds if and only if a = b = c = 1. 368

√ √
Second solution: Choose A, B, C ∈ 0, π2 with a = 2 tan A, b = 2 tan B, √ and c = 2 tan C. Using the well-known trigonometric identity 1 + tan2 θ = 1 , we may rewrite our inequality as cos2 θ 4 ≥ cos A cos B cos C (cos A sin B sin C + sin A cos B sin C + sin A sin B cos C) . 9 We can easily check the following trigonometric identity X cos(A + B + C) = cos A cos B cos C − cos A sin B sin C. Then, the above trigonometric inequality takes the form 4 ≥ cos A cos B cos C [cos A cos B cos C − cos(A + B + C)] . 9 A+B+C . Applying the AM-GM Inequality and Jensen’s InequalLet θ = 3 ity, we have  cos A cos B cos C ≤

cos A + cos B + cos C 3

3

≤ cos3 θ.

We now need to show that 4 ≥ cos3 θ(cos3 θ − cos 3θ). 9 Using the trigonometric identity cos 3θ = 4 cos3 θ − 3 cos θ

or

cos3 θ − cos 3θ = 3 cos θ − 3 cos3 θ,

it becomes


4 ≥ cos4 θ 1 − cos2 θ , 27 which follows from the AM-GM Inequality 


cos2 θ cos2 θ · · 1 − cos2 θ 2 2

 13

 
1 cos2 θ cos2 θ 1 2 ≤ + + 1 − cos θ = . 3 2 2 3

Third solution: Observe that 9(ab + bc + ca) ≤ 3(a + b + c)2 , and so, it suffices to prove that (a2 + 2)(b2 + 2)(c2 + 2) ≥ 3(a + b + c)2 . On the other hand, one has [(a2 −1)(b2 −1)]·[(b2 −1)(c2 −1)]·[(c2 −1)(a2 −1)] = (a2 −1)2 (b2 −1)2 (c2 −1)2 ≥ 0. Therefore, at least one of three expressions (a2 − 1)(b2 − 1), (b2 − 1)(c2 − 1), (c2 − 1)(a2 − 1) must be nonnegative. And since the inequality is symmetric, 369

we can suppose that (a2 − 1)(b2 − 1) ≥ 0. Now, applying the Cauchy Schwarz Inequality, we have (a + b + c)2 = (a · 1 + b · 1 + 1 · c)2 ≤ (a2 + b2 + 1)(12 + 12 + c2 ). According to this inequality, one can see that the above inequality is deduced from (a2 + 2)(b2 + 2) ≥ 3(a2 + b2 + 1). However, this is obviously true since it is equivalent to (a2 − 1)(b2 − 1) ≥ 0. ?F? 04.2. Let x, y, z, t be positive real numbers such that xyzt = 1. Prove that 1 1 1 1 + + + ≥ 1. 2 2 2 (1 + x) (1 + y) (1 + z) (1 + t)2 (China 2004) First solution: From x, y, z, t > 0 and xyzt b positive numbers a, b, c, d satisfying x = , y = a substitution, the desired inequality becomes

= 1, we can find that exist c d a , z = , t = . And by this b c d

a2 b2 c2 d2 + + + ≥ 1. (a + b)2 (b + c)2 (c + d)2 (d + a)2 Now, applying the Cauchy Schwarz Inequality, we have  hX X i hX i2 a2 2 2 ≥ a(a + b) (a + b) (a + d) . (a + b)2 According to this estimation, it suffices to prove that hX i2 X a(a + b) ≥ (a + b)2 (a + d)2 . X Since a(a + b) = the last inequality is

1 [(a + b)2 + (b + c)2 + (c + d)2 + (d + a)2 ], we find that 2 equivalent to

(m + n + p + w)2 ≥ 4(mn + np + pw + wm), where m = (a + b)2 , n = (b + c)2 , p = (c + d)2 , w = (d + a)2 . However, this inequality is clearly true (according to the AM-GM Inequality), so our proof is completed. Note that the equality holds if and only if x = y = z = t = 1. Second solution: According to the Cauchy Schwarz Inequality, we have 1 1 + ≥ 2 (1 + x) (1 + y)2

1 1 1  +  = . y x 1 + xy (1 + xy) 1 + (1 + xy) 1 + x y

And in the same way, we also have 1 1 1 xy + ≥ = . 2 2 (1 + z) (1 + t) 1 + zt 1 + xy 370

Adding up these two inequalities, the result follows immediately. ?F? 04.3. If a, b, c are positive real numbers, then √ a b c 3 2 1< √ +√ +√ ≤ . 2 a2 + b2 b2 + c2 c2 + a2 (China 2004) First solution: By applying the Cauchy Schwarz Inequality, we find that X

a √ 2 a + b2

2

#2 a2 = · (a2 + c2 ) (a2 + b2 )(a2 + c2 ) X  hX i a2 2 2 ≤ (a + c ) (a2 + b2 )(a2 + c2 ) 4(a2 + b2 + c2 )(a2 b2 + b2 c2 + c2 a2 ) = . (a2 + b2 )(b2 + c2 )(c2 + a2 ) s

"

X

On the other hand, the AM-GM Inequality implies (a2 + b2 + c2 )(a2 b2 + b2 c2 + c2 a2 ) = = (a2 + b2 )(b2 + c2 )(c2 + a2 ) + a2 b2 c2 ≤ (a2 + b2 )(b2 + c2 )(c2 + a2 ) +

(a2 + b2 )(b2 + c2 )(c2 + a2 ) 8

9 = (a2 + b2 )(b2 + c2 )(c2 + a2 ). 8 Using this in combination with the above inequality, we get X

a √ a2 + b2

2

9 ≤ , 2

from which it follows that √ a b c 3 2 √ +√ +√ ≤ . 2 a2 + b2 b2 + c2 c2 + a2 This is the right hand side of the desired inequality. We still have to prove the left hand side. However, this is actual easy and we can prove it as follows √

a b c a b c +√ +√ > + + 2 2 2 2 2 a + b b + c c + a +b b +c c +a a b c > + + = 1. a+b+c b+c+a c+a+b

a2

Note that the equality in the right hand side occurs iff a = b = c. 371

Second solution: We provide another solution for the right hand side inc a b equality. With the notations x = , y = , z = , the inequality reduces to a b c proving that xyz = 1 implies r r r 2 2 2 + + ≤ 3. 1 + x2 1 + y2 1 + z2 We presume that x ≤ y ≤ z, which implies xy ≤ 1 and z ≥ 1. We have r

2 + 1 + x2

r

2 1 + y2

so

!2

r

   1 − x2 y 2 2 2 =4 1+ + ≤2 1 + x2 1 + y 2 (1 + x2 )(1 + y 2 )   8 8z 1 − x2 y 2 = = , ≤4 1+ 2 (1 + xy) 1 + xy z+1 

2 + 1 + x2

r

2 ≤2 1 + y2

r

2z , z+1

and we need to prove that r 2 r Because

2z + z+1

r

2 ≤ 3. 1 + z2

2 2 ≤ , we only need to prove that 2 1+z z+1 r 2 2z + ≤ 3. 2 z+1 z+1

This inequality is equivalent to 1 + 3z − 2

p 2z(1 + z) ≥ 0,

or

√

2z −

√

1+z

2

≥ 0,

which is obvious and we are done. Third solution: We give another way to prove the right hand side inequality. Clearly, the inequality asks to prove that if xyz = 1, then r r r 2 2 2 + + ≤ 3. 1+x 1+y 1+z We have two cases. The first and easy one is when xy + yz + zx ≥ x + y + z. In this case, we can apply the Cauchy Schwarz Inequality to get r r X X 2 2 ≤ 3 . 1+x 1+x But

X

2 ≤ 3, since it is equivalent to 1+x X 2(xy + x + y + 1) ≤ 3(2 + x + y + z + xy + yz + zx), 372

or x + y + z ≤ xy + yz + zx, and so in this case, the inequality is proved. The second case is when xy + yz + zx < x + y + z. Thus, (x − 1)(y − 1)(z − 1) = x + y + z − (xy + yz + zx) > 0, and so exactly two of the numbers x, y, z are smaller than 1, let them be x and y. So, we must prove that if x and y are smaller than 1, then r r r 2 2 2xy + + ≤ 3. x+1 y+1 xy + 1 Using the Cauchy Schwarz Inequality, we get r r r r r 1 2 2 2xy 1 2xy + + ≤2 + + , x+1 y+1 xy + 1 x+1 y+1 xy + 1 and so it is enough to prove that this last inequality is at most 3. But this comes down to 1 1 2xy + −1 1− x+1 y+1 xy + 1 r r 2· ≤ . 1 1 2xy 1+ + 1+ x+1 y+1 xy + 1 Because

1 1 + ≥ 1, the left hand side is at most x+1 y+1 1 1 1 − xy + −1= , x+1 y+1 (x + 1)(y + 1)

and so we are left with the inequality 1 − xy ≤ (x + 1)(y + 1)

1 − xy r  , 2xy (1 + xy) 1 + xy + 1

or equivalently, r xy + 1 + (xy + 1)

2xy ≤ xy + 1 + x + y, xy + 1

that is, p x + y ≥ 2xy(xy + 1). p √ This last one follows from 2xy(xy + 1) ≤ 2 xy ≤ x + y, and thus, our inequality is proved. ?F? 04.4. Let n be a positive integer with n greater than one, and let a1 , a2 , . . . , an be positive integers such that a1 < a2 < · · · < an and 1 1 1 + + ··· + ≤ 1. a1 a2 an 373

Prove that, for any real number x, the following inequality holds 

1 1 1 + + ··· + 2 an + x2 a21 + x2 a22 + x2

2 ≤

1 1 . · 2 a1 (a1 − 1) + x2 (China 2004)

Solution: Applying the Cauchy Schwarz Inequality, we have n X i=1

1 2 ai + x2

!2

n X 1 ai

≤ ≤

=

i=1

i=1 n X i=1

=

!" n X

#

n

X ai ai ≤ 2 2 2 2 (ai + x ) (ai + x2 )2 − a2i i=1

n X

(a2i + i=1 n  X 1 2

ai 2 (ai + x2 )2

i=1

x2

ai + ai )(a2i + x2 − ai )

1 1 − a2i − ai + x2 a2i + ai + x2

 .

Denoting an+1 = an + 1, we have that a1 < a2 < · · · < an < an+1 and because ai is a positive integer, we can see that ai ≤ ai+1 − 1 for any i = 1, 2, . . . , n. By this argument, it follows that a2i

1 1 1 1 − 2 ≤ 2 − 2 2 2 2 (ai+1 − 1) + (ai+1 − 1) + x2 − ai + x ai + ai + x ai − ai + x 1 1 = 2 − 2 2 ai − ai + x ai+1 − ai+1 + x2

for any i = 1, 2, . . . , n. Therefore, from the above estimation, we get n X i=1

1 2 ai + x2

!2

 n  1X 1 1 ≤ − 2 a2i − ai + x2 a2i+1 − ai+1 + x2 i=1   1 1 1 − = 2 a21 − a1 + x2 a2n+1 − an+1 + x2 1 1 1 1 < · 2 = · , 2 2 a1 − a1 + x 2 a1 (a1 − 1) + x2

which is just the desired result. ?F? 04.5. Find all real numbers k such that the following inequality a3 + b3 + c3 + d3 + 1 ≥ k(a + b + c + d) holds for all real numbers a, b, c, d ≥ −1. (China 2004) 374

3 Solution: If a = b = c = d = −1, then −3 ≥ k · (−4), and hence k ≥ . If 4 1 1 3 3 a = b = c = d = , then 4 · + 1 ≥ k · 2. Thus k ≤ , and so k = . Now, we 2 8 4 4 want to prove that the inequality 3 a3 + b3 + c3 + d3 + 1 ≥ (a + b + c + d) 4 holds for any a, b, c, d ∈ [−1, ∞). At first, we prove that 4x3 + 1 ≥ 3x for x ∈ [−1, ∞). In fact, from (x + 1)(2x − 1)2 ≥ 0, we have 4x3 + 1 ≥ 3x for all x ∈ [−1, ∞). Therefore 4a3 + 1 ≥ 3a,

4b3 + 1 ≥ 3b,

4c3 + 1 ≥ 3c,

4c3 + 1 ≥ 3d.

By adding theses four inequalities and dividing each side of the resulting inequality by 4, we get 3 a3 + b3 + c3 + d3 + 1 ≥ (a + b + c + d), 4 3 as claimed. Thus, the real number we want to find is k = . 4 ?F? 04.6. Determine the maximum constant λ such that x + y + z ≥ λ, √ √ √ where x, y, z are positive real numbers with x yz + y zx + z xy ≥ 1. (China 2004) √ Solution: The√maximum value of λ is 3. It is not difficult to see that for √ √ 3 √ √ x=y =z = , x yz + y zx + z xy = 1 and x + y + z = 3. Hence, 3 √ the maximum value of λ is less than or equal to 3. It suffices to show that √ x + y + z ≥ 3. Indeed, applying the AM-GM Inequality in combination with (x + y + z)2 the well-known inequalities xy + yz + zx ≤ , we have 3 √ y+z z+x x+y √ √ 1 ≤ x yz + y zx + z xy ≤ x · +y· +z· 2 2 2 (x + y + z)2 = xy + yz + zx ≤ . 3 From this, we deduce that x+y+z ≥

√

3,

as desired. ?F? 04.7. Let a, b, c be positive real numbers. Determine the minimal value of the following expression a + 3c 4b 8c + − . a + 2b + c a + b + 2c a + b + 3c 375

(China 2004) Solution: Setting x = a + 2b + c, y = a + b + 2c and z = a + b + 3c. It is easy to see that c = z − y, b = z + x − 2y, and a = 5y − x − 3z. Now, using the AM-GM Inequality, we have a + 3c 4b 8c + − = a + 2b + c a + b + 2c a + b + 3c (5y − x − 3z) + 3(z − y) 4(z + x − 2y) 8(z − y) = + − x y z     2y z 2x y +4 = −17 + 2 + + y x z y √ √ √ ≥ 4 2 + 8 2 − 17 = 12 2 − 17. √ 2x y 2y z = and = , or 2x = 2y = z. y x z y√ Hence the equality holds if and only if a + b + 2c = 2(a + 2b + c) and a + b + 3c = 2(a + 2b√ + c). √
of equations for b and c in terms
Solving this system of a gives b = 1 + 2 a and c = 4 + 3 2 a. From now, we conclude that the expression a + 3c 4b 8c + − a + 2b + c a + b + 2c a + b + 3c √ √
√

has minimum value 12 2−17 if and only if (a, b, c) = a, 1 + 2 a, 4 + 3 2 a . ?F? The equality holds if and only if

04.8. Determine the largest constant M such that the following inequality holds for all real numbers x, y, z, x4 + y 4 + z 4 + xyz (x + y + z) ≥ M (xy + yz + zx)2 . (Hellenic 2004) 2 Solution: We will show that is the largest value of M. Indeed, take x = 3 2 y = z = 1, we get M ≤ . It suffices to prove that 3 x4 + y 4 + z 4 + xyz (x + y + z) ≥

2 (xy + yz + zx)2 , 3

oor equivalently,

3 x4 + y 4 + z 4 ≥ 2 x2 y 2 + y 2 z 2 + z 2 x2 + xyz (x + y + z) . Applying the well-known inequalities a2 + b2 + c2 ≥ ab + bc + ca when the triple (a, b, c) equals (x2 , y 2 , z 2 ) and (xy, yz, zx), we get 3(x4 + y 4 + z 4 ) ≥ 3(x2 y 2 + y 2 z 2 + z 2 x2 ), and x2 y 2 + y 2 z 2 + z 2 z 2 ≥ xyz(x + y + z). 376

Adding up these two inequalities, we get the result. ?F? 04.9. Let n ≥ 3 be an integer and let t1 , t2 , . . . , tn be positive real numbers such that   1 1 1 2 + + ··· + . n + 1 > (t1 + t2 + · · · + tn ) t1 t2 tn Show that ti , tj , tk are the side lengths of a triangle for all i, j, k with 1 ≤ i < j < k ≤ n. (IMO 2004) Solution: By symmetry, it suffices to show that t1 < t2 + t3 . If n > 3, applying the Cauchy Schwarz Inequality, we have ! n ! ! ! n n n X X1 X X 1 1 1 1 ti n2 + 1 > = t1 + t2 + t3 + ti + + + t t1 t2 t3 t i=1 i=1 i i=4 i=4 i v s  ! 2 ! n   u n X1 u X 1 1 1  ≥  (t1 + t2 + t3 ) + + +t ti t1 t2 t3 ti i=4 i=4 "s # 2   1 1 1 ≥ (t1 + t2 + t3 ) + + +n−3 . t1 t2 t3 Therefore, it follows that s (t1 + t2 + t3 )



1 1 1 + + t1 t2 t3

 <

p n2 + 1 − n + 3.

If n = 3, this inequality holds as well (according to the hypothesis). Since √ √ 1 1 √ √ n2 + 1 − n = ≤ = 10 − 3, the above inequality 3 + 10 n + n2 + 1 implies   1 1 1 (t1 + t2 + t3 ) + + < 10. t1 t2 t3 In order to prove t1 < t2 + t3 , it suffices to prove that if t1 ≥ t2 + t3 , then this inequality cannot hold. Indeed, if t1 ≥ t2 + t3 , we have       1 1 1 t2 + t3 1 1 1 1 (t1 + t2 + t3 ) + + = 1 + t1 + + + (t2 + t3 ) + t1 t2 t3 t2 t3 t1 t2 t3 4t1 t2 + t3 ≥1+ + +4 t2 + t3 t  1  3t1 t2 + t3 t1 =5+ + + t2 + t3 t2 + t3 t1 ≥ 5 + 3 + 2 = 10. This completes our proof. 377

Remark: It is not hard to determine the greatest number f (n) such that, for positive t1 , . . . , tn , the inequality ! n ! n X X1 < f (n) ti ti i=1

i=1

implies that each three of the ti ’s are the side lengths of a triangle. The answer √
2 is f (n) = n + 10 − 3 , and the proof is quite similar to the above solution. ?F? 04.10. For a, b, c positive reals, find the minimum value of b2 + c2 c2 + a2 a2 + b2 + + . a2 + bc b2 + ca c2 + ab (India 2004) Solution: We will show that c2 + a2 a2 + b2 b2 + c2 + + ≥ 3, a2 + bc b2 + ca c2 + ab with equality iff a = b = c. Indeed, after using the AM-GM Inequality, it suffices to prove that (a2 + b2 )(b2 + c2 )(c2 + a2 ) ≥ (a2 + bc)(b2 + ca)(c2 + ab), which is true since by the Cauchy Schwarz Inequality, we have (a2 + b2 )(b2 + c2 )(c2 + a2 ) = p p p = (a2 + b2 )(a2 + c2 ) · (b2 + c2 )(b2 + a2 ) · (c2 + a2 )(c2 + b2 ) ≥ (a2 + bc)(b2 + ca)(c2 + ab). This completes our proof, and so, the searched minimum is 3, which is attained for a = b = c. ?F?   1 . Prove that 04.11. Let x1 , x2 , . . . , xn be real numbers in the interval 0, 2 n Y xi i=1 n X

n Y

!n ≤ "

xi

i=1

i=1 n X

(1 − xi ) #n .

(1 − xi )

i=1

(India 2004) Solution: The desired inequality is equivalent to ! " n # n n n X X X X ln xi − n ln xi ≤ ln(1 − xi ) − n ln (1 − xi ) , i=1

i=1

i=1

378

i=1

or

n X

# ! n n X X [ln(1 − xi ) − ln xi ] ≥ n ln (1 − xi ) − n ln xi .

i=1

i=1

"

i=1

  1 . We have Consider the function f (x) = ln(1 − x) − ln x with x ∈ 0, 2 1 1 1 − 2x − = 2 > 0. 2 2 x (1 − x) x (1 − x)2   1 . And thus, we are allowed to Therefore f (x) is (strictly) convex on 0, 2 apply the Jensen’s Inequality and get   x1 + x2 + · · · + xn f (x1 ) + f (x2 ) + · · · + f (xn ) ≥ nf , n f 0 (x) = −

1 1 − , 1−x x

f 00 (x) =

which is equivalent to n X

# ! n n X X [ln(1 − xi ) − ln xi ] ≥ n ln (1 − xi ) − n ln xi .

i=1

i=1

"

i=1

Our proof is completed. Note that the equality holds if and only if x1 = x2 = · · · = xn . ?F? 04.12. Prove that for all positive real numbers a, b, the following inequality holds p p 2a(a + b)3 + b 2(a2 + b2 ) ≤ 3(a2 + b2 ). (Ireland 2004) First solution: By applying the AM-GM Inequality, we get p 2a(a + b) + (a + b)2 2a(a + b)3 ≤ , 2

p 2b2 + a2 + b2 and b 2(a2 + b2 ) ≤ . 2

Therefore p p 2a(a + b)3 + b 2(a2 + b2 ) ≤ 2a2 + 2b2 + 2ab ≤ 3(a2 + b2 ). The equality holds if and only if a = b. Second solution: Denoting P as the left hand side of the original inequality. According to the Cauchy Schwarz Inequality, we have that hp i2 p √ P2 = 2a(a + b) · (a + b) + 2b2 · a2 + b2    ≤ 2a(a + b) + 2b2 (a + b)2 + a2 + b2 = 4(a2 + ab + b2 )2 , and thus P =

p p 2a(a + b)3 + b 2(a2 + b2 ) ≤ 2a2 + 2b2 + 2ab ≤ 3(a2 + b2 ). 379

?F? 04.13. If a, b, c are positive reals such that a + b + c = 1, prove that   1+a 1+b 1+c b c a . + + ≤2 + + 1−a 1−b 1−c a b c (Japan 2004) First solution: By the Cauchy Schwarz Inequality, it is easy to see that b c a b2 c2 a2 (a + b + c)2 + + = + + ≥ . a b c ab bc ca ab + bc + ca So, the original inequality is deduced from 2(a + b + c)2 1+a 1+b 1+c ≥ + + . ab + bc + ca 1−a 1−b 1−c Noting that

1+a (a + b + c) + a 2a = = + 1, the last inequality is equiv1−a (a + b + c) − a b+c

alent to

2(a + b + c)2 2a 2b 2c ≥ + + + 3. ab + bc + ca b+c c+a a+b Multplying both sides for ab + bc + ca > 0 and using the fact that 2a(ab + bc + ca) 2abc a(b + c) = 2a2 + ≤ 2a2 + , b+c b+c 2 we can see that the above inequality follows from 2(a + b + c)2 ≥ 2a2 +

a(b + c) b(c + a) c(a + b) + 2b2 + + 2c2 + + 3(ab + bc + ca). 2 2 2

which is trivial since it leads to an identity. It is easy to see that the equality 1 occurs if and only if a = b = c = . 3 Second solution: Similar to the above solution, we see that

2a 1+a = + 1−a b+c

1, hence the desired inequality can be written as       2b 2b 2c 2c 2a 2a − + − + − ≥ 3, a c+a b a+b c b+c or

bc ca ab 3 + + ≥ . a(c + a) b(a + b) c(b + c) 2

Now, applying the Cauchy Schwarz Inequality, we have bc ca ab b2 c2 c2 a2 a2 b2 + + = + + a(c + a) b(a + b) c(b + c) abc(c + a) abc(a + b) abc(b + c) (ab + bc + ca)2 ≥ . 2abc(a + b + c) 380

Therefore, in order to prove the original inequality, it suffices to prove that (ab + bc + ca)2 3 ≥ , 2abc(a + b + c) 2 which can be written as (ab + bc + ca)2 ≥ 3abc(a + b + c). We have already proved this nice inequality in the previous problem. ?F? 04.14. Let a, b be real numbers in the interval [0, 1]. Prove that √

b 2 a +√ ≤√ . 2 2 7 2b + 5 2a + 5 (Lithuania 2004)

Solution: For a = b = 0, the inequality is trivial. Let us consider now the case a2 + b2 > 0. Using the Cauchy Schwarz Inequality in combination with the given hypothesis, we have 2    b a2 b2 a √ +√ ≤2 + 2b2 + 5 2a2 + 5 2b2 + 5 2a2 + 5   b2 a2 + . ≤2 2b2 + 5a2 2a2 + 5b2 So it is enough to prove that a2 b2 2 + ≤ . 2 2 2 2 2b + 5a 2a + 5b 7 Because

1 b2 a2 = − , the last inequality can be written as 2b2 + 5a2 5 5(2b2 + 5a2 ) b2 2 a2 + ≥ . 2 2 2 2 2a + 5b 2b + 5a 7

Now, we apply first the Cauchy Schwarz Inequality and then the AM-GM Inequality to get (a2 + b2 )2 a2 b2 (a2 + b2 )2 + ≥ = 2a2 + 5b2 2b2 + 5a2 a2 (2a2 + 5b2 ) + b2 (2b2 + 5a2 ) 2(a2 + b2 )2 + 6a2 b2 2 2 2 (a + b ) 2 ≥ = , 2 2 2 7 (a + b ) 2(a2 + b2 )2 + 6 · 4 as desired. It is easy to see that the equality holds if and only if a = b = 1. ?F? 04.15. Prove that for any positive real numbers a, b, c, 3 a − b3 b3 − c3 c3 − a3 (a − b)2 + (b − c)2 + (c − a)2 . a+b + b+c + c+a ≤ 4 381

(Moldova 2004) Solution: With notice that 3 a − b3 b3 − c3 c3 − a3 (ab + bc + ca) |(a − b)(b − c)(c − a)| , a+b + b+c + c+a = (a + b)(b + c)(c + a) one can write the original inequality as (a + b)(b + c)(c + a) [(a − b)2 + (b − c)2 + (c − a)2 ] ≥ 4 |(a − b)(b − c)(c − a)| . ab + bc + ca We denote with P the left hand side of this inequality, then by applying the AM-GM Inequality, we find that 3(a + b)(b + c)(c + a) p 3 (a − b)2 (b − c)2 (c − a)2 ab + bc + ca p 3 3 (a + b)2 (b + c)2 (c + a)2 ≥ |(a − b)(b − c)(c − a)| . ab + bc + ca

P ≥

Therefore, in order to prove the original inequality, it suffices to show that (a + b)2 (b + c)2 (c + a)2 ≥

64 (ab + bc + ca)3 . 27

This is obviously true because (a + b)2 (b + c)2 (c + a)2 ≥

64 64 (a + b + c)2 (ab + bc + ca)2 ≥ (ab + bc + ca)3 . 81 27

Note that the inequality becomes equality if and only if a = b = c. Second solution: It is easy to see that the inequality is not only cyclic, but symmetric. That is why we may assume that a ≥ b ≥ c > 0. The idea is to use the inequality y x2 + xy + y 2 x x+ ≥ ≥y+ , 2 x+y 2 which is true if x ≥ y > 0. The proof of this inequality is easy and we won’t insist. Now, because a ≥ b ≥ c > 0, we have the three inequalities a+

b a2 + ab + b2 a ≥ ≥b+ , 2 a+b 2

and of course a+

b+

c b2 + bc + c2 b ≥ ≥c+ , 2 b+c 2

c a2 + ac + c2 a ≥ ≥c+ . 2 a+c 2

That is why we can write X a3 − b3

a2 + ab + b2 b2 + bc + c2 a2 + ac + c2 + (b − c) − (a − c) a+b a+b b+ c a+c    a b c ≥ (a − b) b + + (b − c) c + − (a − c) a + 2 2 2 X (a − b)2 =− . 4 = (a − b)

382

In the same manner, we can prove that X a3 − b3 a+b

≤

X (a − b)2 4

,

and the conclusion follows. ?F? 04.16. Prove that if n > 3 and x1 , x2 , . . . , xn > 0 have product 1, then 1 1 1 + + ··· + > 1. 1 + x1 + x1 x2 1 + x2 + x2 x3 1 + xn + xn x1 (Russia 2004) a2 Solution: We use a similar form of the classical substitution x1 = , x2 = a1 a3 a1 , . . . , x3 = . In this case, the inequality becomes a2 an a2 an a1 + + ··· + > 1, a1 + a2 + a3 a2 + a3 + a4 an + a1 + a2 and it is clear, because n > 3 and ai + ai+1 + ai+2 < a1 + a2 + · · · + an for all i. ?F? 04.17. (a) Given real numbers a, b, c with a + b + c = 0, prove that a3 + b3 + c3 > 0

if and only if a5 + b5 + c5 > 0.

(b) Given real numbers a, b, c, d with a + b + c + d = 0, prove that a3 + b3 + c3 + d3 > 0

if and only if a5 + b5 + c5 + d5 > 0. (United Kingdom 2004)

Solution: It is easy to show that a3 + b3 + c3 − (a + b + c)3 = −3(a + b)(b + c)(c + a), and X 5 a5 + b5 + c5 − (a + b + c)5 = − (a + b)(b + c)(c + a) (a + b)2 . 2 We will use these identities to solve our problem. (a) Using the above identities with noting that a + b + c = 0, we have a3 + b3 + c3 = −3(a + b)(b + c)(c + a) = −3(−c)(−a)(−b) = 3abc, and X 5 a5 + b5 + c5 = − (a + b)(b + c)(c + a) (a + b)2 2 X X 5 5 = − (−c)(−a)(−b) (−c)2 = abc a2 . 2 2 383

It follows that (a3 + b3 + c3 )(a5 + b5 + c5 ) =

15 2 2 2 X 2 a b c a . 2

The last quantity is obvious nonnegative, from which we deduce that a5 + b5 + c5 ≥ 0 if and only if a3 + b3 + c3 ≥ 0. Since we need to prove a5 + b5 + c5 > 0 if and only if a3 + b3 + c3 > 0, we have to disprove the following two cases The first case is when it exists some numbers a, b, c such that a5 + b5 + c5 > 0 and a3 + b3 + c3 = 0. Since a3 + b3 + c3 = 0 and a3 + b3 + c3 = 3abc, we get X 5 abc = 0, and hence a5 + b5 + c5 = abc a2 = 0, a contradiction. 2 The second case is when it exists some numbers a, b, c such that a5 +b5 +c5 = 0 X 5 a2 , we and a3 + b3 + c3 > 0. Since a5 + b5 + c5 = 0 and a5 + b5 + c5 = abc 2 X get abc = 0 or a2 = 0. If abc = 0, then we have a3 + b3 + c3 = 3abc = 0, contradiction. If a2 + b2 + c2 = 0, then we have a = b = c = 0 and hence a3 + b3 + c3 = 0, contradiction. From these arguments, we conclude that a3 + b3 + c3 > 0

if and only if a5 + b5 + c5 > 0,

as desired. (b) We use the same arguments to prove this claim. Indeed, we have a3 + b3 + c3 + d3 = a3 + b3 + c3 − (a + b + c)3 = −3(a + b)(b + c)(c + a), and a5 + b5 + c5 + d5 = a5 + b5 + c5 − (a + b + c)5 5 = − (a + b)(b + c)(c + a)[(a + b)2 + (b + c)2 + (c + a)2 ]. 2 It follows that (a3 + b3 + c3 + d3 )(a5 + b5 + c5 + d5 ) =

X 15 (a + b)2 . (a + b)2 (b + c)2 (c + a)2 2

The last quantity is obvious nonnegative, from which we deduce that a5 + b5 + c5 + d5 ≥ 0 if and only if a3 + b3 + c3 + d3 ≥ 0. Since we need to prove a5 + b5 + c5 + d5 > 0 if and only if a3 + b3 + c3 + d3 > 0, we have to disprove the following two cases The first case is when it exists some numbers a, b, c, d such that a5 + b5 + c5 + d5 > 0 and a3 + b3 + c3 + d5 = 0. Since a3 + b3 + c3 + d3 = 0 and a3 + b3 + c3 + d3 = −3(a + b)(b + c)(c + a), we get (a + b)(b + c)(c + a) = 0, and X 5 hence a5 +b5 +c5 +d3 = − (a+b)(b+c)(c+a) (a+b)2 = 0, a contradiction. 2 The second case is when it exists some numbers a, b, c, d such that a5 + b5 + c5 + d5 = 0 and a3 + b3 + c3 + d3 > 0. Since a5 + b5 + c5 + d5 = 0 and X 5 a5 +b5 +c5 +d5 = − (a+b)(b+c)(c+a) (a+b)2 , we get (a+b)(b+c)(c+a) = 0 2 X or (a + b)2 = 0. If (a + b)(b + c)(c + a) = 0, then we have a3 + b3 + c3 + d3 = 384

−3(a+b)(b+c)(c+a) = 0, contradiction. If (a+b)2 +(b+c)2 +(c+a)2 = 0, then we have a = b = c = 0 and hence a3 + b3 + c3 + d3 = −3(a + b)(b + c)(c + a) = 0, contradiction. From these arguments, we conclude that a3 + b3 + c3 + d3 > 0

if and only if a5 + b5 + c5 + d5 > 0,

as desired. ?F? 04.18. Let a, b, c be positive real numbers. Prove that (a5 − a2 + 3)(b5 − b2 + 3)(c5 − c2 + 3) ≥ (a + b + c)3 . (USA 2004) Solution: For any positive number x, the quantities x2 − 1 and x3 − 1 have the same sign. Thus, we have 0 ≤ (x3 − 1)(x2 − 1) = x5 − x3 − x2 + 1, or x5 − x2 + 2 ≥ x3 + 2. It follows that (a5 − a2 + 3)(b5 − b2 + 3)(c5 − c2 + 3) ≥ (a3 + 2)(b3 + 2)(c3 + 2), and hence it suffices to prove that (a3 + 2)(b3 + 2)(c3 + 2) ≥ (a + b + c)3 . Of course, this is true since by the Holder’s Inequality, we have (a3 + 1 + 1)(1 + b3 + 1)(1 + 1 + c3 ) ≥

√ 3

a3 · 1 · 1 +

3 √ √ 3 3 1 · b3 · 1 + 1 · 1 · c3

= (a + b + c)3 . Note that the equality holds if and only if a = b = c = 1. ?F? 04.19. Let x, y, z > 0 such that (x + y + z)3 = 32xyz. Find the minimum and x4 + y 4 + z 4 maximum of . (x + y + z)4 (Vietnam 2004) First solution: We may of course assume that x + y + z = 4 and xyz = 2. x4 + y 4 + z 4 Thus, we have to find the extremal values of . Now, we have 44 x4 + y 4 + z 4 = (x2 + y 2 + z 2 )2 − 2(x2 y 2 + y 2 z 2 + z 2 x2 )  X 2 X 2 = 16 − 2 xy − 2 xy + 4xyz(x + y + z) = 2a2 − 64a + 288, 385

2 where a = xy + yz + zx. Because y + z = 4 − x and yz = , we must have x √ 8 2 (4 − x) ≥ , which implies that 3 − 5 ≤ x ≤ 2. Due to symmetry, we have  x√  x, y, z ∈ 3 − 5, 2 . This means that (x − 2)(y − 2)(z − 2) ≤ 0, and also  √  √  √  x − 3 + 5 y − 3 + 5 z − 3 + 5 ≥ 0. Clearing parenthesis, we deduce that " # √ 5 5−1 a ∈ 5, . 2 x4 + y 4 + z 4 (a − 16)2 − 112 = , we find that the extremal values 44 128 √ 9 383 − 165 5 of the expression are , , attained for the triples (2, 1, 1) and 128 256 √ ! √ √ 1+ 5 1+ 5 , , respectively. 3 − 5, 2 2

But because

Second solution: As in the above solution, we must find the extremal values 1 of x2 + y 2 + z 2 when x + y + z = 1, xyz = , because after that the extremal 32 values of the expression x4 + y 4 + z 4 can be immediately found. Let us make b c a substitution x = , y = , z = , where abc = 1, a + b + 2c = 4. Then 4 4 2 a2 + b2 + 4c2 2 2 2 x +y +z = and so we must find the extremal values of 16 2 a2 + b2 + 4c2 . Now, a2 + b2 + 4c2 = (4 − 2c)2 − + 4c2 and the problem reduces c 1 to find the maximum and minimum of 4c2 − 8c − where there are positive c real numbers a, b, c such that abc "= 1, a + b # + 2c = 4. Of course, this comes √ 4 3− 5 2 down to (4 − 2c) ≥ , or to c ∈ , 1 . But this reduces to the study c 2 " # √ 1 3− 5 2 of the function f (x) = 4x − 8x − defined for , 1 , which is an easy x 2 task. ?F? 05.1. For any positive real numbers a, b, c such that abc = 8, then a2 b2 c2 4 p +p +p ≥ . 3 3 3 3 3 3 3 (a + 1) (b + 1) (b + 1) (c + 1) (c + 1) (a + 1) (APMO 2005) Solution: For any x > 0, the AM-GM Inequality implies
2−x+1 p p (x + 1) + x x2 + 2 x3 + 1 = (x + 1) (x2 − x + 1) ≤ = . 2 2 386

Applying this inequality, we get b2 c2 a2 p +p +p ≥ (a3 + 1) (b3 + 1) (b3 + 1) (c3 + 1) (c3 + 1) (a3 + 1)   a2 b2 c2 ≥4 . + + (a2 + 2) (b2 + 2) (b2 + 2) (c2 + 2) (c2 + 2) (a2 + 2) Therefore, in order to prove the desired inequality, it suffices to show that a2 b2 c2 1 + + ≥ . 2 2 2 2 2 2 (a + 2) (b + 2) (b + 2) (c + 2) (c + 2) (a + 2) 3 By some easy computations (with noting that abc = 8), we can write this inequality as a2 b2 + b2 c2 + c2 a2 + 2(a2 + b2 + c2 ) ≥ 72. This last inequality is obviously true since from the AM-GM Inequality, we have √ 3 a2 b2 + b2 c2 + c2 a2 ≥ 3 a4 b4 c4 = 48, and

√ 3 a2 + b2 + c2 ≥ 3 a2 b2 c2 = 12.

The equality holds if and only if a = b = c = 2. ?F? 05.2. Let a, b, c, d be positive real numbers. Show that 1 1 1 1 a+b+c+d + 3+ 3+ 3 ≥ . 3 a b c d abcd (Austria 2005) 1 1 1 1 Solution: Setting x = , y = , z = , t = , then we can rewrite the a b c d inequality as x3 + y 3 + z 3 + t3 ≥ xyz + yzt + ztx + txy. By the AM-GM Inequality, we have that x3 +y 3 +z 3 ≥ 3xyz,

y 3 +z 3 +t3 ≥ 3yzt,

z 3 +t3 +x3 ≥ 3ztx,

t3 +x3 +y 3 ≥ 3txy.

Adding up these four inequalities and dividing each side of the resulting inequality by 3, we get the result. It is easy to see that the equality holds if and only if a = b = c = d. ?F? 05.3. Let a, b, c be positive real numbers. Prove that a2 b2 c2 4(a − b)2 + + ≥a+b+c+ . b c a a+b+c (Balkan 2005) 387

Solution: The inequality in question is equivalent to  2   2   2  a b c 4(a − b)2 + b − 2a + + c − 2b + + a − 2c ≥ , b c a a+b+c or

(a − b)2 (c − b)2 (a − c)2 4(a − b)2 + + ≥ . b c a a+b+c Now, applying the Cauchy Schwarz Inequality, we have (a − b)2 (c − b)2 (a − c)2 [(a − b) + (c − b) + (a − c)]2 4(a − b)2 + + ≥ = , b c a b+c+a a+b+c which is just the desired inequality. The equality holds if and only if a = b = c. ?F? 05.4. If a, b, c are positive real numbers such that abc = 1, then the inequality holds b c a + 2 + 2 ≤ 1. 2 a +2 b +2 c +2 (Baltic Way 2005) 1 1 1 Solution: Let x = , y = , z = , then xyz = 1. Now, applying the a b c AM-GM Inequality, we have a b c a b c + + ≤ + + a2 + 2 b2 + 2 c2 + 2 2a + 1 2b + 1 2c + 1 1 1 1 + + . = 2+x 2+y 2+z Thus, it suffices to show that 1 1 1 + + ≤ 1, 2+x 2+y 2+z which is equivalent to (2 + x) (2 + y) + (2 + y) (2 + z) + (2 + z) (2 + x) ≤ (2 + x) (2 + y) (2 + z) , or 3 ≤ xy + yz + zx. p However, this is true since xy + yz + zx ≥ 3 3 x2 y 2 z 2 = 3 by the AM-GM Inequality. Therefore, our proof is completed. Note that the equality holds if and only if a = b = c = 1. ?F? 05.5. Let a, b, c be positive real numbers. Prove that       3 1 1 3 2 2 b +a+ ≥ 2a + 2b + . a +b+ 4 4 2 2 (Belarus 2005) 388

Solution: Applying the AM-GM Inequality, we get ! r ! r    3 3 1 1 1 1 a2 + b + b2 + a + ≥ 2 a2 · + b + 2 b2 · + a + 4 4 4 2 4 2 1 1 = (a + b)2 + a + b + ≥ 4ab + a + b + 4 4   1 1 2b + , = 2a + 2 2 1 as desired. The equality holds if and only if a = b = . 2 ?F? 05.6. Given positive real numbers a, b, c such that a + b + c = 1. Prove that √ √ √ 1 a b+b c+c a≤ √ . 3 (Bosnia and Hercegovina 2005) Solution: Using the well-known inequality (a + b + c)2 ≥ 3(ab + bc + ca) and 1 the hypothesise a + b + c = 1, we deduce that ab + bc + ca ≤ . Thus, by the 3 Cauchy Schwarz Inequality, we get  √ √ √ √ √ 2 √ √ √ √ 2 a ab + b bc + c ca a b+b c+c a = ≤ (a + b + c)(ab + bc + ca) ≤ 1 ·

1 1 = . 3 3

This implies that

√ √ √ 1 a b+b c+c a≤ √ , 3 which is just the desired inequality. It is easy to see that the equality holds iff 1 a=b=c= . 3 ?F?

05.7. Let x, y be positive real numbers such that x3 + y 3 = x − y. Prove that x2 + 4y 2 < 1. (China 2005) Solution: From the given hypothesis, we have (x3 + y 3 )(1 − x2 − 4y 2 ) = x3 + y 3 − (x3 + y 3 )(x2 + 4y 2 ) = x3 + y 3 − (x − y)(x2 + 4y 2 )   = y(x2 − 4xy + 5y 2 ) = y (x − 2y)2 + y 2 > 0. Since x3 + y 3 > 0, this inequality implies that x2 + 4y 2 < 1, 389

as desired. ?F? 1 05.8. Let a, b, c be positive real numbers such that ab + bc + ca = . Prove 3 that 1 1 1 + + ≤ 3. a2 − bc + 1 b2 − ca + 1 c2 − ab + 1 (China 2005) Solution: From 3 (ab + bc + ca) = 1, we have a2 − bc + 1 = a2 + 3 (ab + ac) + 2bc = a (a + b + c) + 2 (ab + bc + ca) , and so, we can rewrite our inequality as X This is equivalent to X

2 (ab + bc + ca) ≤ 2. a (a + b + c) + 2 (ab + bc + ca)

1−

 2 (ab + bc + ca) ≥ 1, a (a + b + c) + 2 (ab + bc + ca)

or (a + b + c)

X

a ≥ 1. a (a + b + c) + 2 (ab + bc + ca)

Now, denoting with P the left hand side of the above inequality and applying the Cauchy Schwarz Inequality, we have P = (a + b + c)

X

a2 a2 (a + b + c) + 2a (ab + bc + ca)

(a + b + c)2 . ≥ (a + b + c) · X [a2 (a + b + c) + 2a (ab + bc + ca)] On the other hand, it is easy to verify that X [a2 (a + b + c) + 2a(ab + bc + ca)] = (a + b + c)3 . Therefore, from the above Cauchy Schwarz step, the result follows immedi1 ately. Note that the equality holds if and only if a = b = c = . 3 ?F? 05.9. Given positive real numbers a, b, c such that a + b + c = 1. Prove that the following inequality holds 10(a3 + b3 + c3 ) − 9(a5 + b5 + c5 ) ≥ 1. (China 2005) Solution: Replacing 1 by a + b + c, we can rewrite the original inequality as (10a3 − 9a5 − a) + (10b3 − 9b5 − b) + (10c3 − 9c5 − c) ≥ 0, 390

or equivalently, a(1 − a2 )(9a2 − 1) + b(1 − b2 )(9b2 − 1) + c(1 − c2 )(9c2 − 1) ≥ 0. Now, with noting that for any nonnegative number x, 8 1 (1 + x)(9x2 − 1) − (3x − 1) = (3x + 5)(3x − 1)2 ≥ 0, 3 3 we can find that the above inequality is deduced from a(1 − a)(3a − 1) + b(1 − b)(3b − 1) + c(1 − c)(3c − 1) ≥ 0. Expanding (with noting that a + b + c = 1), one may write it into 4(a2 + b2 + c2 ) − 3(a3 + b3 + c3 ) ≥ 1, or 4(a2 + b2 + c2 )(a + b + c) − 3(a3 + b3 + c3 ) ≥ (a + b + c)3 . Expanding and simplifying once more time, it can be written in the form ab(a + b) + bc(b + c) + ca(c + a) ≥ 6abc. Of course, this last inequality is obviously true by the AM-GM Inequality and 1 so, our proof is completed. The equality holds if and only if a = b = c = . 3 ?F? 05.10. Let ABC be an acute triangle. Determine the least value of the following expression P =

cos2 B cos2 C cos2 A + + . cos A + 1 cos B + 1 cos C + 1 (China 2005)

First solution: Put x = cot A, y = cot B, z = cot C, then x, y, z > 0 and xy + yz + zx = 1. Now, we see that x2 x2 x2 + 1 √  = =√ x cos A + 1 2+1 2+1+x 1+ √ x x x2 + 1 √  x2 x2 + 1 − x x3 √ = = x2 − √ x2 + 1 x2 + 1 x3 x3 = x2 − p = x2 − p (x + y)(x + z) x2 + xy + yz + zx cos2 A

≥ x2 −

x3 x3 − , 2(x + y) 2(x + z) 391

Similarly, we also have cos2 B y3 y3 ≥ y2 − − , cos B + 1 2(y + z) 2(y + x) cos2 C z3 z3 ≥ z2 − − . cos C + 1 2(z + x) 2(z + y) Hence cos2 B cos2 C cos2 A + + cos A + 1 cos B + 1 cos C + 1 x3 + y 3 y3 + z3 z 3 + x3 ≥ x2 + y 2 + z 2 − − − 2(x + y) 2(y + z) 2(z + x) 1 1 1 = x2 + y 2 + z 2 − (x2 − xy + y 2 ) − (y 2 − yz + z 2 ) − (z 2 − zx + x2 ) 2 2 2 1 1 = (xy + yz + zx) = . 2 2

P =

Moreover, we can see that the equality holds for an equilateral triangle. There1 fore, the minimum value of P is . 2 Second solution: Similarly, we need to prove the inequality cos2 A cos2 B cos2 C 1 + + ≥ . cos A + 1 cos B + 1 cos C + 1 2 Put x = cos A, y = cos B, z = cos C then x, y, z > 0 and x2 +y 2 +z 2 +2xyz = 1. Our inequality becomes x2 y2 z2 1 + + ≥ , x+1 y+1 z+1 2 which is equivalent to       2y 2 2z 2 2x2 2 2 2 −x + −y + − z ≥ 1 − x2 − y 2 − z 2 , x+1 y+1 z+1 or

x2 (1 − x) y 2 (1 − y) z 2 (1 − z) + + ≥ 2xyz. 1+x 1+y 1+z

By the AM-GM Inequality, we get s x2 (1 − x) y 2 (1 − y) z 2 (1 − z) x2 y 2 z 2 (1 − x)(1 − y)(1 − z) + + ≥33 . 1+x 1+y 1+z (1 + x)(1 + y)(1 + z) Hence, it suffices to prove that (1 − x)(1 − y)(1 − z) ≥ or



1 −1 x



8 xyz(1 + x)(1 + y)(1 + z), 27

  1 1 8 −1 − 1 ≥ (1 + x)(1 + y)(1 + z). y z 27 392

Now, we can see that this inequality follows from combining the two inequalities 8 (1 + x)(1 + y)(1 + z) ≤ 1, 27 and     1 1 1 −1 −1 − 1 ≥ 1. x y z In addition, it is easy to see that the first inequality follows immediately from the AM-GM Inequality and the well-known x + y + z = cos A + cos B + cos C ≤ 3 . So, it suffices to prove the second. Without loss of generality, we may 2 3 assume that x ≥ y ≥ z. From this assumption and the inequality x+y +z ≤ , 2 we get y + z ≤ 1, and hence it follows that     2 1 1 2 (1 − y − z)(y − z)2 ≥ 0. −1 −1 − −1 = y z y+z yz(y + z)2 On the other hand, we have (y + z)2 − 2(1 − x) = y 2 + z 2 + 2yz + 2x − 2(x2 + y 2 + z 2 + 2xyz) = 2x(1 − x − 2yz) − (y − z)2 2x(1 + x)(1 − x − 2yz) = − (y − z)2 1+x 2x(1 − x2 − 2yz − 2xyz) = − (y − z)2 1+x 2x(y − z)2 (x − 1)(y − z)2 = − (y − z)2 = ≤ 0. 1+x x+1 This implies that 2 2 −1≥ p −1= y+z 2(1 − x)

r

2 − 1 ≥ 0. 1−x

Combining this with the above inequality, we deduce that !2 √
2 √       r 2− 1−x 1 2 1 1 1 −1 −1 −1 ≥ −1 −1 . = x y z x 1−x x √


2 √ 1−x Furthermore, note that the inequality ≥ 1 is equivalent to x
2 √ 1 − 2 − 2x ≥ 0, which is true. So, from the above inequality, can conclude that     1 1 1 −1 −1 − 1 ≥ 1, x y z 2−

as desired. Third solution: We will now give another proof for the inequality cos2 A cos2 B cos2 C 1 + + ≥ . cos A + 1 cos B + 1 cos C + 1 2 393

Since cos A =

b2 + c2 − a2 , we get 2bc

(b2 + c2 − a2 )2 +1 (b2 + c2 − a2 )2 4b2 c2 = = . cos A + 1 2[bc(b + c)2 − a2 bc] b2 + c2 − a2 +1 2bc cos2 A

Therefore, our inequality is equivalent to (b2 + c2 − a2 )2 (c2 + a2 − b2 )2 (a2 + b2 − c2 )2 + + ≥ 1. 2 2 2 2 bc(b + c) − a bc ca(c + a) − b ca ab(a + b)2 − c2 ab By the Cauchy Schwarz Inequality, we have (b2 + c2 − a2 )2 (c2 + a2 − b2 )2 (a2 + b2 − c2 )2 + + ≥ bc(b + c)2 − a2 bc ca(c + a)2 − b2 ca ab(a + b)2 − c2 ab (a2 + b2 + c2 )2 . ≥ ab(a + b)2 + bc(b + c)2 + ca(c + a)2 − abc(a + b + c) So, it is enough to prove that (a2 + b2 + c2 )2 ≥ ab(a + b)2 + bc(b + c)2 + ca(c + a)2 − abc(a + b + c), which is equivalent to a2 (a − b)(a − c) + b2 (b − c)(b − a) + c2 (c − a)(c − b) ≥ 0, It is the Schur’s Inequality (in the special case fourth degree). ?F? 1 1 1 05.11. Let a, b, c be positive real numbers such that + + = 1. Prove a b c that (a − 1) (b − 1) (c − 1) ≥ 8. (Croatia 2005) Solution: From the given hypothesis, we have ab + bc + ca = abc, and r 1 1 1 1 3 1= + + ≥3 , a b c abc using the AM-GM Inequality. This implies that abc ≥ 27, and so we get (a − 1) (b − 1) (c − 1) = abc − (ab + bc + ca) + a + b + c − 1 √ 3 = abc − abc + a + b + c − 1 ≥ 3 abc − 1 ≥ 8, as desired. It is easy to see that the equality holds if and only if a = b = c = 3. ?F? 05.12. Let a, b, c > 0 such that abc = 1. Prove that a a a 3 + + ≥ . (a + 1) (b + 1) (b + 1) (c + 1) (c + 1) (a + 1) 4 394

(Czech-Slovak 2005) Solution: The desired inequality is equivalent to 4a (c + 1) + 4b (a + 1) + 4c (b + 1) ≥ 3 (a + 1) (b + 1) (c + 1) , or ab + bc + ca + a + b + c ≥ 6, which is true since √ 6 ab + bc + ca + a + b + c ≥ 6 a3 b3 c3 = 6, from the AM-GM Inequality. The equality holds if and only if a = b = c = 1. ?F? 05.13. If a, b, c are three positive real numbers such that ab + bc + ca = 1, prove that r r r 1 3 1 3 1 3 1 + 6b + + 6c + + 6a ≤ . a b c abc (Germany 2005) First solution: According to the AM-GM Inequality, we have 1 = ab + bc + √ 1 3 ca ≥ 3 a2 b2 c2 , from which it follows that abc ≤ √ . On the other hand, for 3 3 any real numbers x, y, z, we have 1 (x + y + z)2 − 3(xy + yz + zx) = [(x − y)2 + (y − z)2 + (z − x)2 ] ≥ 0. 2 Setting x = ab, y = bc, z = ca in this inequality, we get 1 = (ab + bc + ca)2 ≥ 1 3(ab · bc + bc · ca + ca · ab) = 3abc(a + b + c), or a + b + c ≤ . Now, applying 3abc the AM-GM Inequality again, we have   √ √ 1 r r 3 3+3 3+ + 6b 1 √ √ 1 a 3 1 3 1 + 6b = · 3 · 3 · + 6b ≤ · a 3 a 3 3   √ 1 1 = 6 3 + + 6b . 9 a Therefore X

r 3

    √ 1 1 1 1 1 + 6b ≤ 18 3 + + + + 6(a + b + c) a 9 a b c   √ 1 ab + bc + ca = 18 3 + + 6(a + b + c) 9 abc   √ 1 1 = 18 3 + + 6(a + b + c) . 9 abc 395

√ 6 Now, from what we have shown above, it is easy to see that 18 3 ≤ and abc 2 6(a + b + c) ≤ . Thus, from the last inequality, we get abc r r r   6 1 1 1 2 3 1 3 1 3 1 = + 6b + + 6c + + 6a ≤ + + , a b c 9 abc abc abc abc which is just the desired inequality. The equality holds if and only if three 1 numbers a, b, c are equal and they are equal to √ . 3 Second solution: By the Power Mean Inequality r 3 3 3 u+v+w 3 u + v + w ≤ , 3 3 the left hand side of the original inequality does not exceed r r 3 3 1 1 1 3 3 ab + bc + ca √ + 6b + + 6c + + 6a = √ + 6(a + b + c). 3 3 b c abc 3 a 3

(1)

The condition ab + bc + ca = 1 enables us to write a+b=

1 − ab ab − (ab)2 = , c abc

b+c=

bc − (bc)2 , abc

c+a=

ca − (ca)2 . abc

Hence ab + bc + ca 1 + 6(a + b + c) = + 3[(a + b) + (b + c) + (c + a)] abc abc 4 − 3[(ab)2 + (bc)2 + (ca)2 ] = . abc Now, we have 3[(ab)2 + (bc)2 + (ca)2 ] ≥ (ab + bc + ca)2 = 1 by the well-known inequality 3(u2 + v 2 + w2 ) ≥ (u + v + w)2 . Hence an upper bound for the 3 3 1 , which is right hand side of (1) is √ . So it suffices to check √ ≤ 3 3 abc abc abc 1 equivalent to (abc)2 ≤ . This follows from the AM-GM Inequality, in view 27 of ab + bc + ca = 1 again 2

(abc) = (ab)(bc)(ca) ≤



ab + bc + ca 3

3

 3 1 1 = = . 3 27

?F? 05.14. Given positive real numbers x, y, z such that xyz ≥ 1. Prove that x5 − x2 y5 − y2 z5 − z2 + + ≥ 0. x5 + y 2 + z 2 y 5 + z 2 + x2 z 5 + x2 + y 2 (IMO 2005) 396

First solution: Note that x5 − x2 x5 − x2 ≥ x5 + y 2 + z 2 x3 (x2 + y 2 + z 2 ) is equivalent to
2
x3 − 1 x2 y 2 + z 2 ≥ 0, x3 (x2 + y 2 + z 2 ) (x5 + y 2 + z 2 ) which is true for all positive x, y, z. Hence 1 x2 − x5 − x2 x . ≥ 2 x5 + y 2 + z 2 x + y2 + z2 Summing the above inequality with its analogous cyclic inequalities, we see that the desired result follows from x2 + y 2 + z 2 −

1 1 1 − − ≥ 0. x y z

Since xyz ≥ 1, x2 + y 2 + z 2 −

1 1 1 − − ≥ x2 + y 2 + z 2 − yz − zx − xy ≥ 0, x y z

so we are done. It is easy to see that the equality holds if and only if x = y = z = 1. Second solution: Note that x5 − x2 x2 + y 2 + z 2 = 1 − , x5 + y 2 + z 2 x5 + y 2 + z 2 and its cyclic analogous forms. The given inequality is equivalent to x2 + y 2 + z 2 x2 + y 2 + z 2 x2 + y 2 + z 2 + + ≤ 3. x5 + y 2 + z 2 y 5 + z 2 + x2 z 5 + x2 + y 2 In view of the Cauchy Schwarz Inequality and the condition xyz ≥ 1, we have
2 √ (x5 + y 2 + z 2 )(yz + y 2 + z 2 ) ≥ x2 xyz + y 2 + z 2 ≥ (x2 + y 2 + z 2 )2 , or

x2 + y 2 + z 2 yz + y 2 + z 2 ≤ 2 . 5 2 2 x +y +z x + y2 + z2

Taking the cyclic sum of the above inequality and analogous forms gives x2 + y 2 + z 2 x2 + y 2 + z 2 x2 + y 2 + z 2 xy + yz + zx + + ≤2+ 2 . x5 + y 2 + z 2 y 5 + z 2 + x2 z 5 + x2 + y 2 x + y2 + z2 It suffices to show that xy + yz + zx ≤ x2 + y 2 + z 2 , which is well-known. Third solution: We may write our inequality as X

X x5 x2 ≥ . x5 + y 2 + z 2 x5 + y 2 + z 2 397

Notice that for all m, n ≥ 0 the function f (t) = Therefore since x ≥

1 , we have yz

x x = 4 ≥ 5 2 2 x +y +z x · x + y2 + z2 It follows that

t is always increasing. mt + n

1 yz x4 ·

1 + y2 + z2 yz

=

x4

1 . + yz(y 2 + z 2 )

x5 x4 ≥ , x5 + y 2 + z 2 x4 + yz(y 2 + z 2 )

and hence X

X X x4 x5 x4 ≥ = 1. ≥ x5 + y 2 + z 2 x4 + yz(y 2 + z 2 ) x4 + y 4 + z 4

(1)

Also, since 1 ≤ xyz, we have x2 x2 · 1 x2 · xyz x2 yz = ≤ = x5 + y 2 + z 2 (y 2 + z 2 ) · 1 + x5 (y 2 + z 2 ) · xyz + x5 x4 + yz(y 2 + z 2 ) x2 yz x2 yz x2 x2 yz = ≤ = . ≤ 4 x + 2y 2 z 2 x4 + y 2 z 2 + y 2 z 2 2x2 yz + y 2 z 2 2x2 + yz From this, we deduce that X

X x2 x2 ≤ . x5 + y 2 + z 2 2x2 + yz

Combining (1) and (2), we see that it is enough to check that y2 z2 x2 + + ≤ 1, 2x2 + yz 2y 2 + zx 2z 2 + xy or equivalently, yz zx xy + + ≥ 1. 2 2 yz + 2x zx + 2y xy + 2z 2 By the Cauchy Schwarz Inequality, we get X

yz (xy + yz + zx)2 ≥ yz + 2x2 yz(yz + 2x2 ) + zx(zx + 2y 2 ) + xy(xy + 2z 2 ) (xy + yz + zx)2 = = 1, (xy + yz + zx)2

as desired. Fourth solution: We have to prove that x2 + y 2 + z 2 x2 + y 2 + z 2 x2 + y 2 + z 2 + + ≤ 3. x5 + y 2 + z 2 y 5 + z 2 + x2 z 5 + x2 + y 2 398

(2)

Because y 2 + z 2 ≥ 2yz and xyz ≥ 1, we have x5

1 1 1 ≤ 4 ≤ , 4 2 2 +y +z x 2x 2 2 2 2 +y +z +y +z yz y2 + z2

and the cyclic analogous forms. Thus it suffices to show that x2 + y 2 + z 2 x2 + y 2 + z 2 x2 + y 2 + z 2 + + ≤ 3. 2x4 2y 4 2z 4 2 2 2 2 2 2 +y +z +z +x +x +y y2 + z2 z 2 + x2 x2 + y 2 However, since this is a homogeneous inequality, the condition xyz ≥ 1 is not relevant anymore. Furthermore, we may assume that x2 + y 2 + z 2 = 3. Then the inequality reduces to 1 2x4 3 − x2 or

+

+ 3 − x2

1 2y 4 3 − y2

+ 3 − y2

+

1 2z 4 3 − z2

≤ 1,

+ 3 − z2

3 − x2 3 − y2 3 − z2 + + ≤ 1, 3x4 − 6x2 + 9 3y 4 − 6y 2 + 9 3z 4 − 6z 2 + 9

where x, y, z are positive real numbers with x2 + y 2 + z 2 = 3. Because 3x4 − 6x2 + 9 = 3(x2 − 1)2 + 6 ≥ 6 and 3 − x2 = y 2 + z 2 ≥ 0, we obtain 3 − x2 3 − x2 ≤ . 3x4 − 6x2 + 9 6 Adding the above inequality and the cyclic analogous forms gives 3 − x2 3 − y2 3 − z2 9 − (x2 + y 2 + z 2 ) + + ≤ = 1, 3x4 − 6x2 + 9 3y 4 − 6y 2 + 9 3z 4 − 6z 2 + 9 6 as desired. 1 Fifth solution: From the given condition, we have ≤ yz. Using it and the x AM-GM Inequality, we find that 4(x5 − x2 ) 5x5 − 4x2 + y 2 + z 2 3x5 + (2x5 + y 2 + z 2 − 4x2 ) + 1 = = x5 + y 2 + z 2 x5 + y 2 + z 2 x5 + y 2 + z 2 p 3x5 + 4 4 x10 y 2 z 2 − 4x2 3x5 ≥ ≥ x5 + y 2 + z 2 x5 + y 2 + z 2 4 3x 3x4 3x4 = ≥ 4 ≥ . 1 x + yz(y 2 + z 2 ) x4 + y 4 + z 4 4 2 2 x + · (y + z ) x From this, it follows that 3 x4 1 x5 − x2 ≥ · − . 5 2 2 4 4 4 x +y +z 4 x +y +z 4 399

Adding the above inequality and the cyclic analogous forms gives y5 − y2 z5 − z2 3 x4 + y 4 + z 4 3 x5 − x2 + + ≥ − = 0, · x5 + y 2 + z 2 y 5 + z 2 + x2 z 5 + x2 + y 2 4 x4 + y 4 + z 4 4 as desired. ?F? 05.15. Let a1 ≤ a2 ≤ · · · ≤ an be positive real numbers such that a21 + a22 + · · · + a2n = 1, n

a1 + a2 + · · · + an = m, n

where 1 ≥ m > 0. Prove that for all i satisfying ai ≤ m, we have n − i ≥ n(m − ai )2 . (Iran 2005) First solution: By the Cauchy Schwarz Inequality, we have a1 + a2 + · · · + ai + ai+1 + ai+2 + · · · + an n a1 + a2 + · · · + ai ai+1 + ai+2 + · · · + an + = n n q

m=

a1 + a2 + · · · + ai ≤ + q n (n −

≤ ai + = ai +

i)(a2i+1

(n − i)(a2i+1 + a2i+2 + · · · + a2n ) n

+

a2i+2

+ ··· +

a2n )

n q 2 (n − i)(n − a1 − a22 − · · · − a2i ) n

p n(n − i) ≤ ai + , n and therefore

r m − ai ≤

n−i . n

This means that n − i ≥ n(m − ai )2 , which yields our conclusion. Second solution: Let bk = m − ak for all k = 1, 2, . . . , n, we then have that m ≥ b1 ≥ b2 ≥ · · · ≥ bn ,

n X

bk = 0,

k=1

n X b2k = n(1 − m2 ). k=1

Since ai ≤ m, we can note that m ≥ b1 ≥ b2 ≥ · · · ≥ bi ≥ 0, and thus it remains to prove that n−i . b2i ≤ n By the Cauchy Schwarz Inequality, we get 1 b21 + b22 + · · · + b2i ≥ (b1 + b2 + · · · + bi )2 , i 400

and 1 (bi+1 + bi+2 + · · · + bn )2 n−i 1 = (−b1 − b2 − · · · − bi )2 n−i 1 (b1 + b2 + · · · + bi )2 . = n−i

b2i+1 + b2i+2 + · · · + b2n ≥

Summing up these two inequalities, it follows that     1 nib2i 1 1 1 2 2 (b1 + b2 + · · · + bi ) ≥ · i2 b2i = n(1 − m ) ≥ + + , i n−i i n−i n−i and therefore

(n − i)(1 − m2 ) . i On the other hand, we also have b2i ≤ m2 , and so we can write that   (n − i)(1 − m2 ) . b2i ≤ min m2 , i b2i ≤

Note that this proves our inequality, since (it is easy to check that)   2 n−i 2 (n − i)(1 − m ) ≤ . min m , i n ?F? 05.16. If three nonnegative real numbers a, b, c satisfy the condition a2

1 1 1 + 2 + 2 = 2, +1 b +1 c +1

prove that 3 ab + bc + ca ≤ . 2 (Iran 2005) 1 1 ,y = 2 , and a2 + 1 b +1 , then 1 > x, y, z > 0 and x + y + z = 2. We furthermore get

First solution: Set the following substitutions x = z=

1 c2 + 1

1−x y+z−x = ≥ 0, x 2x 1−y z+x−y b2 = = ≥ 0, y 2y 1−z x+y−z c2 = = ≥ 0, z 2z

a2 =

and therefore x, y, z are the sidelengths of a triangle. Next, put m = z+x−y x+y−z ,p = . We then have that m + n + p = 2 and 2 2 r r r m n p a= , b= , c= . n+p p+m m+n 401

y+z−x ,n = 2

Now, by using the AM-GM Inequality, r r n n m n m m · =2 · ≤ + , 2ab = 2 n+p p+m m+p n+p m+p n+p and similarly, we get 2bc ≤

n p + , m+n m+p

2ca ≤

m p + . m+n n+p

Thus 2(ab + bc + ca) ≤

m n n p m p + + + + + = 3. m+p n+p m+n m+p m+n n+p

The equality holds if and only if a = b = c =

√1 . 2

Second solution: From the given hypothesis, we get       1 1 1 a2 b2 c2 1= 1− 2 + 1− 2 + 1− 2 = 2 + 2 + 2 . a +1 b +1 c +1 a +1 b +1 c +1 On the other hand, the Cauchy Schwarz Inequality yields a2 b2 c2 (a + b + c)2 + + ≥ . a2 + 1 b2 + 1 c2 + 1 a2 + b2 + c2 + 3 Thus, we have a2 + b2 + c2 + 3 ≥ (a + b + c)2 , which leads us to 3 ab + bc + ca ≤ . 2 ?F? 05.17. If x, y, z are real numbers satisfying xyz = −1, prove that x4 + y 4 + z 4 + 3(x + y + z) ≥

y 2 + z 2 z 2 + x2 x2 + y 2 + + . x y z (Iran 2005)

Solution: According to the condition from the hypothesis, we have 3(x + y + z) = −3xyz(x + y + z), and thus y 2 + z 2 z 2 + x2 x2 + y 2 + + = −yz(y 2 + z 2 ) − zx(z 2 + x2 ) − xy(x2 + y 2 ) x y z = −x3 (y + z) − y 3 (z + x) − z 3 (x + y). The inequality in question is now equivalent to x4 + y 4 + z 4 − 3xyz(x + y + z) ≥ −x3 (y + z) − y 3 (z + x) − z 3 (x + y), which can be rewritten as [x4 + y 4 + z 4 + x3 (y + z) + y 3 (z + x) + z 3 (x + y)] − 3xyz(x + y + z) ≥ 0, 402

or (x + y + z)(x3 + y 3 + z 3 − 3xyz) ≥ 0, which is obviously true according to (x + y + z)(x3 + y 3 + z 3 − 3xyz) = (x + y + z)2 (x2 + y 2 + z 2 − xy − yz − zx), and to the well-known x2 + y 2 + z 2 ≥ xy + yz + zx. ?F? 05.18. Let a, b, c be positive real numbers such that a + b + c = 1. Prove that √ √ √ 3 3 a 1 + b − c + b 3 1 + c − a + c 1 + a − b ≤ 1. (Japan 2005) Solution: Because 1 + b − c > 0, we can apply the AM-GM Inequality and get   p √ 1+1+1+b−c ab − ac 3 3 a 1 + b − c = a 1 · 1 · (1 + b − c) ≤ a =a+ . 3 3 Adding this to the two analogous inequalities and using the fact that a+b+c = 1, we get the result. It is easy to see that the equality holds if and only if 1 a=b=c= . 3 ?F? 05.19. For any real numbers x1 , x2 , . . . , xn with x21 + x22 + · · · + x2n = 1, prove that r x1 x2 xn n < . + + ··· + 2 1 + x21 1 + x21 + x22 1 + x21 + x22 + · · · + x2n (Korea 2005) Solution: For convenience, let us denote x0 = 0. Then, by applying the Cauchy Schwarz Inequality, we can see that the left hand side of the original inequality does not exceed v v u n u n 2 X u u X xi x2i tn tn ≤ (1 + x20 + · · · + x2i )2 (1 + x20 + · · · + x2i )2 i=1 i=1 v u n u X x2i ≤ tn (1 + x20 + · · · + x2i−1 )(1 + x20 + · · · + x2i ) i=1 v u n   u X 1 1 t = n − 1 + x20 + · · · + x2i−1 1 + x20 + · · · + x2i i=1 s   r 1 1 n = n − = . 2 2 2 2 1 + x0 1 + x0 + · · · + xn 403

It is easy to see that the equality does not occur. So, we must have n X

xi 2 + · · · + x2 < 1 + x 1 i i=1

r

n . 2

The proof is completed. ?F? 05.20. Given positive real numbers a, b, c such that a4 + b4 + c4 = 3. Prove that 1 1 1 + + ≤ 1. 4 − ab 4 − bc 4 − ca (Moldova 2005) Solution: Since a4 + b4 + c4 = 3, it is easy to see that a2 , b2 , c2 < 2. Now, applying the Cauchy Schwarz Inequality and the AM-GM Inequality, we find that 1 4 4 2 1 + ≥ ≥ = . 2 2 2 2 4−a 4−b 4−a +4−b 8 − 2ab 4 − ab Adding this to the two analogous inequalities and dividing each side of the resulting inequality by 2, we get 1 1 1 1 1 1 + + ≤ + + . 2 2 4 − ab 4 − bc 4 − ca 4−a 4−b 4 − c2 Therefore, it suffices to prove that 1 1 1 + + ≤ 1. 2 2 4−a 4−b 4 − c2 To prove it, we observe that for each x2 < 2, 1 x4 + 5 ≤ , 4 − x2 18 since it is equivalent to (x2 − 1)2 (2 − x2 ) ≥ 0, true. So we have 1 1 1 a4 + b4 + c4 + 15 + + ≤ = 1, 4 − a2 4 − b2 4 − c2 18 as desired. The proof is completed. Note that the equality holds if and only if a = b = c = 1. ?F? 05.21. Let a, b, c ∈ [0, 1]. Prove that a b c + + ≤ 2. bc + 1 ca + 1 ab + 1 (Poland 2005) 404

Solution: Since a, b, c ∈ [0, 1], we have (1 − a)(1 − b) + (1 − ab)(1 − c) ≥ 0. It follows that 2 + abc ≥ a + b + c. Using this inequality, we get a b c a b c + + ≤ + + bc + 1 ca + 1 ab + 1 abc + 1 abc + 1 abc + 1 a+b+c 2 + abc 2 + 2abc = ≤ ≤ = 2, 1 + abc 1 + abc 1 + abc as desired. Note that the equality holds if and only if (a, b, c) equals (1, 1, 0), or (1, 0, 1), or (0, 1, 1). ?F? 05.22. Let a, b, c be positive real numbers such that a + b + c = 1. Prove that r p p p ab (1 − c) + bc (1 − a) + ca (1 − b) ≤

2 . 3

(Republic of Srpska 2005) Solution: Applying the Cauchy Schwarz Inequality in combination with the well-known inequality 3(ab + bc + ca) ≤ (a + b + c)2 , we get hX p i2 ab (1 − c) ≤ (ab + bc + ca) (1 − c + 1 − a + 1 − b) 2 (a + b + c)2 = , 3 3   1 1 1 as desired. The equality holds iff (a, b, c) = , , . 3 3 3 ?F? ≤2·

05.23. Let a, b, c be positive real numbers such that abc ≥ 1. Prove that 1 1 1 + + ≤ 1. 1+a+b 1+b+c 1+c+a (Romania 2005) Solution: By expanding, we can rewrite the inequality as X

(1 + b + c) (1 + c + a) ≤ (1 + a + b) (1 + b + c) (1 + c + a) ,

or equivalently, 2 + 2 (a + b + c) ≤ 2abc + a2 (b + c) + b2 (c + a) + c2 (a + b) . Since abc ≥ 1, it suffices to show that a2 (b + c) + b2 (c + a) + c2 (a + b) ≥ 2 (a + b + c) . 405

Now, applying the AM-GM Inequality and the Chebyshev’s Inequality, we have √ √ √ a2 (b + c) + b2 (c + a) + c2 (a + b) ≥ 2a2 bc + 2b2 ca + 2c2 ab  √ √ √  ≥2 a a+b b+c c √ √ √  2 ≥ (a + b + c) a+ b+ c 3 q 3 √ 2 ≥ ·3 abc (a + b + c) ≥ 2(a + b + c). 3 Therefore, the last inequality is obviously true. And so, our proof is completed. Note that the equality holds if and only if a = b = c = 1. ?F? 05.24. Let n is a positive integer. Prove that if x is a positive real numbers, then (2x)n 1 + xn+1 ≥ . (1 + x)n−1 (Russia 2005) Solution: By the AM-GM Inequality, we have √ n+1 1 + xn+1 ≥ 2 xn+1 = 2x 2 ,

√ 1 1 + x ≥ 2 x = 2x 2 .

Therefore  1 n−1
n+1 1 + xn+1 (1 + x)n−1 ≥ 2x 2 · 2x 2 = (2x)n . Dividing both sides of this inequality by (1 + x)n−1 , we get the result. It is easy to see that the equality holds if and only if x = 1. ?F? 05.25. Given positive real numbers x, y, z such that x2 + y 2 + z 2 = 1. Prove the following inequality x3

x y z + 3 + 3 > 3. + yz y + zx z + xy (Russia 2005)

Solution: Setting a =

yz zx xy ,b = ,c = , then we have ab + bc + ca = 1 and x y z x3

x 1 = . + yz a + bc

Therefore, the original inequality is equivalent to 1 1 1 + + > 3. a + bc b + ca c + ab 406

Without loss of generality, we may assume that a = max{a, b, c}. If b + c > 1, then we have a + b ≥ c + b > 1 and a + c ≥ b + c > 1, from which it follows that 1 1 > = 1, a + bc a(b + c) + bc 1 1 > = 1, b + ca b(c + a) + ca 1 1 > = 1. c + ab c(a + b) + ab Therefore, it is clear that 1 1 1 + + > 3. a + bc b + ca c + ab Let us consider now the case b + c ≤ 1. In this case, we apply the Cauchy Schwarz Inequality and obtain 1 1 4 4 4 4 + ≥ = = > . b + ca c + ab b + ca + c + ab a(b + c) + b + c 1 − bc + b + c 1+b+c On the other hand, we have a + bc =

1 − bc 1 bc(b + c − 1) 1 + bc = + ≤ , b+c b+c b+c b+c

1 ≥ b + c. Using this in combination with the above a + bc inequality, it suffices to prove that

which yields that

(b + c) +

4 ≥ 3. b+c+1

However, this inequality is obvious since from the AM-GM Inequality, we get (b + c) +

4 4 = (b + c + 1) + − 1 ≥ 4 − 1 = 3. b+c+1 b+c+1 ?F?

05.26. Let a, b, c be positive real numbers. Prove that r b c a 3 √ +√ +√ ≥ (a + b + c). 2 c+a b+c a+b (Serbia and Montenegro 2005) Solution: Without loss of generality, we may assume that a ≥ b ≥ c. From this assumption, we have √

1 1 1 ≥√ ≥√ . c+a b+c a+b 407

Therefore, applying the Chebyshev’s Inequality and the AM-GM Inequality, we get   a b c 1 1 1 1 √ +√ +√ ≥ (a + b + c) √ +√ +√ 3 c+a c+a b+c a+b b+c a+b a+b+c 3 ≥ · qp 3 3 (a + b) (b + c) (c + a) r a+b+c 3 ≥r = (a + b + c), 2 (a + b) + (b + c) + (c + a) 3 as desired. The equality holds if and only if a = b = c. ?F? 05.27. If a, b, c are positive real numbers such that ab + bc + ca = 1. Prove that r √ 3 1 3 3 + 6 (a + b + c) ≤ . 3 abc abc (Slovenia 2005) Solution: The original inequality can be written as 1 a2 b2 c2 [1 + 6abc (a + b + c)] ≤ . 9

√ 3 Now, using the AM-GM Inequality, we find that 1 = ab + bc + ca ≥ 3 a2 b2 c2 , 1 and hence a2 b2 c2 ≤ . On the other hand, applying the well-known inequality 27 (x + y + z)2 ≥ 3(xy + yz + zx) for the triple (x, y, z) = (ab, bc, ca), we get 1 = (ab + bc + ca)2 ≥ 3abc(a + b + c). From these inequalities, we conclude that a2 b2 c2 [1 + 6abc (a + b + c)] ≤

1 1 (1 + 2) = , 27 9

1 as desired. Note that the equality holds if and only if a = b = c = √ . 3 ?F? 05.28. Let a1 , a2 , . . . , a95 be positive real numbers. Prove that 95 X

ak ≤ 94 +

k=1

95 Y

max {1, ak } .

k=1

(Taiwan 2005) Solution: Let bk = max{ak , 1}, then we have

95 Y k=1

95 95 X X ak ≤ bk . k=1

k=1

408

max {1, ak } =

95 Y

bk , and

k=1

So it suffices to prove that 95 X

bk ≤ 94 +

k=1

95 Y

bk .

k=1

This inequality can be written as (1 − b1 )(1 − b2 ) + (1 − b1 b2 )(1 − b3 ) + · · · + (1 − b1 b2 · · · b94 )(1 − b95 ) ≥ 0, which is obviously true because bk ≥ 1 for all k = 1, 2 . . . , 95. Our proof is completed. ?F? 05.29. Let a, b, c be positive real numbers. Prove that     1 1 1 a b c 2 + + + + . ≥ (a + b + c) b c a a b c (United Kingdom 2005) First solution: Due to the cyclicity, we may assume that c is the smallest numbers among a, b, c. Then, applying the AM-GM Inequality, we have     1 1 1 a+b+c b b (a + b + c) + + = +1+ a b c b a c  2 1 a+b+c b b ≤ + + +1 . 4 b a c Using this inequality, it suffices to prove that   a+b+c b b a b c + + ≥ + + + 1, 2 b c a b a c which is equivalent to a

c − + b b



or

b b − c a

 (a − c)



 +

 2c − 2 ≥ 0, a

b 2 1 + − b ac a

 ≥ 0,

1 b 2 2 + ≥ √ ≥ . It is easy to see that b ac a ac the equality holds if and only if a = b = c. and this is true since a − c ≥ 0 and

a b c Second solution: Denote x = , y = , z = , then we have xyz = 1. Now, b c a with noting that   1 1 1 a b c b c a (a + b + c) + + = + + + + + +3 a b c b c a a b c 1 1 1 = x + y + z + + + + 3, x y z 409

we can rewrite the original inequality in the following form (x + y + z)2 ≥ 3 + x + y + z + Because

1 1 1 + + . x y z

1 1 1 + + = xy + yz + zx, it is equivalent to x y z (x + y + z)2 ≥ 3 + x + y + z + xy + yz + zx,

or x2 + y 2 + z 2 + xy + yz + zx ≥ 3 + x + y + z. By the AM-GM Inequality, we have p xy + yz + zx ≥ 3 3 x2 y 2 z 2 = 3, and x2 + y 2 + z 2 = (x2 + 1) + (y 2 + 1) + (z 2 + 1) − 3 ≥ 2(x + y + z) − 3 √ ≥ x + y + z + 3 3 xyz − 3 ≥ x + y + z. Therefore, the last inequality is obviously true and so, our proof is completed. ?F? 05.30. Let a, b, c be positive real numbers. Prove that  3  3  3 a b c 3 + + ≥ . a+b b+c c+a 8 (Vietnam 2005) First solution: By the Cauchy Schwarz Inequality, we get  hX X i X a2 2 a3 a(a + b) ≥ , (a + b)3 a+b X X X X and since a(a + b) = a2 + ab ≤ 2 a2 , it suffices to prove that X

a2 a+b

2 ≥

3X 2 a , 4

or equivalently, q X X 2a2 ≥ 3 a2 . a+b X 2a2 X a2 + b2 X a2 − b2 X a2 + b2 = + = , the above inequality Since a+b a+b a+b a+b can be rewritten as X a2 + b2 q X ≥ 3 a2 , a+b or equivalently, X  a2 + b2 a + b  q X X − ≥ 3 a2 − a, a+b 2 410

i.e.

X X (a − b)2 (a − b)2 ≥q X X . 2(a + b) 3 a2 + a q X X From the Cauchy Schwarz Inequality, we have 3 a2 ≥ a, hence it is enough to show that X X (a − b)2 (a − b)2 X ≥ , 2(a + b) 2 a

which is obviously valid since it can be written as X

c(a − b)2 ≥ 0. (a + b)(a + b + c)

Note that the equality holds if and only if a = b = c. Second solution: By the Cauchy Schwarz Inequality, we get X  hX 2 i X a3 3/2 3/2 3 3 , ≥ c a c (a + b) (a + b)3 and therefore we are left to prove that 8

X

c3/2 a3/2

2

≥3

X

We continue by setting the substitutions x = case, the inequality to prove becomes

c3 (a + b)3 . √

ab, y =

√

bc, z =

√

ca. In this

8(x3 + y 3 + z 3 )2 ≥ 3(x2 + y 2 )3 + 3(y 2 + z 2 )3 + 3(z 2 + x2 )3 , or equivalently, X

(x6 + y 6 ) − 9

X

x2 y 2 (x2 + y 2 ) + 16

X

x3 y 3 ≥ 0,

which is valid because x6 + y 6 − 9x2 y 2 (x2 + y 2 ) + 16x3 y 3 = (x2 + 4xy + y 2 )(x − y)4 ≥ 0.

Third solution: Using the Power Mean Inequality, we get  X 1/3  X 1/2 1 a3 1 a2 ≥ , 3 (a + b)3 3 (a + b)2 hence we must show that a2 b2 c2 3 + + ≥ . 2 2 2 (a + b) (b + c) (c + a) 4 411

b c a Consider the substitutions x = , y = , z = . Then xyz = 1, and thus the a b c previous inequality becomes 1 1 3 1 + + ≥ . 2 2 2 (1 + x) (1 + y) (1 + z) 4 Since 1 1 1 xy(x − y)2 + (xy − 1)2 + − ≥ 0, = (1 + x)2 (1 + y)2 1 + xy (1 + xy)(1 + x)2 (1 + y)2 we have that

1 1 1 z + ≥ = , (1 + x)2 (1 + y)2 1 + xy z+1

and therefore 1 1 1 z 1 (z − 1)2 3 3 + + ≥ + = + ≥ . 2 2 2 2 2 (1 + x) (1 + y) (1 + z) z + 1 (1 + z) 4(z + 1) 4 4

Fourth solution: Like in the third solution, we reach the point where we must show that b2 c2 3 a2 + + ≥ . 2 2 2 (a + b) (b + c) (c + a) 4 This time we proceed as follows: from the Cauchy Schwarz Inequality, we have X  hX i hX i2 1 hX i2 a2 2 2 2 ≥ a(a + b) (a + b) (a + c) = (a + b) , (a + b)2 4 and therefore, we are left to prove that i2 hX X (a + b)2 ≥ 3 (a + b)2 (a + c)2 , which is immediate, according to the AM-GM Inequality. ?F? 06.1. Let x1 , x2 , . . . , xn be positive real numbers such that x1 +x2 +· · ·+xn = 1. Prove that ! n ! n X X √ 1 n2 √ xi ≤√ . 1 + x n + 1 i i=1 i=1 (China 2006) Firs solution: From the AM-GM Inequality, we get ! n ! √ !2 n n n X X X √ 1 1 n+1 n X√ √ √ √ xi ≤ xi + , 4n 1 + x 1 + x n + 1 i i i=1 i=1 i=1 i=1 and thus it is enough to show that r  n  X n √ 1 n √ xi + √ ≤ 2n . n+1 1 + xi n+1 i=1 412

We shall now make use of the following auxiliary result Lemma. For any positive real number x, the following inequality holds p
n(n + 1) 2 n √ 1 √ x+ √ ≤ n x + 3n + 4 . 2 2(n + 1) n+1 1+x Proof. This rewrites as 2  n √ n 1 √ ≤ x+ √ (n2 x + 3n + 4)2 , 4(n + 1)3 n+1 1+x and since from the Cauchy Schwarz Inequality, we have 2    n n √ 1 1 √ ≤ (nx + 1) x+ √ + n+1 x+1 n+1 1+x (nx + 1)(nx + 2n + 1) = , (n + 1)(x + 1) the problem reduces to show that 4(n + 1)2 (nx + 1)(nx + 2n + 1) ≤ n(x + 1)(n2 x + 3n + 4)2 . On the other hand, the AM-GM Inequality gives us that (n2 x + 3n + 4)2 ≥ 4(n2 x + n + 2)(2n + 2) = 8(n + 1)(n2 x + n + 2), and therefore, we are left to prove that (n + 1)(nx + 1)(nx + 2n + 1) ≤ 2n(x + 1)(n2 x + n + 2), which is valid, since it reduces to (1 − n)(t − 2)2 ≤ 0. Returning to our inequality, we now conclude that  p n n  X
n(n + 1) X 2 1 n √ √ xi + √ ≤ n xi + 3n + 4 2 2(n + 1) 1 + xi n+1 i=1 i=1 r n = 2n . n+1 The equality holds if and only if x1 = x2 = · · · = xn =

1 . n

Second solution: Put yi = xi + 1, for all i = 1, 2, . . . , n. In this case, we have yi ≥ 1 and y1 + y2 + · · · + yn = n + 1, and therefore the inequality in question becomes ! n ! n X 1 X p n2 yi − 1 ≤√ , √ yi n+1 i=1

i=1

or equivalently, n X i=1

n X p yj − 1 j=1

√

≤√

yi 413

n2 . n+1

From the Cauchy Schwarz Inequality, we get n X p yi − 1 i=1

= y1   √ √ √ 1 √ 1 1 n √ · y1 − 1 + y2 − 1 · √ + · · · + yn − 1 · √ n n n = √ y1 s   √ 1 1 1 n + (y2 − 1) + · · · + (yn − 1) (y1 − 1) + + · · · + n n n ≤ √ y1 r 2n + 1 = −ny1 + 2(n + 1) − , ny1 √

and similarly, one can prove that n X p yj − 1 j=1

√

r 2n + 1 ≤ −nyi + 2(n + 1) − nyi

yi

for any i = 1, 2, . . . , n. This yields

n X i=1

n X p yj − 1 j=1

√

yi

≤

n X i=1

r 2n + 1 −nyi + 2(n + 1) − nyi

v u n   u X 2n + 1 t ≤ n −nyi + 2(n + 1) − nyi i=1 v " # u n X u 2n + 1 1 = tn n(n + 1) − n y i=1 i v   u u u   u  2n + 1 n2  u   ≤ un n(n + 1) − · n X  u  n  t yi i=1

s   2n + 1 n2 n2 = n n(n + 1) − · =√ . n n+1 n+1 ?F? 06.2. Let a, b, c be positive real numbers satisfying a + b + c = 1. Prove that √ ab bc ca 2 √ +√ +√ ≤ . 2 ab + bc bc + ca ca + ab 414

(China 2006) First solution: From the Cauchy Schwarz Inequality, we get √ √ X a b X X a 2b ab √ √ √ √ , = ≤ c+ a c+a ab + bc hence it suffices to prove that √ √ √ 2a b 2c a 2b c √ √ +√ √ +√ √ ≤ 1. c+ a a+ b b+ c √ √ √ Furthermore, set a = x, b = y, c = z. We thus have to prove that 2x2 y 2y 2 z 2z 2 x + + ≤ x2 + y 2 + z 2 , z+x x+y y+z or equivalently,  X 2x2 y x +y +z + ≥ 2(xy + yz + zx). 2xy − z+x 2

2

2

In other words, 2

2

2



x + y + z + 2xyz

1 1 1 + + x+y y+z z+x

 ≥ 2(xy + yz + zx).

Now since the Cauchy Schwarz Inequality gives us that   1 1 9xyz 1 + + ≥ , 2xyz x+y y+z z+x x+y+z the problem reduces to proving that x2 + y 2 + z 2 +

9xyz ≥ 2(xy + yz + zx), x+y+z

which is simply a particular case of the Schur’s Inequality written for the third 1 degree. Note that the equality occurs iff a = b = c = . 3 Second solution: By the Cauchy Schwarz Inequality, we have #2 r 2 "X r X ab a(ab + bc + ca) ab √ = · ab + bc + ca c+a ab + bc X  X  ab a(ab + bc + ca) ≤ ab + bc + ca c+a X a(ab + bc + ca) = , c+a and thus, it is enough to prove that X a(ab + bc + ca) c+a 415

1 ≤ (a + b + c)2 , 2

or equivalently, 2ab2 2bc2 2ca2 + + ≤ a2 + b2 + c2 . a+b b+c c+a This can be rewritten into       2bc2 2ca2 2ab2 2 2 2 + 2c − + 2a − ≥ a2 + b2 + c2 , 2b − a+b b+c c+a or in other words,  2

b3 c3 c3 + + a+b b+c c+a



≥ a2 + b2 + c2 .

This last inequality is valid, since from the Cauchy Schwarz Inequality, we get   3 2(a2 + b2 + c2 )2 b c3 c3 ≥ 2 2 + + a+b b+c c+a a + b2 + c2 + ab + bc + ca 2(a2 + b2 + c2 )2 = a2 + b2 + c2 . ≥ 2(a2 + b2 + c2 )

Third solution: By same Cauchy Schwarz Inequality, we have X

ab √ ab + bc

2

#2 a2 b = a+b· (a + b)(a + c)  i X hX a2 b (a + b) ≤ (a + b)(a + c) 2 X a b , =2 (a + b)(a + c) "

X√

s

hence it suffices to prove that 4

X

a2 b ≤ a + b + c, (a + b)(a + c)

which is equivalent to 4a2 b(b + c) + 4b2 c(c + a) + 4c2 a(a + b) ≤ (a + b)(b + c)(c + a)(a + b + c), this last one being true, because it can be written as ab(a − b)2 + bc(b − c)2 + ca(c − a)2 ≥ 0. ?F? 06.3. Suppose that a1 , a2 , . . . , an are real numbers with sum 0. Prove that the following inequality holds n−1

max a2i ≤

1≤i≤n

nX (ai − ai+1 )2 . 3 i=1

416

(China 2006) Solution: It is sufficient to prove that for every 1 ≤ i ≤ n, we have 3 2 a . n i

(a1 − a2 )2 + (a2 − a3 )2 + · · · + (an−1 − an )2 ≥

We will the Cauchy Schwarz Inequality to prove it. Indeed, from the Cauchy Schwarz Inequality, we have that for all m0 , m1 , m2 , . . . , mn−1 , mn ∈ R (m0 = mn = 0), "n−1 # n ! "n−1 # n−1 ! X X X X (ak − ak+1 )2 m2k = (ak − ak+1 )2 m2k k=1

k=0

k=1

≥

"n−1 X

k=1

#2 mk (ak − ak+1 )

k=1 n X = (mk − mk−1 )ak

"

#2 .

k=1

Now, choosing mk = k for k = 0, . . . , i − 1 and mk = k − n for k = i, . . . , n, we have n i−1 n X X X (k − n)2 , k2 + m2k = k=0

k=i

k=0

and n X

(mk − mk−1 )ai =

n i−1 X X (mk − mk−1 )ak (mk − mk−1 )ak + (mi − mi−1 )ai + k=i+1

k=1

k=1

=

i−1 X

ak + (1 − n)ai +

k=1

n X

ak = −nai .

k=i+1

So, from the above inequality, we get "n−1 # " i−1 # n X X X 2 2 2 (ak − ak+1 ) k + (k − n) ≥ n2 a2i . k=1

k=0

k=i

On the other hand, it is clear that i−1 X k=0

n X i(i − 1)(2i − 1) + (n − i)(n − i + 1)(2n − 2i + 1) n3 ≤ . k + (k −n)2 = 6 3 2

k=i

Combining this with the above inequality, we get the result. ?F? 06.4. Let a, b, c be the sidelengths of a triangle. Prove that √ √ √ b+c−a c+a−b a+b−c √ √ +√ √ √ √ +√ √ √ ≤ 3. b+ c− a c+ a− b a+ b− c 417

(IMO Shortlist 2006) First solution: a, b, c are the sidelengths of a triangle, the numbers √ Since √ √ x = a, y = b, z = c are also the sidelengths of a triangle (note that the numbers −x+y +z, x−y +z and x+y −z are positive). In this case, according to the Cauchy Schwarz Inequality, !2 p X −x2 + y 2 + z 2 X −x2 + y 2 + z 2 , ≤3 −x + y + z (−x + y + z)2 and thus, it is sufficient to prove that X −x2 + y 2 + z 2 (−x + y + z)2

≤ 3.

Moving all into one side and distributing the 3 to each of the fractions equally, the last inequality can be rewritten as  X y 2 + z 2 − x2 1− ≥ 0, (y + z − x)2 or equivalently, 2

X

1 (x − y)(x − z) ≥ 0. (y + z − x)2

Without loss of generality, we can assume that x ≥ y ≥ z, then 0 < y +z −x < z + x − y, from which it follows that 1 1 ≥ . 2 (y + z − x) (z + x − y)2 Therefore, with notice that (x − y)(x − z) ≥ 0, we get 1 1 (x − y)(x − z) + (y − z)(y − x) ≥ 2 (y + z − x) (z + x − y)2 1 1 ≥ (x − y)(x − z) + (y − z)(y − x) 2 (z + x − y) (z + x − y)2 (x − y)2 = ≥ 0. (z + x − y)2 On the other hand, it is clear that X

1 (z − x)(z − y) ≥ 0, so (x + y − z)2

1 (x − y)(x − z) ≥ 0, (y + z − x)2

and it completes our proof. Note that the equality holds if and only if a = b = c. Second solution: Since the inequality is symmetric in the three variables, we may assume that a ≥ b ≥ c. We claim that √ a+b−c √ √ √ ≤ 1, a+ b− c 418

and

√ b+c−a c+a−b √ √ ≤ 2. √ √ +√ √ b+ c− a c+ a− b √

It is clear that the denominators are positive. So, the first inequality is equivalent to √ √ √ √ a + b ≥ a + b − c + c. Squaring both sides, we can rewrite it as √ or equivalently,

a+

√ 2 √ √ 2 b ≥ a+b−c+ c , √

ab ≥

p c(a + b − c).

This inequality follows immediately from the inequality√ (a − c)(b − c) ≥√0. √ √ Now, we prove the second inequality. Setting p = a + b and q = a − b, √ we obtain a − b = pq and p ≥ 2 c. It now becomes √ √ c − pq c + pq √ + √ ≤ 2. c−q c+q We now apply the Cauchy Schwarz Inequality to deduce √     √ c − pq c + pq 2 c − pq c + pq 1 1 √ √ + √ ≤ √ +√ +√ c−q c+q c−q c+q c−q c+q
√ 2
√ √ 2 2 4 c − cpq 2 c c − pq 2 c · = = 2 2 c−q c−q (c − q 2 )2

4 c2 − 2cq 2 4 c2 − 2cq 2 + q 4 ≤ ≤ = 4. (c − q 2 )2 (c − q 2 )2 ?F? 06.5. Determine the least real number M such that the inequality ab(a2 − b2 ) + bc(b2 − c2 ) + ca(c2 − a2 ) ≤ M (a2 + b2 + c2 )2 holds for all real numbers a, b, c. (IMO 2006) Solution: The given inequality can be rewritten as |(a − b)(b − c)(c − a)(a + b + c)| ≤ M (a2 + b2 + c2 )2 . √ 3 3 9 2 Now, let a = 1 − √ , b = 1, c = 1 + √ , we get M ≥ . We will show that 32 2√ 2 9 2 the constant M = works. Indeed, put x = a − b, y = b − c, z = c − a and 32 s = a + b + c, then x + y + z = 0, and it becomes |xyzs| ≤

M 2 (x + y 2 + z 2 + s2 )2 . 9 419

Since x + y + z = 0, there exist two numbers, supposed x and y, having the same sign. Now, denote x + y = 2m, then z = −2m. Applying the AM-GM Inequality with noting that xy ≥ 0, we have  |xyzs| = xy|zs| ≤

x+y 2

2

|zs| = 2m3 s .

On the other hand, from the Cauchy Schwarz Inequality and the AM-GM Inequality, we deduce that 2 (x + y)2 2 2 (x + y + z + s ) ≥ +z +s = (2m2 + 2m2 + 2m2 + s2 )2 2 2  √ √ 4 ≥ 4 8m6 s2 = 32 2 m3 s . 2

2

2

2 2



Combining these two inequalities, we get |xyzs| ≤

1 M 2 √ (x2 + y 2 + z 2 + s2 )2 = (x + y 2 + z 2 + s2 )2 . 9 16 2

We have equality if and only if x = y and 2m2 = s2 , i.e. when 2b = a + c and (c − a)2 = 18b2 . There are many triples (a, b, c) satisfying this system of 3 3 equations (for example, we can take a = 1 − √ , b = 1, c = 1 + √ as above). 2 2 √ 9 2 From now, we conclude that the minimum value of M is . 32 ?F? 06.6. Let x1 , x2 , x3 , y1 , y2 , y3 , z1 , z2 , z3 be positive real numbers. Find the maximum value of real number A if M = (x31 + x32 + x33 + 1)(y13 + y23 + y33 + 1)(z13 + z23 + z33 + 1), and N = A(x1 + y1 + z1 )(x2 + y2 + z2 )(x3 + y3 + z3 ), then M ≥ N always holds, and find the condition that the equality holds. (Japan 2006) 1 Solution: Let x1 = x2 = x3 = y1 = y2 = y3 = z1 = z2 = z3 = √ , we get 3 6 3 3 A ≤ . From this, if we can show that the constant A = works, then we can 4 4 also conclude that it is value we need to find. However, this constant indeed works. It suffices to prove (x31 + x32 + x33 + 1)(y13 + y23 + y33 + 1)(z13 + z23 + z33 + 1) 3 ≥ . (x1 + y1 + z1 )(x2 + y2 + z2 )(x3 + y3 + z3 ) 4 Put a =

x1 + x2 + x3 y1 + y2 + y3 z 1 + z 2 + z3 ,b = and c = . Using the 3 3 3 420

Holder’s Inequality, we have (x31 + x32 + x33 + 1)(y13 + y23 + y33 + 1)(z13 + z23 + z33 + 1) ≥ ≥ (3a3 + 1)(3b3 + 1)(3c3 + 1)     1 1 1 1 1 1 3 3 3 + 3b + + + 3c = 3a + + 2 2 2 2 2 2 !3 r r r 1 1 1 1 1 1 3 3 3 3a3 · · + · 3b3 · + · · 3c3 ≥ 2 2 2 2 2 2 3 = (a + b + c)3 . 4 On the other hand, from the AM-GM Inequality, we get (x1 + y1 + z1 )(x2 + y2 + z2 )(x3 + y3 + z3 ) ≤ [(x1 + y1 + z1 ) + (x2 + y2 + z2 ) + (x3 + y3 + z3 )]3 27 [(x1 + x2 + x3 ) + (y1 + y2 + y3 ) + (z1 + z2 + z3 )]3 = 27 = (a + b + c)3 .

≤

Combining these two inequalities, we deduce that (x31 + x32 + x33 + 1)(y13 + y23 + y33 + 1)(z13 + z23 + z33 + 1) 3 ≥ , (x1 + y1 + z1 )(x2 + y2 + z2 )(x3 + y3 + z3 ) 4 3 as desired. And so, we conclude that the maximum value of A is . 4 ?F? 06.7. Let a, b, c, d be real numbers with sum 0. Prove the inequality (ab + ac + ad + bc + bd + cd)2 + 12 ≥ 6(abc + abd + acd + bcd). (Kazakhstan 2006) First solution: Replacing d = −a − b − c, our inequality becomes (a2 + b2 + c2 + ab + bc + ca)2 + 12 + 6(a + b + c)(ab + bc + ca) ≥ 6abc, or equivalently, (a2 + b2 + c2 + ab + bc + ca)2 + 12 + 6(a + b)(b + c)(c + a) ≥ 0. This rewrites as 2 1 (a + b)2 + (b + c)2 + (c + a)2 + 12 + 6(a + b)(b + c)(c + a) ≥ 0. 4 Setting x =

b+c c+a a+b y= ,z = , then the previous inequality goes into 2 2 2 (x2 + y 2 + z 2 )2 + 48 + 24xyz ≥ 0. 421

Now, by the AM-GM Inequality, we have (x2 + y 2 + z 2 )2 ≥ 9 |xyz|4/3 . and because 24xyz ≥ −24 |xyz|, it suffices to prove that 9t4 + 48 − 24t3 ≥ 0, where t = |xyz|1/3 . This is true, since 9t4 + 48 − 24t3 = 3(3t2 + 4t + 4)(t − 2)2 ≥ 0.

Secdon solution: From the Rolle’s theorem, we know that there exist real numbers x, y, z such that  3   x + y + z = (a + b + c + d) = 0 4 ab + ac + ad + bc + bd + cd = 2(xy + yz + zx)   abc + abd + acd + bcd = 4xyz. Therefore, the inequality in question is equivalent to (xy + yz + zx)2 + 3 ≥ 6xyz. Without loss of generality, assume now that z = min {x, y, z}; then since x + y + z = 0, we have z = −x − y ≤ 0, and so our inequality becomes (x2 + xy + y 2 )2 + 3 + 6xy(x + y) ≥ 0. Let us now put s = x + y ≥ 0, p = xy. In this case, the previous inequality rewrites as (s2 − p)2 + 3 + 6sp ≥ 0, or f (p) = p2 + 2(3s − s2 )p + s4 + 3 ≥ 0. We have now that ∆f = (3s − s2 )2 − s4 − 3 = −3(2s + 1)(s − 1)2 ≤ 0, and we thus conclude that f (p) ≥ 0. ?F? 06.8. Let a, b, c be sides of the triangle. Prove that   c  a  2 b − 1 + b2 − 1 + c2 − 1 ≥ 0. a c a b (Moldova 2006) First solution: The original inequality is equivalent to a2 b b2 c c2 a + + ≥ a2 + b2 + c2 . c a b 422

By the Cauchy Schwarz Inequality, we have a2 b b2 c c2 a a4 b2 b4 c2 c4 a2 (a2 b + b2 c + c2 a)2 + + = 2 + 2 + 2 ≥ . c a b a bc b ca c ab abc(a + b + c) Therefore, it suffices to prove that (a2 b + b2 c + c2 a)2 ≥ abc(a + b + c)(a2 + b2 + c2 ). Due to the cyclicity, we can suppose without loss of generality that b is the number between a and c, hence (a − b)(b − c) ≥ 0. Now, applying the AM-GM Inequality, we obtain 4abc(a + b + c)(a2 + b2 + c2 ) ≤ [ac(a + b + c) + b(a2 + b2 + c2 )]2 . On the other hand, 2(a2 b + b2 c + c2 a) − ac(a + b + c) − b(a2 + b2 + c2 ) = (a − b)(b − c)(a + b − c), which is obviously nonnegative. Therefore the inequality (a2 b + b2 c + c2 a)2 ≥ abc(a + b + c)(a2 + b2 + c2 ) is valid and it completes our proof. The equality holds if and only if a = b = c.

Second solution: Using the substitutions a =

1 1 1 ,b = and c = , the x y z

inequality becomes 1 x2



   z 1 x 1 y −1 + 2 −1 + 2 − 1 ≥ 0, y y z z x

or yz 2 (z − y) + zx2 (x − z) + xy 2 (y − x) ≥ 0. Without loss of generality, we can assume that a = min{a, b, c}, and hence x = max{a, b, c}. Denoting the left hand side of the last inequality as E(x, y, z), we will show that E(x, y, z) ≥ E(y, y, z) ≥ 0. We have E(x, y, z) − E(y, y, z) = z(x3 − y 3 ) − z 2 (x2 − y 2 ) + y 3 (x − y) − y 2 (x2 − y 2 ) = (x − y)(x − z)(xz + yz − y 2 ). Since (x − y)(x − z) ≥ 0 and xz + yz − y 2 ≥ y(2z − y) =

2b − c (b − a) + (a + b − c) = > 0, 2 b c b2 c

it follows that E(x, y, z) − E(y, y, z) ≥ 0. On the other hand, we have E(y, y, z) = yz(y − z)2 ≥ 0. 423

So our statement is proved. Third solution: Writing the inequality as E(a, b, c) ≥ 0, where E(a, b, c) = a3 b2 + b3 c2 + c3 a2 − abc(a2 + b2 + c2 ). X X X X Since 2E(a, b, c) = a3 (b−c)2 − a2 (b3 −c3 ) and a2 (b3 −c3 ) = a2 (b− c)3 , we have directly 2E(a, b, c) =

X

a2 (b − c)2 (a − b + c) ≥ 0. ?F?

06.9. Let a, b, c be positive real numbers with ab + bc + ca = abc. Prove that b4 + c4 c4 + a4 a4 + b4 + + ≥ 1. ab(a3 + b3 ) bc(b3 + c3 ) ca(c3 + a3 ) (Poland 2006) Solution: We first notice that the constraint can be written as 1 1 1 + + = 1. a b c It is now enough to establish the auxiliary inequality   x4 + y 4 1 1 1 ≥ + , xy(x3 + y 3 ) 2 x y or

2 x4 + y 4 ≥ x3 + y 3 (x + y) , where x, y > 0. However, we obtain


2 x4 + y 4 − x3 + y 3 (x + y) = x4 + y 4 − x3 y − xy 3 = x3 − y 3 (x − y) ≥ 0.   x4 + y 4 1 1 1 Therefore, we have established the inequality ≥ + . xy(x3 + y 3 ) 2 x y And using it, we can get the result. Note that the equality holds if and only if a = b = c = 3. ?F? 06.10. Find the maximum value of (x3 + 1)(y 3 + 1), for all real numbers x, y, satisfying the condition that x + y = 1. (Romania 2006) Solution: Put xy = t, as x + y = 1 we get (x3 + 1)(y 3 + 1) = t3 − 3t + 2.   1 x+y 2 From x + y = 1, we obtain t = xy ≤ = . It is easy to prove that 2 4 424

1 , with equality if and only if t = −1. We infer that 4 3 3 (x + 1)(y + 1) ≤ 4 for x, y ∈ R with x + y = 1 and (ψ 3 + 1)(−1/ψ 3 + 1) = 4, where ψ is one of the roots of z 2 − z − 1 = 0. ?F?

t3 − 3t + 2 ≤ 4 for t ≤

06.11. Let a, b, c be three positive real numbers with sum 3. Prove that 1 1 1 + + ≥ a2 + b2 + c2 . a2 b2 c2 (Romania 2006) First solution: From the AM-GM Inequality, we get 1 1 1 1 1 1 3 + 2+ 2 ≥ + + = , 2 a b c ab bc ca abc hence it suffices to prove that 3 ≥ abc(a2 + b2 + c2 ). Again, the AM-GM Inequality gives us that (ab + bc + ca)2 ≥ 3abc(a + b + c) = 9abc, and therefore, we are left to show that (a2 + b2 + c2 )(ab + bc + ca)2 ≤ 27, which is obviously valid, since a2 + b2 + c2 + 2(ab + bc + ca) (a + b + c )(ab + bc + ca) ≤ 3 6 (a + b + c) = = 27. 27 2

2

2

2



3

Note that the equality holds iff a = b = c = 1. Second solution: Write the inequality in the form  X 1 2 − a + 4a − 4 ≥ 0, a2 which is equivalent to X (a − 1)2 (1 + 2a − a2 ) a2

≥ 0.

Without loss of generality, we may assume that a ≥ b ≥ c. We have to consider two cases √ √ Case 1. a ≤ 1 + 2. Since c ≤ b ≤ a ≤ 1 + 2, we have 1 + 2a − a2 ≥ 0, 1 + 2b − b2 ≥ 0 and 1 + 2c − c2 ≥ 0. Therefore, the inequality is obviously true. 425

Case 2. a > 1 +

√

2. Since b + c = 3 − a < 2 − bc ≤

√ 2 2 < , we have 3

1 (b + c)2 < , 4 9

and hence 1 1 1 1 2 1 + 2+ 2 > 2+ 2 ≥ > 18 > (a + b + c)2 > a2 + b2 + c2 . 2 a b c b c bc

Remark: Actually, we can show more: For any two positive integers n, p satisfying, n ≥ 4 and p ≥ 4, the proposition P(n, p) is false: n n X X 1 ≥ xpi xpi i=1

for xi ∈ R,

xi > 0,

i=1

i = 1, . . . , n ,

n X

xi = n.

i=1

This variation was considered as a problem in the Romanian IMO Team Selection Tests from 2007. We continue with its solution Solution: Notice first that it is enough to find a set of values xi for n = 4 such that n n X X 1 E= − xpi < 0, xpi i=1

i=1

as then for any n > 4 we can extend this set of values by taking the extra n − 4 ones to be equal to 1. Now that we have reduced it to the case n = 4, it makes sense to look for ”simple” cases: • some xi very small - it yields E > 0, no good; • all xi equal - it yields common value 1, for which E = 0, no good; • let’s then try taking the smallest three xi equal to some value 0 < x < 1, the fourth one, denoted by y, will be 1 < y < 4, y = 4 − 3x. Then,   p  3 1 1 x p p 2p p E = p + p − 3x − y = p 3 + − 3x − (xy) x y x y 1 1 = p [3 − (xy)p ] + p [1 − 3(xy)p ]. x y It seems natural now to look for the maximum possible value for xy, it is not 1 4 1 difficult to see that xy = · (3x) · (4 − 3x) ≤ · (2)2 = (by the AM-GM 3 3 3 2 and y = 2. Then, as Inequality), with equality for 3x = 4 − 3x i.e. x = 3  p  4 4 4 4 256 ≥ = > 1 and p ≥ 4, we have > 3, hence E < 0 for the set 3 3 3 81 2 2 2 of values ( , , , 2 and 1 repeated n − 4 times). 3 3 3 You might now wonder what can we say about the propositions P(4, 3) and P(3, 4). As a matter of fact, they are true. However, we will omit their proof here. 426

?F?  06.12. Consider real numbers a, b, c contained in the interval that 2≤

 1 , 1 . Prove 2

a+b b+c c+a + + ≤ 3. 1+c 1+a 1+b (Romania 2006)

Solution: We begin by proving the left hand side of the inequality. Since 1 a, b ≥ , we have a + b ≥ 1, and thus 2 a+b a+b ≥ . 1+c a+b+c By adding the other two similar relations to the inequality from above, we obtain 2=

(a + b) + (b + c) + (c + a) a+b b+c c+a ≤ + + . a+b+c 1+c 1+a 1+b

For the second inequality, note that the considered expression can be written as  X a c + . 1+c 1+a As a, c ≤ 1, we have a a ≤ , 1+c a+c

and

c c ≤ , 1+a c+a

and so

a c a c + ≤ + = 1. 1+c 1+a a+c c+a The other two cyclic relations occur (again) similarly. Summing up the all three, we get the desired result. ?F?

06.13. Let a, b be positive real numbers. Determine the largest constant M such that for all k ∈ [0, π] , we have 1 1 M + ≥ . ka + b kb + a a+b (Thailand 2006) 4 4 . We claim that is our π+1 π+1 answer. Indeed, from the Cauchy Schwarz Inequality, we have

Solution: Let a = b, k = π, we get M ≤

1 1 4 4 1 4 1 + ≥ = · ≥ · . ka + b kb + a ka + b + kb + a k+1 a+b π+1 a+b 4 . And since the equality π+1 4 can occur, we conclude that the maximum value of M is . π+1

This proves that the inequality holds for M =

427

?F? 06.14. If x, y, z are positive numbers satisfying the condition xy +yz +zx = 1, show that √
2 √ √ √ 27 (x + y)(y + z)(z + x) ≥ x + y + y + z + z + x ≥ 6 3. 4 (Turkey 2006) Solution: By the Cauchy Schwarz Inequality,
2 √ √ √ x + y + y + z + z + x ≤ 3 (x + y + y + z + z + x) = 6(x + y + z), and hence, in order to prove the left inequality, it suffices to show that 8 (x + y)(y + z)(z + x) ≥ (x + y + z). 9 For this, we proceed as follows (x + y)(y + z)(z + x) = = (x + y + z)(xy + yz + zx) − xyz 1 ≥ (x + y + z)(xy + yz + zx) − (x + y + z)(xy + yz + zx) 9 8 8 = (x + y + z)(xy + yz + zx) = (x + y + z). 9 9 For the right hand side of the inequality, we can make use of the AM-GM Inequality in combination with the Minkowsky’s Inequality
2 √ √ √ x+y+ y+z+ z+x ≥ hp i p p ≥3 (x + y)(x + z) + (y + z)(y + x) + (z + x)(z + y)  p p p x2 + 1 + y 2 + 1 + z 2 + 1 =3 p ≥ 3 (x + y + z)2 + (1 + 1 + 1)2 p √ ≥ 3 3(xy + yz + zx) + 9 = 6 3. ?F? 06.15. Let a, b, c be real numbers. Prove that the following inequality holds Xp (a2 − ab + b2 )(b2 − bc + c2 ) ≥ a2 + b2 + c2 . (VMEO 2006) Solution: By the Cauchy Schwarz Inequality, we have    a 2 3a2  c 2 3c2 2 2 2 2 (a − ab + b )(b − bc + c ) = b − + b− + 2 4 2 4   2 a  c  3ac ≥ b− b− + . 2 2 4 428

It follows that    3ac p a c (a2 − ab + b2 )(b2 − bc + c2 ) ≥ b − b− + 2 2 4  a  c  3ac ≥ b− b− + 2 2 4 1 = b2 + (2ca − ab − bc). 2 Therefore  X Xp 1 2 2 2 2 2 b + (2ca − ab − bc) (a − ab + b )(b − bc + c ) ≥ 2 = a2 + b2 + c2 . Our proof is completed. It is easy to see that the equality holds if and only if (a, b, c) equals (t, t, t), or (t, 0, 0), or (0, t, 0), or (0, 0, t), where t is an arbitrary real numbers. ?F? √ √ √ 07.1. Let x, y, z be positive real numbers such that x + y + z = 1. Prove that the following inequality holds x2 + yz y 2 + zx z 2 + xy p +p +p ≥ 1. 2x2 (y + z) 2y 2 (z + x) 2z 2 (x + y) (APMO 2007) Solution: According to the Cauchy Schwarz Inequality and the well-known inequality (a + b + c)2 ≥ 3(ab + bc + ca), we have X

(xy + yz + zx)2 yz 1 X y2z2 1
p √ √ =√ ≥√ · √ √ y+z y+z+ z+x+ x+y 2xyz 2 xyz 2x2 (y + z) 1 (xy + yz + zx)2 1 3xyz(x + y + z) √ √ ≥ √ · ≥ √ · 2 3 xyz x + y + z 2 3 xyz x + y + z √
1 1p 1 √ √ 3(x + y + z) ≥ x+ y+ z = , = 2 2 2

and X

x2 1 X x 1 (x + y + z)2 p √ √ =√ ≥√ · √ √ y+z 2 2 x y+z+y z+x+z x+y 2x2 (y + z) 1 (x + y + z)2 ≥√ ·p 2 (x + y + z) [x(y + z) + y(z + x) + z(x + y)] s s 3 1 (x + y + z) 1 (x + y + z) · 3(xy + yz + zx) = ≥ 2 xy + yz + zx 2 xy + yz + zx √
1 1p 1 √ √ = 3(x + y + z) ≥ x+ y+ z = . 2 2 2

Adding up these two inequalities, we get the desired result. It is easy to see 1 that the equality holds if and only if x = y = z = . 9 429

?F? 07.2. If a, b, c ∈ R such that abc = 1, then     1 1 1 b+c c+a a+b 1 1 1 2 2 2 ≥ 6+2 . + + a +b +c + 2 + 2 + 2 +2 a + b + c + + + a b c a b c a b c (Brazil 2007) Solution: We have 2

2

2



a +b +c +2

1 1 1 + + a b c



= a2 + b2 + c2 + 2(ab + bc + ca) = (a + b + c)2 ,

1 1 1 + + + 2 (a + b + c) = a2 b2 + b2 c2 + c2 a2 + 2abc(a + b + c) a2 b2 c2 = (ab + bc + ca)2 , and  6+2

b+c c+a a+b + + a b c



2(a + b + c)(ab + bc + ca) abc = 2(a + b + c)(ab + bc + ca). =

Therefore, our inequality is equivalent to (a + b + c)2 + (ab + bc + ca)2 ≥ 2(a + b + c)(ab + bc + ca), which is obviously true by the AM-GM Inequality. ?F? 07.3. Given an integer n ≥ 2, find the largest constant C(n) for which the inequality n X X
√ xi ≥ C(n) 2xi xj + xi xj i=1

1≤j 0, pentru a ar˘ata c˘a A A ≥ B, este suficient s˘ a ar˘ at˘am c˘a B ≥ 1. In cadrul metodei compar˘ arii, inegalitatea de demonstrat trebuie rearanjat˘a ˆıntr-o form˘ a convenabil˘ a, evident˘a sau demonstrabil˘a printr-o metod˘a cunoscut˘ a, utilizˆ and operat¸ii de reducere, combinare, factorizare, spargere ¸si completare cu p˘ atrate, care necesit˘a abilitate, experient¸˘a ¸si bun˘a observat¸ie din partea rezolvitorului. Ex. 1 Pentru numerele reale x, y, z care satisfac xy + yz + zx = −10, ar˘atat¸i c˘ a: x2 + 5y 2 + 8z 2 ≥ 40. 489

Proof. Deoarece x2 + 5y 2 + 8z 2 − 40 = x2 + 5y 2 + 8z 2 + 4(xy + yz + zx) = (x + 2y + 2z)2 + (y − 2z)2 ≥ 0, rezult˘ a x2 + 5y 2 + 8z 2 ≥ 40. Egalitatea are loc atunci cˆand xy + yz + zx = −10, x + 2y + 2z = 0 ¸si y − 2z = 0, deci cˆand x = 4, y = −2, z = −1 sau x = −4, y = 2, z = 1. Ex. 2 Fie a, b, c ∈ R+ . Demonstrat¸i c˘a pentru orice numere reale x, y, z, avem: s   abc a+b b+c c+a 2 2 2 xy + yz + zx . x +y +z ≥ 2 (a + b)(b + c)(c + a) c a b [Observat¸ie] Deoarece pentru a = b = c se obt¸ine cunoscuta inegalitate x2 + y 2 + z 2 ≥ xy + yz + zx, echivalent˘a cu (x − y)2 + (y − z)2 + (z − x)2 ≥ 0, putem apela la o metod˘a similar˘a pentru a demonstra inegalitatea cerut˘ a. Proof. Scriem inegalitatea sub forma "

s # b a 2 ab 2 x + y −2 xy b+c c+a (b + c)(c + a)

s # c b 2 bc 2 + y + z −2 yz c+a a+b (c + a)(a + b)   r a 2 c ca 2 + z + x −2 zx ≥ 0, a+b b+c (b + c)(a + b) "

sau " X

ab p

x

y

a(b + c)

#2

−p b(c + a)

≥ 0,

P unde este simbolul ˆınsum˘arii ciclice. Ultima inegalitate este evident adev˘ arat˘ a. Egalitatea are loc atunci cˆand x y z p =p =p . a(b + c) b(c + a) c(a + b)

Ex.3 Fie a, b, c ∈ R+ . Ar˘ atat¸i c˘a: a2a b2b c2c ≥ ab+c bc+a ca+b . 490

Proof. Imp˘ art¸ind ambii membri prin ab+c bc+a ca+b , inegalitatea poate fi scris˘ a astfel  a a−b  b b−c  c c−a ≥ 1. b c a Luˆ and pe rˆ and a ≥ b ¸si a ≤ b, constat˘am c˘a  a a−b b

≥ 1.

Din aceast˘ a inegalitate ¸si din celelelte dou˘a similare rezult˘a imediat inegalitatea cerut˘ a. Egalitatea are loc pentru a = b = c. [Not˘ a] Putem scrie inegalitatea de mai sus sub forma a3a b3b c3c ≥ aa+b+c ba+b+c ca+b+c , sau aa bb cc ≥ (abc)

a+b+c 3

.

In general, dac˘ a xi ∈ R+ , i = 1, 2, · · · , n, avem xx1 1 · xx2 2 · · · · · xxnn ≥ (x1 x2 · · · xn )

x1 +x2 +···+xn n

.

Demonstrat¸ia este similar˘a, scriind inegalitatea sub forma Y  xi xi −xj i 0, deoarece a + b + c ≥ b + c − 2a ¸si (b − c)2 ≥ 0, este suficient s˘a demonstr˘am c˘a 2[(a − b)2 + (c − a)2 ] ≥ (b + c − 2a)2 . Intr-adev˘ ar, cu notat¸iile x = b − a ¸si y = c − a, avem 2[(a − b)2 + (c − a)2 ] − (b + c − 2a)2 = 2(x2 + y 2 ) − (x + y)2 = (x − y)2 = (b − c)2 ≥ 0. Egalitatea are loc atunci cˆand a = b = c, precum ¸si atunci cˆand unul dintre numerele a, b, c este 0, iar ceilelalte sunt egale. Ex. 6 Fie a, b, c numere reale nenegative astfel ˆıncˆat a2 + b2 + c2 = 3. Ar˘atat¸i c˘ a ab2 + bc2 + ca2 ≤ abc + 2. Proof. F˘ ar˘ a a pierde din generalitate, presupunem c˘a b este cuprins ˆıntre a ¸si c, adic˘ a (b − a)(b − c) ≤ 0. Avem 2 − ab2 − bc2 − ca2 + abc = 2 − ab2 − b(3 − a2 − b2 ) − ca2 + abc = b3 − 3b + 2 − a(b2 − ab + ca − bc) = (b − 1)2 (b + 2) − a(b − a)(b − c) ≥ 0. Egalitatea are loc atunci cˆand a = b = c = 1, precum ¸si atunci cˆand √ (a, b, c) este o permutatre ciclic˘a a tripletului (0, 1, 2) . Ex. 7 Dac˘ a a, b, c sunt numere reale pozitive, atunci a b c a2 + b2 + c2 + + −2≥ . b c a ab + bc + ca Proof. Deoarece inegalitatea este ciclic˘a, f˘ar˘a a pierde din generalitate presupunem c˘ a c = min{a, b, c}. Avem     a b c a b b c b (a − b)2 (a − c)(b − c) + + −3 = + −2 + + − −1 = + b c a b a c a a ab ac ¸si a2 + b2 + c2 (a − b)2 + (a − c)(b − c) −1= . ab + bc + ca ab + bc + ca Rezult˘ a

a b c a2 + b2 + c2 + + −2− b c a ab + bc + ca 492

 =

1 1 − ab ab + bc + ca =



2

(a − b) +



1 1 − ac ab + bc + ca

 (a − c)(b − c)

c2 (a + b)(a − b)2 + b2 (a + c)(a − c)(b − c) ≥ 0. abc(ab + bc + ca)

Egalitatea are loc atunci cˆand a = b = c. Ex. 8 Fie a, b, c ∈ R+ astfel ˆıncˆat a2 + b2 + c2 = 1. G˘asit¸i minimul pentru S=

1 1 2(a3 + b3 + c3 ) 1 + + − . a2 b2 c2 abc

[Observat¸ie] Nu ezitat¸i s˘a ghicit¸i r˘aspunsul corect al unei probleme de extremum, deoarece acest lucru v˘a poate ajuta la rezolvarea problemei! Operat¸ia de ”ghicire” a valorii extreme este strˆans legat˘a de valorile variabilelor pentru care inegalitatea devine egalitate, de care trebuie s˘a ¸tinem seama ¸si ˆın etapa de demonstrare a inegalit˘a¸tii. Proof. Pentru a = b = c = c˘ a S ≥ 3. Intr-adev˘ ar,

√1 , 3

rezult˘a S = 3. Vom ˆıncerca s˘a ar˘at˘am

1 1 1 2(a3 + b3 + c3 ) + + − 3 − a2 b2 c2 abc  2  2 2 2 2 2 2 2 2 2 a +b +c a +b +c a +b +c a b2 c2 = + + −3−2 + + a2 b2 c2 bc ca ab       2   1 1 1 1 1 a b2 c2 1 2 2 2 + +b + +c + −2 + + =a b2 c2 a2 c2 a2 b2 bc ca ab       1 2 1 2 1 2 2 1 2 1 2 1 =a − +b − +c − ≥ 0. b c c a a b S−3=

Prin urmare, S are valoarea minim˘a 3, care este atins˘a atunci cˆand a = b = c = √13 . Ex. 9 Dac˘ a x ∈ R, y ≥ 0 ¸si y(y + 1) ≤ (x + 1)2 , atunci y(y − 1) ≤ x2 . Proof. In cazul 0 ≤ y ≤ 1, avem evident y(y − 1) ≤ 0 ≤ x2 . In cazul y > 1, scriem ipoteza y(y + 1) ≤ (x + 1)2 sub forma r 1 1 y ≤ (x + 1)2 + − , 4 2 iar inegalitatea cerut˘ a y(y − 1) ≤ 0 ≤ x2 sub forma echivalent˘a r 1 1 y ≤ x2 + + . 4 2 493

Prin urmare, este suficient s˘a ar˘at˘am c˘a r r 1 1 1 1 2 (x + 1) + − ≤ x2 + + , 4 2 4 2 adic˘ a

r (x +

1)2

1 + ≤ 4

r x2 +

1 + 1. 4

Prin ridicare la p˘ atrat, obt¸inem inegalitatea r 1 2x ≤ 2 x2 + , 4 care este ˆın mod clar adev˘arat˘a. Cu aceasta, demonstrat¸ia este ˆıncheiat˘a. Avem egalitate pentru x = 0 ¸si y ∈ {0, 1}. Ex. 10 Fie a, b, c numere reale pozitive. Ar˘atat¸i c˘a a b c (a − c)2 + + ≥3+ . b c a ab + bc + ca Proof. Prin dezvoltare ¸si reducere, inegalitatea devine astfel b2 +

ab2 a2 c bc2 + + ≥ 2ab + 2bc. c b a

Scriem aceast˘ a inegalitate sub forma echivalent˘a  2   2  bc ab a2 c 2 + ab − 2bc + + + b − 3ab ≥ 0. a c b Din inegalitatea mediilor a dou˘a ¸si respectiv trei numere , avem r bc2 bc2 + ab ≥ 2 · ab = 2bc a a ¸si r 2 a2 c ab2 a2 c 3 ab 2 + +b ≥3 · · b2 = 3ab, c b c b iar prin ˆınsumarea acestor dou˘a inegalit˘a¸ti obt¸inem inegalitatea cerut˘a. Avem egalitate dac˘ a ¸si numai dac˘a a = b = c. Ex. 11 Dac˘ a a, b, c sunt numere reale pozitive, atunci a(a − b) b(b − c) c(c − a) + + ≥ 0. b+c c+a a+b Proof. Deoarece   a(a − b) a a−b = + 1 (a − b) − a + b = (a + b + c) − a + b, b+c b+c b+c 494

rezult˘ a X a(a − b) b+c

= (a + b + c)

Xa−b b+c

= (a + b + c)

X a + c b+c

 −1 .

Prin urmare, inegalitatea dorit˘a este echivalent˘a cu X  a+c (a + b + c) − 3 ≥ 0. b+c Aceast˘ a inegalitate rezult˘a imediat din inegalitatea mediilor r Xa+c a+c b+a c+b ≥33 · · = 3. b+c b+c c+a a+b Avem egalitate dac˘ a ¸si numai dac˘a a = b = c. Ex. 12 Pentru n ∈ N+ , demonstrat¸i c˘a 1 1 1 1 1 1 1 (1 + + · · · + ) ≥ ( + + ··· + ). n+1 3 2n − 1 n 2 4 2n Proof. Scriem inegalitatea sub forma echivalent˘a n(1 +

1 1 1 1 1 + ··· + ) ≥ (n + 1)( + + · · · + ), 3 2n − 1 2 4 2n

sau

n 1 1 1 + n( + + · · · + ) 2 3 5 2n − 1 1 1 1 1 1 1 ≥ n( + + · · · + ) + ( + + ··· + ). 4 6 2n 2 4 2n Ultima inegalitate este adev˘arat˘a, deoarece n 1 1 1 ≥ + + ··· + 2 2 4 2n

¸si 1 1 1 1 1 1 + + ··· + ≥ + ··· + . 3 5 2n − 1 4 6 2n Avem egalitate numai pentru n = 1. Ex. 13 Fie a, b, c ∈ R+ astfel ˆıncˆat abc = 1. Ar˘atat¸i c˘a (a + b)(b + c)(c + a) ≥ 4(a + b + c − 1). Proof. F˘ ar˘ a a pierde din generalitate, presupunem c˘a a ≥ 1. Inegalitatea dorit˘ a este echivalent˘a cu a2 (b + c) + b2 (c + a) + c2 (a + b) + 6 ≥ 4(a + b + c), sau (a2 − 1)(b + c) + b2 (c + a) + c2 (a + b) + 6 ≥ 4a + 3(b + c). 495

√ √ Deoarece (a + 1)(b + c) ≥ 2 a · 2 bc = 4, este suficient s˘a ar˘at˘am c˘a 4(a − 1) + b2 (c + a) + c2 (a + b) + 6 ≥ 4a + 3(b + c), adic˘ a a(b2 + c2 ) + bc(b + c) − 3(b + c) + 2 ≥ 0. 1 Aplicˆ and inegalitatea mediilor ¸si inegalitatea b2 + c2 ≥ (b + c)2 , avem 2 p p a(b2 + c2 ) + bc(b + c) ≥ 2 abc(b2 + c2 )(b + c) = 2 (b2 + c2 )(b + c) p ≥ (b + c) 2(b + c). Prin urmare, este suficient s˘a demonstr˘am c˘a p (b + c) 2(b + c) − 3(b + c) + 2 ≥ 0. Cu notat¸ia x =

b+c , inegalitatea poate fi scris˘a astfel 2 2x3 − 3x2 + 1 ≥ 0.

Deoarece 2x3 − 3x2 + 1 = (x − 1)2 (2x + 1) ≥ 0, demonstrat¸ia este complet˘a. Egalitatea se atinge atunci cˆand a = b = c = 1. Ex. 14 Fie a, b, c numere pozitive astfel ˆıncˆat abc = 1. Ar˘atat¸i c˘a a b 1 + + ≥ a + b + 1. b c a Proof. Scriem inegalitatea sub forma echivalent˘a       a b b 1 1 2· + + + + + a ≥ 3a + 2b + 2. b c c a a Aplicˆ and inegalitatea mediilor, rezult˘a r r       2 a b b 1 1 b 3 a 2· + + + + +a ≥3 +2 + 2 = 3a + 2b + 2, b c c a a bc ca adic˘ a inegalitatea cerut˘a. Avem egalitate pentru a = b = c = 1. [Not˘ a] In acelea¸si condit¸ii, urm˘atoarea inegalitate mai ”tare” are loc: a b 1 p 2 + + ≥ 3(a + b2 + 1). b c a Aceast˘ a inegalitate este echivalent˘a cu   1 1 p 2 a + b + ≥ 3(a2 + b2 + 1), b a 496

care, prin ridicare la p˘atrat, devine   1 1 2 2 4 a b + 2b − 3 + 2 + 2 ≥ b2 + 3 − . b a b Deoarece b4 + 2b − 3 +

1 (b − 1)2 (2b + 1) 1 > 2b − 3 + = ≥ 0, b2 b2 b2

din inegalitatea mediilor rezult˘a r   1 1 1 4 2 a b + 2b − 3 + 2 + 2 ≥ 2 b4 + 2b − 3 + 2 , b a b r˘ amˆ anˆ and de ar˘ atat c˘a r 2

b4 + 2b − 3 +

1 2 ≥ b2 + 3 − . 2 b b

Efectuˆ and o nou˘ a ridicare la p˘atrat, obt¸inem b5 − 2b3 + 4b2 − 7b + 4 ≥ 0, adic˘ a b(b2 − 1)2 + 4(b − 1)2 ≥ 0. Ex. 15 Dac˘ a a, b, c sunt numere reale pozitive, atunci 1 1 1 8 1 + + + ≥ . a(a + b) b(b + c) c(c + d) d(d + a) (a + c)(b + d) Proof. Scriem inegalitatea sub forma X a(b + d) + c(b + d) a(a + b) echivalent˘ a cu

Xb+d a+b

Vom ar˘ ata c˘ a

+

≥ 8,

X c(b + d) ≥ 8. a(a + b)

Xb+d a+b

≥4

¸si X c(b + d) ≥ 4. a(a + b) Tinˆ and seama de inegalitatea 1 1 4 + ≥ , x y x+y echivalent˘ a cu (x − y)2 ≥ 0 pentru x, y > 0, avem     Xb+d 1 1 1 1 = (b + d) + + (a + c) + a+b a+b c+d b+c a+d 497

≥

4(b + d) 4(a + c) + = 4. (a + b) + (c + d) (b + c) + (a + d)

√ De asemenea, aplicˆ and de dou˘a ori inegalitatea mediilor x + y ≥ 2 xy, √ √ 2 echivalent˘ a cu ( x − y) ≥ 0, avem     X c(b + d) d c a b +(a+c) = (b+d) + + a(a + b) a(a + b) c(c + d) b(b + c) d(d + a) 2(a + c) 4(b + d) 2(b + d) 4(a + c) +p ≥ ≥p + = 4. (a + b) + (c + d) (b + c) + (d + a) (a + b)(c + d) (b + c)(d + a) Egalitatea are loc atunci cˆand a = c ¸si b = d. Ex. 16 (Cˆırtoaje) Dac˘ a a, b, c sunt numere reale, atunci a2 − bc b2 − ca c2 − ab + + ≥ 0. 4a2 + b2 + 4c2 4b2 + c2 + 4a2 4c2 + a2 + 4b2 Proof. Deoarece 4(a2 − bc) (b + 2c)2 = 1 − , 4a2 + b2 + 4c2 4a2 + b2 + 4c2 putem scrie inegalitatea sub forma echivalent˘a (c + 2a)2 (a + 2b)2 (b + 2c)2 + + ≤ 3. 4a2 + b2 + 4c2 4b2 + c2 + 4a2 4c2 + a2 + 4b2 Aplicˆ and inegalitatea Cauchy-Schwarz (vezi Ex. 38), avem (b + 2c)2 (b + 2c)2 b2 2c2 = ≤ + . 4a2 + b2 + 4c2 (2a2 + b2 ) + 2(2c2 + a2 ) 2a2 + b2 2c2 + a2 In mod similar, (c + 2a)2 c2 2a2 ≤ + , 4b2 + c2 + 4a2 2b2 + c2 2a2 + b2 (a + 2b)2 a2 2b2 ≤ + . 4c2 + a2 + 4b2 2c2 + a2 2b2 + c2 Prin adunarea acestor inegalit˘a¸ti rezult˘a imediat inegalitatea dorit˘a. Avem egalitate pentru a = b = c, precum ¸si pentru a = 2b = 4c, sau b = 2c = 4a, sau c = 2a = 4b. [Not˘ a] In mod similar putem demonstra c˘a pentru orice k > 0, avem

2ka2

a2 − bc b2 − ca c2 − ab + + ≥ 0, 2 2 2 2 2 2 2 2 +b +k c 2kb + c + k a 2kc + a2 + k 2 b2

cu egalitate pentru a = b = c, precum ¸si pentru a = kb = k 2 c, sau b = kc = k 2 a, sau c = ka = k 2 b. 498

4.2

Metoda major˘ arii ¸si minor˘ arii

Dac˘ a ne-am blocat ˆın demonstrarea direct˘a a inegalit˘a¸tii A ≤ B, putem ˆıncerca s˘ a g˘ asim o cantitate C care act¸ioneaz˘a ca o punte ˆıntre A ¸si B: dac˘ a avem A ≤ C ¸si C ≤ B, atunci ˆın mod natural avem A ≤ B. Cu alte cuvinte, putem majora pe A cu C, apoi pe C cu B (aceea¸si idee se aplic˘ a la minorare). Important este de a g˘asi un nivel corespunz˘ator de majorare sau de minorare. Ex. 17 Demonstrat¸i c˘ a pentru toate numerele reale pozitive a, b, c, avem 1 1 1 1 + + ≤ . a3 + b3 + abc b3 + c3 + abc c3 + a3 + abc abc [Observat¸ie] Cˆ and demonstr˘am inegalit˘a¸ti cu fract¸ii, adesea este bine s˘ a aducem fract¸iile la un numitor comun. Aceast˘a operat¸ie se poate realiza inclusiv prin majorarea sau minorarea numitorilor. Proof. Intrucˆ at a3 + b3 = (a + b)(a2 + b2 − ab) ≤ (a + b)ab, avem a3

1 1 c ≤ = . 3 + b + abc ab(a + b) + abc abc(a + b + c)

Similar, 1 b3

+

c3

+ abc

≤

a , abc(a + b + c)

¸si 1 b ≤ , 3 + a + abc abc(a + b + c) Adunˆ and cele trei inegalit˘a¸ti de mai sus, obt¸inem c3

a3

1 1 1 1 + 3 + 3 ≤ , 3 3 3 + b + abc b + c + abc c + a + abc abc

cu egalitate dac˘ a ¸si numai dac˘a a = b = c. Ex. 18 Fie a, b, c numere reale pozitive astfel ˆıncˆat a + b + c = 3. Ar˘atat¸i c˘a 1 + 8abc ≥ 9 min{a, b, c}. Proof. F˘ ar˘ a a pierde din generalitate, presupunem c˘a a ≤ b ≤ c. Trebuie s˘ a ar˘ at˘ am c˘ a 1 + 8abc ≥ 9a. Din a + b + c = 3 ¸si a ≤ b ≤ c rezult˘a a ≤ 1, iar din (a − b)(a − c) ≥ 0 obt¸inem bc ≥ a(b + c) − a2 = a(3 − a) − a2 = a(3 − 2a). Prin urmare, avem 1 + 8abc − 9a ≥ 1 + 8a2 (3 − 2a) − 9a = (1 − a)(1 − 4a)2 ≥ 0. Egalitatea are loc atunci cˆand a = b = c =1, precum  ¸si atunci cˆand 1 1 5 (a, b, c) este o permutare ciclic˘a a tripletului , , . 4 4 2 499

Ex. 19 Fie a, b, c numere reale pozitive astfel ˆıncˆat a ≤ b ≤ c ¸si ab + bc + ca = 3. Ar˘ atat¸i c˘ a 1 1 1 + + ≥ 1. 2a + b 2b + c 2c + a Proof. Vom utiliza cunoscuta inegalitate (x + y + z)2 ≥ 3(xy + yz + zx), echivalent˘ a cu (x − y)2 + (y − z)2 + (z − x)2 ≥ 0. 1 1 1 ,y= ¸si z = , avem 2a + b 2b + c 2c + a  2 1 1 1 9(a + b + c) + + . ≥ 2a + b 2b + c 2c + a (2a + b)(2b + c)(2c + a)

Astfel, pentru x =

R˘ amˆ ane s˘ a ar˘ at˘ am c˘ a 9(a + b + c) ≥ (2a + b)(2b + c)(2c + a). Scriem aceast˘ a inegalitate sub forma omogen˘a 3(a + b + c)(ab + bc + ca) ≥ (2a + b)(2b + c)(2c + a), echivalent˘ a cu ab2 + bc2 + ca2 ≥ a2 b + b2 c + c2 a, sau (a − b)(b − c)(c − a) ≥ 0. Pentru a ≤ b ≤ c, ultima inegalitate este evident adev˘arat˘a. Egalitatea are loc pentru a = b = c. Ex. 20 (Schur) Dac˘ a a, b, c sunt numere reale nenegative, atunci (1) (2)

a(a − b)(a − c) + b(b − c)(b − a) + c(c − a)(c − b) ≥ 0; a2 (a − b)(a − c) + b2 (b − c)(b − a) + c2 (c − a)(c − b) ≥ 0.

Proof. Deoarece inegalitatea este simetric˘a ˆın raport cu a, b, c, f˘ar˘a a pierde din generalitate vom considera c˘a a ≥ b ≥ c. (1) In ipotez a ≥ b ≥ c, este suficient s˘a ar˘at˘am c˘a a(a − b)(a − c) + b(b − c)(b − a) ≥ 0. Intr-adev˘ ar, avem a(a − b)(a − c) + b(b − c)(b − a) = (a − b)2 (a + b − c) ≥ 0. Egalitatea are loc atunci cˆand a = b = c, precum ¸si atunci cˆand unul dintre numerele a, b, c este 0, iar celelalte sunt egale. 500

(2) In mod similar, este suficient s˘a ar˘at˘am c˘a a2 (a − b)(a − c) + b2 (b − c)(b − a) ≥ 0. Intr-adev˘ ar, a2 (a − b)(a − c) + b2 (b − c)(b − a) = (a − b)2 (a2 + ab + b2 − bc − ca) ≥ 0. Egalitatea are loc atunci cˆand a = b = c, precum ¸si atunci cˆand unul dintre numerele a, b, c este 0, iar celelalte sunt egale.

[Nota 1] Inegalitatea Schur de gradul trei are urm˘atoarele forme echivalente, frecvent utilizate ˆın demonstrarea altor inegalit˘a¸ti: a3 + b3 + c3 + 3abc ≥ ab(a + b) + bc(b + c) + ca(c + a) ¸si (a + b + c)3 + 9abc ≥ 4(a + b + c)(ab + bc + ca). [Nota 2] Inegalitatea Schur de gradul patru este adev˘arat˘a pentru orice numere reale a, b, c ¸si are formele echivalente a4 + b4 + c4 + abc(a + b + c) ≥ ab(a2 + b2 ) + bc(b2 + c2 ) + ca(c2 + a2 ) ¸si 6abcp ≥ (p2 − q)(4q − p2 ), unde p = a + b + c ¸si q = ab + bc + ca. Ex. 21 Fie a, b, c numere reale pozitive astfel ˆıncˆat a ≤ 1 ≤ b ≤ c ¸si a+b+c=

1 1 1 + + . a b c

Ar˘ atat¸i c˘ a 1 1 1 + 2 + 2 ≥ a2 + b2 + c2 . 2 a b c Proof. Scriem inegalitatea sub forma   1 1 2 2 (a + c ) − 1 ≥ b2 − 2 , a2 c2 b sau 2

2

(a + c )



1 −1 ac



1 +1 ac



 ≥

1 b− b

  1 b+ . b

1 1 1 + + , rezult˘a a b c   1 1 (a + c) − 1 = b − ≥ 0. ac b

Din condit¸ia a + b + c =

501

Astfel, trebuie s˘ a ar˘ at˘am c˘a     1 1 2 2 . (a + c ) + 1 ≥ (a + c) b + ac b   1 Din inegalitatea evident˘a (b − c) − 1 ≥ 0, rezult˘a bc c+

1 1 ≥b+ . c b

Prin urmare, este suficient s˘a demonstr˘am c˘a     1 1 2 2 (a + c ) . + 1 ≥ (a + c) c + ac c Intr-adev˘ ar, avem     1 1 (1 − a2 )(c − a) (a2 + c2 ) + 1 − (a + c) c + = ≥ 0. ac c a Egalitatea are loc pentru b = 1 ¸si ac = 1. Ex. 22 Fie ai ≥ 1 (i = 1, 2, · · · , n). Demonstrat¸i c˘a (1 + a1 )(1 + a2 ) · · · (1 + an ) ≥

2n (1 + a1 + a2 + · · · + an ). n+1

[Observat¸ie] Examinˆand cele dou˘a p˘art¸i ale inegalit˘a¸tii de mai sus, cum putem proceda pentru a obt¸ine 2n ¸si ˆın membrul stˆang? Proof. Avem (1 + a1 )(1 + a2 ) · · · (1 + an ) a1 − 1 a2 − 1 an − 1 )(1 + ) · · · (1 + ). 2 2 2 Intrucˆ at ai − 1 ≥ 0, rezult˘a = 2n (1 +

(1 + a1 )(1 + a2 ) · · · (1 + an ) a1 − 1 a2 − 1 an − 1 + + ··· + ) 2 2 2 a1 − 1 a2 − 1 an − 1 ≥ 2n (1 + + + ··· + ) n+1 n+1 n+1 ≥ 2n (1 +

=

2n [n + 1 + (a1 − 1) + (a2 − 1) + · · · + (an − 1)] n+1 =

2n (1 + a1 + a2 + · · · + an ). n+1

Am obt¸inut astfel inegalitatea init¸ial˘a. Pentru n ≥ 2, egalitatea are loc atunci cˆ and a1 = a2 = · · · = an = 1. 502

Ex. 23 G˘ asit¸i cel mai mare num˘ar real α, astfel ˆıncˆat x y z p +√ +p ≥α 2 2 2 2 2 x +z y +z x + y2 pentru orice numere reale nenegative x, y, z, dintre care cel put¸in dou˘a nenule. Proof. Pentru x = 0 ¸si y = z, membrul stˆang al inegalit˘a¸tii este egal cu 2. Prin urmare, α nu poate fi mai mare ca 2. S˘a consider˘am a¸sadar α = 2. F˘ ar˘ a a pierde din generalitate, presupunem c˘a x ≤ y ≤ z. Vom ar˘ata c˘a p √ x y x2 + y 2 x2 + z 2 z p +√ ≥ √ ≥ 2. +p +p x2 + z 2 x2 + z 2 y2 + z2 x2 + y 2 x2 + y 2 Inegalitatea din dreapta rezult˘a imediat din inegalitatea mediilor. Scriem acum inegalitatea din stˆanga sub forma p √ x x2 + y 2 − y x2 + z 2 − z p + p ≥ √ , x2 + z 2 y2 + z2 x2 + y 2 sau x x2 x2 p p ≥√ +p . √ y2 + z2 x2 + z 2 ( x2 + y 2 + y) x2 + y 2 ( x2 + z 2 + z) p √ Deoarece x2 + y 2 + y > 2y ¸si x2 + z 2 + z > 2z, este suficient s˘a ar˘ at˘ am c˘ a 1 x x p ≥ √ + p . 2 2 2y x + z y2 + z2 2z x2 + y 2 In cazul netrivial x > 0, avem √

x 1 1 y =p ≤q =p z 2 2 2 1 + (x) +z y + z2 1 + ( yz )2

x2

¸si x 1 1 p =q ≤√ . 2 2 y 2 x +y 1 + ( x )2 Prin urmare, este suficient s˘a demonstr˘am inegalitatea 1 1 1 p ≥ p + √ , y2 + z2 2 y 2 + z 2 2 2z care este echivalent˘ a cu inegalitatea evident˘a p √ 2z ≥ y 2 + z 2 . Astfel, inegalitatea cerut˘a este demonstrat˘a. Prin urmare, αmax = 2. In ipoteza x ≤ y ≤ z, egalitatea are loc pentru x = 0 ¸si y = z. 503

Ex. 24 Dac˘ a x, y, z ∈ [1, 4], atunci     y x y z z x ≥5 + 9. 8 + + + + y z x x y z Proof. Fie  E(x, y, z) = 8

x y z + + y z x



 −5

y z x + + x y z

 − 9.

Deoarece inegalitatea este ciclic˘a, f˘ar˘a a pierde din generalitate, presupunem c˘ a x = max{x, y, z}. Vom ar˘ata c˘a √ E(x, y, z) ≥ E(x, xz, z) ≥ 0. Avem √



E(x, y, z) − E(x, xz, z) = 8

r  r   y z x y x z + −2 −5 + −2 y z z x y x

√

xz)2 (8x − 5z) ≥ 0. xyz r x , 1 ≤ t ≤ 2, avem De asemenea, cu notat¸ia t = z =

(y −

√

 r   r  x z z x E(x, xz, z) = 8 2 + −3 −5 2 + −3 z x x z     2 1 8 5 2 = 8 2t + 2 − 3 −5 + t − 3 = 2 (t−1)2 (2t+1)− (t−1)2 (t+2) t t t t =

(t − 1)2 (4 + 5t)(2 − t) ≥ 0. t2

Avem egalitate atunci cˆand x = y = z, precum ¸si atunci cˆand (x, y, z) este o permutare ciclic˘a a tripletului (4, 2, 1).

4.3

Metoda coeficient¸ilor nedeterminat¸i

Demonstrarea unor inegalit˘a¸ti se poate face prin spargerea acestora ˆın inegalitˆıt¸i mai simple, cu coeficient¸i nedeterminat¸i, dar care urmeaz˘a a fi calculat¸i astfel ˆıncˆ at inegalit˘a¸tile respective s˘a fie adev˘arate, sau s˘a se ajung˘ a la rezultatul dorit. In multe cazuri, succesul metodei depinde de igeniozitatea ¸si experient¸a rezolvitorului. Ex. 25 Dac˘ a x, y, z ∈ [1, 2], atunci     1 1 1 1 1 1 3 + + ≥2 + + . x + 2y y + 2z z + 2x x+y y+z z+x 504

Proof. Presupunem c˘a exist˘a o constant˘a real˘a α astfel ˆıncˆat   3 2 1 1 ≥0 − +α − x + 2y x + y x y pentru orice x, y ∈ [1, 2]. Atunci, ˆınsumˆand aceast˘a inegalitate cu celelalte dou˘ a similare, obt¸inem inegalitatea dorit˘a. Deoarece   (x − y)[xy − α(x + y)(x + 2y)] 3 2 1 1 = − +α − , x + 2y x + y x y xy(x + y)(x + 2y) 1 alegˆ and α = , rezult˘a 6 3 2 1 − + x + 2y x + y 6



1 1 − x y

 =

(x − y)2 (2y − x) ≥ 0. 6xy(x + y)(x + 2y)

Avem egalitate dac˘ a ¸si numai dac˘a x = y = z. Ex. 26 Fie x, y, z numere reale, nu toate nule. G˘asit¸i valoarea maxim˘a a fract¸iei xy + 2yz . x2 + y 2 + z 2 Proof. Pentru a g˘ asi valoarea maxim˘a cerut˘a, este suficient s˘a ar˘at˘am c˘ a exist˘ a o constant˘ a c, astfel ˆıncˆat xy + 2yz ≤ c, + y2 + z2

x2

¸si c˘ a semnul egal se obt¸ine cel put¸in pentru un triplet x, y, z dat. Deoarece ambii termeni ai sumei xy + 2yz cont¸in variabila y, vom sparge termenul y 2 al sumei x2 + y 2 + z 2 ˆın αy 2 ¸si (1 − α)y 2 . Din inegalitatea mediilor, avem √ x2 + αy 2 ≥ 2 α · xy ¸si

√ (1 − α)y 2 + z 2 ≥ 2 1 − α · yz,

iar prin ˆınsumarea acestor inegalit˘a¸ti obt¸inem √ √ x2 + y 2 + z 2 ≥ 2 α · xy + 2 1 − α · yz, ! r √ 1 − α x2 + y 2 + z 2 ≥ 2 α xy + · yz . α r 1−α 1 Pentru a avea = 2, trebuie s˘a alegem α = . Rezult˘a α 5 √ xy + 2yz 5 , ≤ 2 2 2 x +y +z 2 y z xy + 2yz cu egalitate pentru x = √ = . Prin urmare, fract¸ia 2 are 2 x + y2 + z2 5 √ 5 valoarea maxim˘ a . 2 505

Ex. 27 Fie x, y, z numere reale nenegative, dintre care cel put¸in dou˘a nenule. Demonstrat¸i c˘ a x z y p +p +√ ≥ 2. 2 2 x +z y2 + z2 x2 + y 2 Proof. Dac˘ a exist˘ a o constant˘a real˘a α astfel ˆıncˆat x 2xα p ≥ α x + yα + zα y2 + z2 pentru orice numere reale nenegative x, y, z, atunci obt¸inem inegalitatea dorit˘ a prin ˆınsumarea acestei inegalit˘a¸ti cu celelalte dou˘a similare. Deoarece p xα + y α + z α ≥ 2 xα (y α + z α ), inegalitatea de mai sus este adev˘arat˘a dac˘a x xα p ≥p , xα (y α + z α ) y2 + z2 adic˘ a y α + z α ≥ xα−2 (y 2 + z 2 ). Este u¸sor de observat c˘a ultima inegalitate este verificat˘a ca identitate pentru α = 2. Cu aceasta, demonstrat¸ia este ˆıncheiat˘a. Inegalitatea init¸ial˘ a devine egalitate cˆand unul dintre numerele x, y, z este 0, iar celelalte dou˘ a sunt egale. Ex. 28 (Cirtoaje) Fie x1 , x2 , ..., xn numere reale pozitive astfel ˆıncˆat x1 x2 ...xn = 1. Ar˘ atat¸i c˘ a 1 1 1 + + ... + ≥ 1. 1 + (n − 1)x1 1 + (n − 1)x2 1 + (n − 1)xn Proof. Presupunem c˘a exist˘a o constant˘a real˘a k astfel ˆıncˆat 1 xki ≥ k 1 + (n − 1)xi x1 + xk2 + ... + xkn pentru i ∈ {1, 2, ..., n}. Atunci, ˆınsumˆand toate aceste inegalit˘a¸ti pentru i = 1, 2, ..., n, obt¸inem inegalitatea dorit˘a. Scriem inegalitatea de mai sus sub forma xk1 + ... + xki−1 + xri+1 + ... + xkn ≥ (n − 1)xk+1 . i Din inegalitatea mediilor, avem q xk1 + ... + xki−1 + xki+1 + ... + xkn ≥ (n − 1) n−1 (x1 ...xi−1 xi+1 ...xn )k −k

= (n − 1)xin−1 , 1 iar pentru k = − 1 obt¸inem chiar inegalitatea dorit˘a. Cu aceasta, n inegalitatea este demonstrat˘a. Avem egalitate dac˘a ¸si numai dac˘a x1 = x2 = ... = xn = 1. 506

Ex. 29 [Cˆırtoaje] Fie a, b, c, d numere reale pozitive astfel ˆıncˆat abcd = 1. Ar˘atat¸i c˘ a 1 1 1 1 + + + ≤ 1. a+b+2 b+c+2 c+d+2 d+a+2 Proof. Dac˘ a exist˘ a o constant˘a real˘a k astfel ˆıncˆat 2 ck + dk , ≤ k a+b+2 a + bk + ck + dk atunci putem obt¸ine inegalitatea dorit˘a prin ˆınsumarea acestei inegalit˘a¸ti cu celelalte inegalit˘ a¸ti similare. Inegalitatea de mai sus este echivalent˘a cu (a + b)(ck + dk ) ≥ 2(ak + bk ). Deoarece ck + dk ≥ 2(cd)k/2 =

2 , (ab)k/2

este suficient s˘ a ar˘ at˘ am c˘a a + b ≥ (ab)k/2 (ak + bk ). 1 Aceast˘ a inegalitate este omogen˘a pentru k = . In acest caz, inegalitatea 2 devine √ √ √ 4 a + b ≥ ab( a + b), sau

√ √ √ √ 4 4 4 ( a3 − b3 )( 4 a − b) ≥ 0,

ultima form˘ a fiind evident adev˘arat˘a. Egalitatea are loc pentru a = b = c = d = 1. [Not˘ a] In general, pentru a1 , a2 , ..., an numere reale pozitive satisf˘acˆand a1 a2 ...an = 1, are loc inegalitatea X

1 ≤ 1, a1 + a2 + ... + ak + n − k

unde 1 ≤ k ≤ n − 1. Ex. 30 Fie a, b, c numere reale pozitive astfel ˆıncˆat a + b + c = 3. Ar˘atat¸i c˘a 1 1 1 + 2 + 2 ≥ a2 + b2 + c2 . 2 a b c Proof. Dac˘ a exist˘ a o constant˘a real˘a k astfel ˆıncˆat 1 − a2 + k(a − 1) ≥ 0, a2 atunci putem obt¸ine inegalitatea dorit˘a prin ˆınsumarea acestei inegalit˘a¸ti cu celelalte inegalit˘ a¸ti similare. Deoarece 1 (1 − a)[1 + a + (1 − k)a2 + a3 ) 2 − a + k(a − 1) = , a2 a2 507

alegˆ and k = 4, rezult˘ a 1 (1 − a)2 (1 + 2a − a2 ) 2 − a + 4(a − 1) = . a2 a2 √ Prin urmare, pentru a ≤ 1 + 2, avem 1 − a2 + 4(a − 1) ≥ 0 a2 In continuare, f˘ ar˘ a a pierde din generalitate, consider˘am a ≥ b ≥ c. Avem dou˘ a cazuri. √ Cazul 1. a ≤ 1 + 2. Prin ˆınsumarea inegalit˘a¸tilor 1 − a2 + 4(a − 1) ≥ 0, a2 1 − b2 + 4(b − 1) ≥ 0, b2 1 − c2 + 4(c − 1) ≥ 0, c2 obt¸inem inegalitatea dorit˘a. √ √ 2 Cazul 2. a > 1 + 2. Deoarece b + c = 3 − a < 2 − 2 < , avem 3 1 1 bc ≤ (b + c)2 < , 4 9 deci 1 1 1 1 2 1 + 2+ 2 > 2+ 2 ≥ > 18 > (a + b + c)2 > a2 + b2 + c2 . 2 a b c a b bc Cu aceasta, inegalitatea este demonstrat˘a. Egalitatea are loc numai ˆın cazul a = b = c = 1. Ex. 31 Fie a, b, c, d numere reale nenegative astfel ˆıncˆat a2 + b2 + c2 + d4 = a + b + c + d. Ar˘ atat¸i c˘ a a4 + b4 + c4 + d4 ≥ a + b + c + d. Proof. Pentru orice α ∈ R, inegalitatea dorit˘a este echivalent˘a cu fiecare din inegalit˘ a¸tile de mai jos: a4 + b4 + c4 + d4 + α(a + b + c + d) ≥ a + b + c + d + α(a2 + b2 + c2 + d2 ), X a(a3 − αa + α − 1) ≥ 0, X a(a − 1)(a2 + a − α + 1) ≥ 0. Alegˆ and α = 3, inegalitatea devine astfel X a(a + 2)(a − 1)2 ≥ 0, fiind evident adev˘ arat˘a. Egalitatea are loc atunci cˆand a, b, c, d ∈ {0, 1}.

508

Ex. 32 (Ostrowski) Fie trei seturi de numere reale a1 , a2 , · · · , an , b1 , b2 , · · · , bn ¸si x1 , x2 , · · · , xn care satisfac relat¸iile a1 x1 + a2 x2 + · · · + an xn = 0, b1 x1 + b2 b2 + · · · + bn xn = 1. Demonstrat¸i c˘ a x21 + x22 + · · · + x2n ≥ unde A=

n X

a2i

6= 0, B =

i=1

n X

b2i ,

A , AB − C 2 n X C=( ai bi )2 .

i=1

i=1

Proof. Oricare ar fi numerele reale α ¸si β, avem n X

x2i =

i=1 n X i=1

deci

x2i =

n X

x2i + α

i=1 n X

n X

ai xi + β(

i=1

n X

bi xi − 1),

i=1 n

(xi +

i=1

i=1 n X

αai + βbi 2 X (αai + βbi )2 ) − − β, 2 4

x2i ≥ −

i=1

n X (αai + βbi )2 i=1

4

− β,

(4.1)

cu egalitate dac˘ a ¸si numai dac˘a αai + βbi , i = 1, 2, · · · , n. (4.2) 2 P P Inlocuind (4.2) ˆın relat¸iile ni=1 ai xi = 0 ¸si ni=1 bi xi = 1, obt¸inem xi = −

αA + βC = 0, αC + βB = −1, deci α=

2C 2A , β=− . 2 AB − C AB − C 2

(4.3)

Astfel, din (4.1) ¸si (4.3) rezult˘a n X

x2i ≥

i=1

A , AB − C 2

adic˘ a inegalitatea cerut˘a. Din (4.2) ¸si (4.3) rezult˘a c˘a egalitatea are loc atunci cˆ and Abi − Cai xi = , i = 1, 2, · · · , n. AB − C 2

509

[Nota] Inegalitatea de mai sus poate fi mai simplu demonstrat˘a utilizˆand inegalitatea Cauchy-Schwarz (vezi Ex. 38). Astfel, pentru t ∈ R, avem n n n X X X 2 2 [ (ai t + bi ) ] · ( xi ) ≥ [ (ai t + bi )xi ]2 = 1, i=1

i=1

i=1

(At2 + 2Ct + B)(x21 + x22 + · · · x2n ) − 1 ≥ 0, At2 + 2Ct + B −

1 ≥ 0. x21 + x22 + · · · + x2n

Deoarece A > 0, ultima inegalitate este adev˘arat˘a pentru orice t real dac˘ a ¸si numai dac˘ a ∆ ≤ 0, unde ∆ = 4C 2 − 4A(B − x21 − x22 − · · · − x2n ) este discriminantul trinomului de gradul doi ˆın t din membrul stˆang al inegalit˘ a¸tii. Din condit¸ia ∆ ≤ 0 obt¸inem inegalitatea dorit˘a.

4.4

Metoda normaliz˘ arii

Cˆ and o inegalitate f (x1 , x2 , ..., xn ) ≥ 0 este omogen˘a de ordinul k, adic˘a f (tx1 , tx2 , ..., txn ) = tk f (x1 , x2 , ..., xn ), putem presupune, f˘ar˘a a pierde din generalitate, c˘ a suma variabilelor este constant˘a, ˆın scopul de a aduce inegalitatea la o form˘a mai simpl˘a, mai u¸sor demonstrabil˘a. In general, putem considera c˘a g(x1 , x2 , ..., xn ) = const, unde g(x1 , x2 , ..., xn ) este o expresie omogen˘a de ordin arbitrar, sau chiar f1 (x1 , x2 , ..., xn ) = f2 (x1 , x2 , ..., xn ), unde f1 (x1 , x2 , ..., xn ) ¸si f2 (x1 , x2 , ..., xn ) sunt expresii omogene de ordin diferit. Ex. 33 Dac˘ a a, b, c sunt numere reale pozitive, atunci (2a + b + c)2 (2b + c + a)2 (2c + a + b)2 + + ≤ 8. 2a2 + (b + c)2 2b2 + (c + a)2 2c2 + (a + b)2 Proof. Deoarece inegalitatea este omogen˘a, presupunem c˘a a + b + c = 3. Trebuie s˘ a ar˘ at˘ am c˘ a (a + 3)2 (b + 3)2 (c + 3)2 + + ≤ 8, 2a2 + (3 − a)2 2b2 + (3 − b)2 2c2 + (3 − c)2 adic˘ a f (a) + f (b) + f (c) ≤ 8, unde f (x) = Avem f (x) =

(x + 3)2 , x ∈ R+ . 2x2 + (3 − x)2

x2 + 6x + 9 1 8x + 6 = (1 + ) 2 3(x − 2x + 3) 3 (x − 1)2 + 2 510

1 8x + 6 4(x + 1) ≤ (1 + )= . 3 2 3 A¸sadar, f (a) + f (b) + f (c) ≤

4(a + 1 + b + 1 + c + 1) = 8, 3

ceea ce trebuia demonstrat. Avem egalitate pentru a = b = c. Ex. 34 Dac˘ a a, b, c ¸si x, y, z sunt numere reale nenegative, atunci [(b + c)x + (c + a)y + (a + b)z]2 ≥ 4(ab + bc + ca)(xy + yz + zx). Proof. Scriem inegalitatea sub forma p 2 (ab + bc + ca)(xy + yz + zx) ≤ (b + c)x + (c + a)y + (a + b)z. Deoarece inegalitatea este omogen˘a ˆın raport cu x, y, z, putem considera c˘ a x + y + z = a + b + c. In aceast˘a ipotez˘a, inegalitatea cerut˘a se obt¸ine din inegalitatea mediilor, astfel: p 2 (ab + bc + ca)(xy + yz + zx) ≤ (ab + bc + ca) + (xy + yz + zx) (a + b + c)2 − a2 − b2 − c2 (x + y + z)2 − x2 − y 2 − z 2 + 2 2 2 2 2 a +x b + y 2 c2 + z 2 = (a + b + c)(x + y + z) − − − 2 2 2 ≤ (a + b + c)(x + y + z) − ax − by − cz = (b + c)x + (c + a)y + (a + b)z. y z x In cazul abc 6= 0, egalitatea are loc dac˘a ¸si numai dac˘a = = . a b c =

Ex. 35 Fie a, b, c numere reale astfel ˆıncˆat a + b + c > 0 ¸si b2 ≥ 4ac. Ar˘atat¸i c˘a 4 min{a, b, c} ≤ a + b + c ≤

9 max{a, b, c}. 4

Proof. Deoarece condit¸iile date ¸si inegalitatea cerut˘a sunt omogene ˆın raport cu a, b, c, presupunem, f˘ar˘a a pierde din generalitate, c˘a a+b+c = 1. (A) S˘ a ar˘ at˘ am c˘ a 4 max{a, b, c} ≥ . 9

(4.4)

Consider˘ am cazul netrivial b < 49 . Din a + c = 1 − b > 59 , rezult˘a c˘a 1 a ≤ 9 implic˘ a c > 49 , iar c ≤ 19 implic˘a a > 49 , inegalitatea cerut˘a fiind adev˘ arat˘ a ˆın ambele cazuri. R˘amˆane de analizat cazul b < 94 ¸si 4 4 , deci ( 59 − c) · c < ac < 81 , iar din a, c > 19 . Din b2 ≥ 4ac rezult˘a ac < 81 5 4 4 ( 9 − c) · c < 81 , obt¸inem c > 9 . Avem egalitate ˆın (4.4) pentru a = b = 49 ¸si c = 19 , sau b = c = 49 ¸si a = 19 . (B) S˘ a ar˘ at˘ am c˘ a 1 min{a, b, c} ≤ . 4 511

(4.5)

Consider˘ am cazul netrivial a > √ b > c. Prin urmare, √

1 4

¸si b, c > 0. Din b2 ≥ 4ac > c, rezult˘a

3 c+c 0, din considerente de omegenitate, vom presupune c = 1. In plus, cu notat¸ia a = x3 , x ≥ 1, inegalitatea devine succesiv astfel: 2(x3 − 1)2 ≥ (x3 + 5)(2x3 − 3x2 + 1), (x − 1)2 [2(x2 + x + 1)2 − (x3 + 5)(2x + 1)] ≥ 0, (x − 1)2 (x3 + 2x2 − 2x − 1) ≥ 0, (x − 1)3 (x2 + 3x + 1) ≥ 0, ultima inegalitate fiind evident adev˘arat˘a. Scriem acum inegalitatea din stˆ anga sub forma √ √ √ 3 a − b − 3 3 ac( 3 a − b) ≥ 0, sau

√ √ √ √ √ √ 3 3 3 3 ( 3 a − b)( a2 + ab + b2 − 3 3 ac) ≥ 0.

Ultima inegalitate este adev˘arat˘a, deoarece √ √ √ √ √ √ √ √ √ 3 3 3 3 3 a2 + ab + b2 − 3 3 ac ≥ 3 ab − 3 3 ac = 3 3 a( b − 3 c) ≥ 0. Cu aceasta, inegalitatea este demonstrat˘a. Avem egalitate pentru a = b = c, precum ¸si pentru a = b ¸si c = 0.

4.5

Metoda omogeniz˘ arii

Metoda omogeniz˘ arii poate fi aplicat˘a la demonstrarea inegalit˘a¸tilor normalizate, atunci cˆ and acestea devin mai simple prin omogenizare. Uneori, pentru demonstrarea unei inegalit˘a¸ti normalizate se poate utiliza mai ˆıntˆ ai metoda omogeniz˘arii, apoi metoda normaliz˘arii (a doua normalizare fiind, desigur, diferit˘a de prima). 515

Ex. 41 Fie a, b, c numere reale nenegative astfel ˆıncˆat a + b = a4 + b4 . Ar˘atat¸i c˘ a a3 + b3 ≥ a2 + b2 . Proof. Scriem inegalitatea sub forma omogen˘a (a + b)(a3 + b3 )3 ≥ (a4 + b4 )(a2 + b2 )3 , care este echivalent˘ a cu A − 3B ≥ 0, unde A = (a + b)(a9 + b9 ) − (a4 + b4 )(a6 + b6 ), B = a2 b2 (a4 + b4 )(a2 + b2 ) − a3 b3 (a + b)(a3 + b3 ). Deoarece A = ab(a3 − b3 )(a5 − b5 ) ¸si B = a2 b2 (a − b)(a5 − b5 ), rezult˘ a A − 3B = ab(a − b)3 (a5 − b5 ) ≥ 0. Avem egalitate dac˘ a ¸si numai dac˘a a = b = 1. Ex. 42 Fie a, b, c numere reale nenegative astfel ˆıncˆat ab + bc + ca = 3. Ar˘atat¸i c˘ a 1 1 3 1 + + ≥ . a2 + 1 b2 + 1 c2 + 1 2 Proof. In urma dezvolt˘arii, inegalitatea devine astfel a2 + b2 + c2 + 3 ≥ a2 b2 + b2 c2 + c2 a2 + 3a2 b2 c2 . Din cunoscuta inegalitate (a + b + c)(ab + bc + ca) ≥ 9abc, echivalent˘ a cu a(b − c)2 + b(c − a)2 + c(a − b)2 ≥ 0, rezult˘ a a + b + c ≥ 3abc. Astfel, este suficient s˘a ar˘at˘am c˘a a2 + b2 + c2 + 3 ≥ a2 b2 + b2 c2 + c2 a2 + abc(a + b + c). Scriem aceast˘ a inegalitate sub forma omogen˘a (ab + bc + ca)(a2 + b2 + c2 ) + (ab + bc + ca)2 ≥ 3(a2 b2 + b2 c2 + c2 a2 ) +3abc(a + b + c), 516

echivalent˘ a cu ab(a2 + b2 ) + bc(b2 + c2 ) + ca(c2 + a2 ) ≥ 2(a2 b2 + b2 c2 + c2 a2 ), sau ab(a − b)2 + bc(b − c)2 + ca(c − a)2 ≥ 0. Avem egalitate atunci cˆand a = b = c = 1, ¸si atunci cˆand (a, b, c) √ precum √ este o permutare ciclic˘a a tripletului (0, 3, 3). Ex. 43 Fie a, b, c, d numere reale pozitive astfel ˆıncˆat a + b + c + d = 4. Ar˘atat¸i c˘ a a b c d + + + ≥ 2. a2 + 1 c2 + 1 d2 + 1 a2 + 1 Proof. Avem ab2 ab2 ab a = a − ≥ a − =a− . b2 + 1 b2 + 1 2b 2 In mod similar, b bc c cd d da ≥= b − , 2 ≥= c − , 2 ≥= d − . c2 + 1 2 d +1 2 a +1 2 Prin urmare, este suficient s˘a ar˘at˘am c˘a a+b+c+d−

ab + bc + cd + da ≥ 2, 2

adic˘ a 4 ≥ ab + bc + cd + da. Scriem aceast˘ a inegalitate sub forma omogen˘a (a + b + c + d)2 ≥ 4(ab + bc + cd + da), echivalent˘ a cu (a − b + c − d)2 ≥ 0. Avem egalitate pentru a = b = c = d = 1. Ex. 44 Fie a ¸si b numere reale nenegative astfel ˆıncˆat 2a2 + b2 = 2a + b. Ar˘atat¸i c˘ a a−b 1 − ab ≥ . 3 Proof. Pentru a = b = 0, inegalitatea este adev˘arat˘a. Altfel, scriem inegalitatea ˆın forma omogen˘a (2a2 + b2 )2 (a − b)(2a2 + b2 ) − ab ≥ . (2a + b)2 3(2a + b) Deoarece

(2a2 + b2 )2 (a − b)(4a3 − b3 ) − ab = , (2a + b)2 (2a + b)2 517

r˘ amˆ ane s˘ a ar˘ at˘ am c˘ a (a − b)[3(4a3 − b3 ) − (2a + b)(2a2 + b2 )] ≥ 0. Aceasta este echivalent˘a cu inegalitatea evident˘a (a − b)2 (4a2 + 3ab + 2b2 ) ≥ 0. Egalitatea are loc pentru a = b = 1. Ex. 45 Fie a, b, c numere reale pozitive astfel ˆıncˆat a + b + c = 3. Ar˘atat¸i c˘a a2 + 3b b2 + 3c c2 + 3a + + ≥ 6. a+b b+c c+a Proof. Prin omogenizare, inegalitatea devine succesiv astfel: X a2 + (a + b + c)b a+b

≥ 2(a + b + c),

X  a2 + ab + b2 + bc a+b X b(c − a) a+b X

 − a − b ≥ 0, ≥ 0,

b(b + c)(c2 − a2 ) ≥ 0,

ab3 + bc3 + ca3 − abc(a + b + c) ≥ 0. Ultima inegalitate rezult˘a din inegalitatea Cauchy-Schwarz, astfel √ √ √ ab3 c + abc3 + a3 bc 3 3 3 ab + bc + ca ≥ = abc(a + b + c). c+a+b Cu aceasta, demonstrat¸ia este ˆıncheiat˘a. Avem egalitate pentru a = b = c = 1. Ex. 46 Fie a ¸si b numere reale astfel ˆıncˆat 9a2 + 8ab + 7b2 ≤ 24. Demonstrat¸i c˘a 7a + 5b + 6ab ≤ 18. Proof. Observ˘ am c˘ a pentru a = b = 1 avem egalitate. Tinˆand seama de acest fapt, pentru a avea numai expresii omogene de gradul doi ˆın a ¸si b vom utliza inegalit˘ a¸tile 2a ≤ a2 + 1 and 2b ≤ b2 + 1, echivalente respectiv 2 cu (a − 1) ≥ 0 ¸si (b − 1)2 ≥ 0. Atunci, avem 2(7a + 5b + 6ab − 18) ≤ 7(a2 + 1) + 5(b2 + 1) + 12ab − 36 = 7a2 + 5b2 + 12ab − 24 = (9a2 + 8ab + 7b2 − 24) − 2(a2 + b2 − 2ab) ≤ −2(a − b)2 ≤ 0, de unde rezult˘ a imediat inegalitatea dorit˘a. Egalitatea are loc dac˘a ¸si numai dac˘ a a = b = 1. 518

Ex. 47 Fie a, b, c numere reale pozitive care satisfac abc = 1. Demonstrat¸i c˘a 81(a2 + 1)(b2 + 1)(c2 + 1) ≤ 8(a + b + c)4 . Proof. Deoarece a2 + 1 = a2 +

√ 3

a2 b2 c2 =

√ 3

a2

√ 3

a4 +

√ 3

 b2 c2 ,

scriem inegalitatea ˆın forma omogen˘a √  √  √  √ √ √ 3 3 3 3 3 3 81 a4 + b2 c2 b4 + c2 a2 c4 + a2 b2 ≤ 8(a + b + c)4 . Pentru a demonstra noua inegalitate omogen˘a, presupunem c˘a a + b + c = 3. Inegalitatea devine astfel r  √  √  √ √ √ √ 3 3 3 3 3 3 3 4 2 2 4 2 2 4 2 2 a + b c b + c a c + a b ≤ 2. Utilizˆ and inegalitatea mediilor, este suficient s˘a ar˘at˘am c˘a  X √ √ 3 3 a4 + b2 c2 ≤ 6. Utilizˆ and din nou inegalitatea mediilor, avem  X  a2 + a + a bc + bc + 1  X √ √ 3 3 4 2 2 a + b c ≤ + 3 3 P a)2 + 2 a + 3 = 6. = 3 Cu aceasta, inegalitatea este demonstrat˘a. Avem egalitate dac˘a ¸si numai dac˘ a a = b = c = 1. (

P

Ex. 48 Fie a, b, c numere reale nenegative care satisfac realat¸ia (a + b)(b + c)(c + a) = 2. Demonstrat¸i c˘ a (a2 + bc)(b2 + ca)(c2 + ab) ≤ 1. Proof. Scriem inegalitatea dorit˘a sub forma omogen˘a 4(a2 + bc)(b2 + ca)(c2 + ab) ≤ (a + b)2 (b + c)2 (c + a)2 . Deoarece inegalitatea este simetric˘a ˆın a, b, c, presupunem, f˘ar˘a a pierde din generalitate, c˘ a a ≥ b ≥ c. Avem a2 + bc ≤ a2 + ac ≤ (a + c)2 519

¸si, din inegalitatea mediilor, 4(b2 + ca)(c2 + ab) ≥ (b2 + ca + c2 + ab)2 . Atunci, este suficient s˘a demonstr˘am inegalitatea b2 + c2 + ab + ac ≥ (a + b)(b + c), echivalent˘ a cu inegalitatea evident˘a c(c−b) ≤ 0. Cu aceasta, demonstrat¸ia este ˆıncheiat˘ a. Avem egalitate atunci cˆand unul dintre numerele a, b, c este 0, iar celelalte dou˘a sunt egale cu 1. Exercise 1 1 Fie x, y, z ∈ R. Ar˘ atat¸i c˘a (x2 + y 2 + z 2 )[(x2 + y 2 + z 2 )2 − (xy + yz + zx)2 )] ≥ (x + y + z)2 [(x2 + y 2 + z 2 ) − (xy + yz + zx)]2 . 2 Fie m, n ∈ N+ , m > n. Ar˘atat¸i c˘a (1 +

1 n 1 ) < (1 + )m n m

3 Fie a, b, c ∈ R+ . Ar˘ atat¸i c˘a a8 + b8 + c8 1 1 1 + + ≤ . a b c a3 b3 c3 4 Fie numerele reale a1 , a2 , · · · , a100 care satisfac: (1) a1 ≥ a2 ≥ · · · ≥ a100 ≥ 0; (2) a1 + a2 ≤ 100; (3) a3 + a4 + · · · + a100 ≤ 100. G˘asit¸i maximul pentru a21 + a22 + · · · + a2100 . 5 Fie k, n numere ˆıntregi pozitive, 1 ≤ k < n. Dac˘a x1 , x2 , · · · , xk sunt numere reale pozitive astfel ˆıncˆat x1 + x2 + ... + xk = x1 x2 ...xk , atunci xn−1 + x2n−1 + · · · + xn−1 ≥ kn. 1 k 6 Dac˘ a a, b, c ∈ R, ar˘ atat¸i c˘a (a2 + ab + b2 )(b2 + bc + c2 )(c2 + ca + a2 ) ≥ (ab + bc + ca)3 . 7 Fie a, b numere reale pozitive date, iar θ ∈ (0, π2 ). G˘asit¸i maximul pentru √ √ y = a sin θ + b cos θ. 520

8 Fie a1 , a2 , ..., an numere reale astfel ˆıncˆat |ai | ≤ 2 pentru i = 1, 2, ..., n ¸si a31 + a32 + ... + a3n = 0. Ar˘atat¸i c˘a a1 + a2 + ... + an ≤

2n . 3

9 Pentru n ≥ 3, fie x1 , x2 , · · · , xn ∈ [−1, 1] astfel ˆıncˆat x51 + x52 + ... + x5n = 0. Demonstrat¸i c˘ a x1 + x2 + ... + xn ≤

8n . 15

10 Fie x1 , x2 , · · · , xn numere reale astfel ˆıncˆat x1 +x2 +...+xn = na. Ar˘atat¸i c˘ a n n X 1 X 2 (xk − a) ≤ ( |xk − a|)2 . 2 k=1

k=1

11 Dac˘ a x, y, z ≥ 0, atunci x(y + z − x)2 + y(z + x − y)2 + z(x + y − z)2 ≥ 3xyz. 12 Dac˘ a x, y, z sunt numere reale pozitive, atunci 1 1 9 1 + + ≥ . 2 2 2 (x + y) (y + z) (z + x) 4(xy + yz + zx) 13 (Cˆırtoaje, Ji Chen) Dac˘a a, b, c ¸si x, y, z sunt numere reale nenegative, atunci 2 2 2 9 + + ≥ . (a + b)(x + y) (b + c)(y + z) (c + a)(z + x) (b + c)x + (c + a)y + (a + b)z 14 Fie x, y, z numere pozitive astfel ˆıncˆat xyz = 1. Demonstrat¸i c˘a x3

x y z + 3 + 3 ≤ 1. +y+z y +z+x z +x+y

15 Dac˘ a a, b, c sunt numere reale pozitive astfel ˆıncˆat a + b + c = 3, atunci a b c 6( + + + 3 ≥ 7(a2 + b2 + c2 ). b c a 16 Dac˘ a a, b, c,d sunt numere reale nenegative astfel ˆıncˆat a + b + c + d = 4, atunci a2 bc + b2 cd + c2 da + d2 ab ≤ 4. 17 Dac˘ a a, b, c sunt numere reale nenegative, cel mult unul egal cu 0, atunci b2

1 1 1 12 + 2 + 2 ≥ . 2 2 2 − bc + c c − ca + a a − ab + b (a + b + c)2 521

18 Dac˘ a a, b, c sunt numere reale nenegative, cel mult unul egal cu 0, atunci ab − bc + ca bc − ca + ab ca − ab + bc 3 + + ≥ . 2 2 2 2 2 2 b +c c +a a +b 2 19 (Cˆırtoaje) Dac˘ a a ≥ b ≥ c > 0, atunci √ 3(a − c)2 4(a − c)2 3 ≤ a + b + c − 3 abc ≤ . 4(a + b + c) a+b+c 20 Fie a, b, c numere reale pozitive astfel ˆıncˆat abc = 1. Ar˘atat¸i c˘a 3+a 3+b 3+c + + ≥ 3. 2 2 (1 + a) (1 + b) (1 + c)2 21 Dac˘ a a, b, c, d sunt numere reale pozitive astfel ˆıncˆat a + b + c + d = 4, atunci 1 1 1 1 + + + ≥ a2 + b2 + c2 + d2 . ab bc cd da 22 Dac˘ a 0 < a ≤ b ≤ c ≤ 0, atunci a b c 2a 2b 2c + + ≥ + + . b c a b+c c+a a+b 23 Dac˘ a a, b, c, d 6=

1 3

sunt numere reale pozitive care satisfac relat¸ia abcd = 1,

atunci 1 1 1 1 + + + ≥ 1. (3a − 1)2 (3b − 1)2 (3c − 1)2 (3d − 1)2 24 Dac˘ a a, b, c sunt numere reale pozitive care satisfac relat¸ia ab+bc+ca = 3, atunci bc + 2 ca + 2 ab + 2 + + 2 ≥ 3. a2 + 2 b2 + 2 c +2 25 Fie a, b, c numere reale nenegative, dintre care cel mult unul egal cu 0. Demonstrat¸i c˘ a b(c + a) c(a + b) a(b + c) + + ≥ 2. b2 + bc + c2 c2 + ca + a2 a2 + ab + b2 26 Fie a, b, c, d, e numere reale astfel ˆıncˆat a + b + c + d + e = 0. Demonstrat¸i c˘ a a2 + b2 + c2 + d2 + e2 ≥ 3(ab + bc + cd + de + ea). 27 Dac˘ a a1 , a2 , ...an ∈ [1, 2] , atunci n X i=1

n

X 3 2 ≥ , ai + 2ai+1 ai + ai+1 i=1

unde an+1 = a1 .

522