My Inequality Project M.Ramchandran No.33,Belfast Apts,Ramachandran Street,T.Nagar E-mail address:
[email protected] URL: http://mathematicaldreams.wordpress.com/ Dedicated to Professor XY
Contents Preface
v
Chapter 1. AM-GM Inequality 1. Arithmetic Mean - Geometric Mean 2. Proof 3. Beginners’ Practice Problems 4. Geometric Interpretations 5. AM-GM Tautogrid Technique 6. Nessbit’s Inequality 7. The Reverse Technique 8. The Weighted AM-GM Inequality 9. Method Of Balancing Co-efficients by AM-GM 10. Quasiliearisation 11. Equivalent Summation Technique 12. The G function 13. Problem Set
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1 1 2 4 7 7 9 10 15 16 18 19 21 22
Preface This is the preface and it is created using a TeX field in a paragraph by itself containing \chapter*{Preface}. When the document is loaded, this appears if it were a normal chapter, but it is actually an unnumbered chapter.
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CHAPTER 1
AM-GM Inequality 1. Arithmetic Mean - Geometric Mean Lets go from scratch, Definition 1 (Arithmetic Mean). Arithmetic mean of two non-negative real numbers a and b is defined as the average of the two numbers and is mathematically expressed as a+b A.M. = 2 where ofcourse, A.M. stands for the arithmetic mean of the two concerned non-negative real numbers - a and b. Extending this idea to n-variables - a1 , a2 , a3 , .., an we get that Pn ai a1 + a2 + a3 + ... + an A.M. = = i=1 n n
Definition 2 (Geometric Mean). Geometric Mean of two real numbers is the collection of positive real numbers is the nth root of the product of the numbers. Note that if it is even, we take the positive nth root and it is mathematically expressed as √ G.M. = ab where, G.M. stands for Geometric Mean of the two concerned non-negative real numbers - a and b Extending this idea to n-variables - a1 , a2 , a3 , .., an we get that n Y 1 1 G.M. = (a1 a2 a3 ...an ) n = ( ai ) n i=1
Recall the fact that for any real number x , we know that x2 ≥ 0. √ √ therfore we know that for all non-negative real numbers - a and b √ √ ( a − b)2 ≥ 0 √ a + b − 2 ab ≥ 0 a+b √ ≥ ab 2 ring any bells? yes you have probably spotted that LHS of the inequality is the arithmetic mean we discussed and ofcourse the RHS is the geometric mean. Now this encourages the following proposition Theorem 1. Arithmetic Mean of some ’n’ non-negative real numbers is always greatern than or equal to the Geometric Mean of the same
1
2
1. AM-GM INEQUALITY
is that true? if so where is the validity? by seeing that it is true for certaina two positive reals doesnt imply a wider truth for any number of variables.
2. Proof Proof. Lets proceede by proving the above statement for smaller numbers and eventually the general caseSince we know that A.M. ≥ G.M. for two variables we have p+q √ ≥ pq 2 r+s √ ≥ rs 2 also we have √ √ pq + rs q√ 1 pqrs = (pqrs) 4 ≥ 2 combining we get that√ √ p+q + r+s pq + rs 1 p+q+r+s 2 = 2 ≥ ≥ (pqrs) 4 4 2 2 or,.. 1 p+q+r+s ≥ (pqrs) 4 4 which is AM-GM inequality for 4 variables!!! With a similar idea we can do the same as above for 8 - variables by splitting into 4-4 and using AM-GM for 4 variables. Now something should be irking in your mind.... Can’t we prove in the same way for any n?? - the answer is NO. Why not? - well this covers only 2-powers not even even integers.. so this proof is incomplete considering the general case. But a proof with the same idea can be given by induction for any 2-powers.It goes as follows Consider 2k+1 variables - a1 , a2 , a3 , ..., a2k+1 Assume the truth of the statement for 2k , we shall prove it for 2k+1 Since it is true for 2k , we have 1 a1 + a2 + a3 + ... + a2k ≥ (a1 a2 a3 ..a2k ) 2k −→ (1) k 2 and, 1 a2k +1 + a2k +2 + a2k +3 + ... + a2k+1 ≥ (a2k +1 a2k +2 a2k +3 ...a2k+1 ) 2k −→ (2) 2k again, 1
1
1 a1 + a2 + ... + a2k+1 (a1 a2 a3 ..a2k ) 2k + (a2k +1 a2k +2 a2k +3 ...a2k+1 ) 2k (1) + (2) = ≥ ≥ (a1 a2 ...a2k+1 ) 2k+1 k+1 2 2 2 or, 1 a1 + a2 + a3 + ... + a2k+1 ≥ (a1 a2 a3 ..a2k+1 ) 2k+1 2k+1 thus by induction it is proved for all powers of 2!!! this is a bit closer to the full proof.. But then what about 6, 10.. and 3, 5, 7... how do we prove them..? with a little bit of trickery we get away with 3 - since the inequality is true for 4 we have 1
4 1 1 a + b + c + (abc) 3 ≥ (abc) 3 · 4 = (abc) 3 4
and this re-arranges to 1 a+b+c ≥ (abc) 3 3
2. PROOF
3
which is the AM-GM inequality for 3 reals!!!! this can also be done by the following way a + b + c + ( a+b+c ) a+b+c 1 3 ≥ [abc · ( ]4 4 3 that is..
a+b+c a+b+c 1 ≥ [abc · ( ]4 3 3
which re-arranges to 1 a+b+c ≥ (abc) 3 3 The above two proofs crucially modifies our way of looking at AM- GM inequality. Let us re-define ourselves the meaning of AM,GM into A.M. = sum equaliser G.M. = product equaliser what does that mean ? It means that say A is the A.M. of 3 variables x, y, z and G the G.M. then -
x + y + z = 3A and, x.y.z = G3 If we know that the inequality is true for a certain m , then for any n such that m > n we can prove the validity of AM-GM inequality as followsm−n 1 a1 + a2 + ... + an + (m − n)A ≥ (a1 a2 a3 ..an ) m · A m m m−n n mA ≥ (G) m · A m m Am ≥ Gn · Am−n
or, An ≥ Gn =⇒ A ≥ G thus we have prove the AM-GM inequality for any ’n’ as we know it to be true for any 2k .. or A.M. ≥ G.M. for any n This is trivial looking inequality is probably the most celebrated of all that we all shall discuss in this book.It has far and wide applications.. Proof. The inequality is true for 2n if it is true for n or it is true for all powers of 2 (already proved) Suppose that the inequality is true for n numbers.We then choose s an = n−1 where, s = a1 + a2 + a3 + .... + an According to the inductive hypothesis, we get a1 a2 a2 a3 ...an · s 1 s s+ ≥ n( )n n−1 n−1 1
⇐⇒ s ≥ (n − 1)(a1 a2 a3 ..an ) n−1 Therfore if the inequality is true for n numbers than it will be true for n − 1 numbers and by induction (Cauchy Induction),the inequality is true for every natural number n .Equality occurs if and only if a1 = a2 = a3 = ... = an ∇
4
1. AM-GM INEQUALITY
3. Beginners’ Practice Problems 1. For a, b, c ∈ R+ 0 .Prove that (a + b)(b + c)(c + a) ≥ 8abc Solution Note that by AM-GM,
√ a + b ≥ 2 ab √ b + c ≥ 2 bc √ c + a ≥ 2 ca
multiplying the above inequalities we get the desired with equality for a = b = c ∇ 2.For a, b, c ∈ R+ 0 . Prove that a2 + b2 + c2 ≥ ab + bc + ca Solution a2 + b2 ≥ 2ab b2 + c2 ≥ 2bc c2 + a2 ≥ 2ac adding this we get the desired with equality for a = b = c ∇ 3. For a, b, c ∈ R+ such that - a + b + c = 2.Prove that abc ≥ 8(1 − a)(1 − b)(1 − c) Solution Set 1 − a = x, 1 − b = y, 1 − c = z therfore by condition, c = x + y, b = x + z, a = y + z substituting this in the inequality we get that it is equivalent to (x + y)(y + z)(z + x) ≥ 8xyz which we just proved! with equality for a = b = c =
2 3
∇ 4. For a, b, c ∈ R+ .Prove that (a2 b + b2 c + c2 a)(a2 c + b2 a + c2 b) ≥ 9a2 b2 c2 Solution Observe that by AM-GMa2 b + b2 c + c2 a ≥ 3abc
3. BEGINNERS’ PRACTICE PROBLEMS
5
and, a2 c + b2 a + c2 b ≥ 3abc multiply the above two inequalities to get the desired with equality for a = b = c ∇ 5.For a, b, c ∈ R+ 0 .Prove that -
√ √ √ a4 + b4 + c4 ≥ abc( ab + bc + ca)
(Source:Nagasaki University 1970) Solution This problem seems a step tougher to novice. Expanding the RHS will not lead in the correct direction. Let us try to transform this inequality into an equivalent one that for convenient sake looks simpler. One way of this being done is by taking abc to the LHS √ √ √ a4 + b4 + c4 ≥ abc( ab + bc + ca)
⇐⇒
√ √ √ a3 b3 c3 + + ≥ ab + bc + ca bc ac ab a3 a3 b3 c3 + + + ≥ 4a bc bc ca ab b3 c3 a3 b3 + + + ≥ 4b ac ac ba cb c3 a3 b3 c3 + + + ≥ 4c ab ba bc ca
the three inequalities are true by AM-GM, add these inequality to get √ √ √ a3 b3 c3 + + ≥ a + b + c ≥ ab + bc + ca bc ca ab notice that the last inequality is true and is equivalent to Problem 2 thus we have proved the inequality with equality for a = b = c (by M.Ramchandran) aliter: Alternatively another ingenious AM-GM solution will be notice by AM-GM that √ 3a4 + 3b4 + 2c4 ≥ abc ab 8 √ 3b4 + 3c4 + 2a4 ≥ abc bc 8 √ 3c4 + 3a4 + 2b4 ≥ abc ca 8 adding these inequalities we get the desired. (by Mathias Tejs) ∇
6
1. AM-GM INEQUALITY
Note It is not expected of the reader to get the above two proofs (if he/she is a newbie). Such proofs come due to some strong observation and several wrong tries(like mine.. :P). Both involved splitting the terms into terms with suitable co-efficients which shall come as time goes. So dont get disheartened or awed. 6. For a, b, c ∈ R+ 0 such that a + b + c + d = 1.Prove that 1 4 Solution Write the Inequality using the given condition as ab + bc + cd ≤
ab + bc + cd ≤
(a + b + c + d)2 4
By AM-GM, (a + b + c + d)2 (a + c) + (b + d) 2 =( ) ≥ (a + c)(b + d) = ab + bc + cd + da ≥ ab + bc + cd 4 2 with the equality for a = b = c, d = 0 or d = c = b, a = 0 (by M.Ramchandran) ∇ 7. For a, b, c ∈ R+ 0 .Prove that b2 + c2 c2 + a2 a2 + b2 + + ≥a+b+c a+b b+c c+a Solution Note that by AM-GM, 2(x2 + y 2 ) ≥ (x + y)2 for all non-negative reals x, y thus, a2 + b2 b2 + c2 c2 + a2 a+b b+c c+a + + ≥ + + =a+b+c a+b b+c c+a 2 2 2 with equality for a = b = c ∇ 8. For a, b, c ∈ R+ such that a + b + c + d = 1.Prove that 0 a2 b2 c2 d2 1 + + + ≥ a+b b+c c+d d+a 2 Solution We have (a − b) + (b − c) + (c − d) + (d − a) = 0 ⇐⇒ ⇐⇒
b2 − c 2 c2 − d2 d2 − a 2 a 2 − b2 + + + =0 a+b b+c c+d d+a
b2 c2 d2 b2 c2 d2 a2 a2 + + + = + + + a+b b+c c+d d+a a+b b+c c+d d+a
thus multiplying the original inequality to be proven by 2 we get that a2 + b2 b2 + c2 c2 + d2 d2 + a2 + + + ≥1=a+b+c+d a+b b+c c+d d+a which can be proved by proceeding similar to the previous question. (by R.Keerthan) ∇ Thus we close this section in the notion that the reader has atleast become familiar with the concepts that have been explained.
5. AM-GM TAUTOGRID TECHNIQUE
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4. Geometric Interpretations 5. AM-GM Tautogrid Technique While dealing with inequalities sometimes we might be having the denominators as the trouble terms and applying AM-GM in a tricky way we suprisingly are able to solve some rather difficult-looking inequalities. I shall demonstrate the method through the following inequalities. It certainly isnt rocket-science just some common sense which probably the reader might have arrived at with some 5 minutes of genuine thinking. 1. For a, b, c ∈ R+ .Prove that -
a2 b2 c2 + + ≥a+b+c b c a
Solution Note that by AM-GM, a2 + b ≥ 2a b b2 + c ≥ 2b c c2 + a ≥ 2c a adding the above inequalities we get the desired with equality for a = b = c ∇ 2.For a, b, c ∈ R+ .Prove that 0 √ √ √ a4 + b4 + c4 ≥ abc( ab + bc + ca) (Source:Nagasaki University 1970) Solution Proceed as in my solution until you get √ √ √ b3 c3 a3 + + ≥ a + b + c ≥ ab + bc + ca bc ca ab The last but one inequality can also be proved in a simple way as follows a3 + b + c ≥ 3a bc b3 + c + a ≥ 3b ca c3 + a + b ≥ 3c ab Note: What is the trick? - the idea is the remove the trouble terms and here they are present in the denominator so we do it by adding suuitably. 3. For a, b, c ∈ R+ 0 such that a + b + c + d = 1.Prove that b2 c2 d2 1 a2 + + + ≥ a+b b+c c+d d+a 2 Solution BY AM-GM, a2 a+b + ≥a a+b 4 2 b b+c + ≥b b+c 4 c2 c+d + ≥c c+d 4 d+a d2 + ≥d d+a 4
8
1. AM-GM INEQUALITY
adding these we get the desired. with equality for a = b = c = d =
1 4
(by M.Ramchandran) ∇ 4. For a, b, c ∈ R+ such that ab + bc + cd + da = 1.Prove that0 3 X 1 a ≥ b+c+d 3 cyclic
Solution By AM-GM, a3 b+c+d b3 a+c+d c3 b+a+d d3 b+c+a
b+c+d 1 + ≥ 18 12 a+c+d 1 + + ≥ 18 12 b+a+d 1 + + ≥ 18 12 b+c+a 1 + + ≥ 18 12
+
a 2 b 2 c 2 d 2
Add these inequalities to get LHS ≥
a+b+c+d−1 3
and also frolm AM-GM, (a + b + c + d)2 ≥ 4(a + c)(b + d) = 4 or a+b+c+d≥2 and the conclusion follows. With equallity for a = b = c = d = 41 . (by mathlinks user : quykhtn-qa1) ∇ 5. For a, b, c ∈ R+ such that a + b + c = 2.Prove thatb c a + + >2 b(a + b) c(b + c) a(a + c) (Source:Own Inequality) Solution By AM-GM,
a + (a + b)a + ab ≥ 3a b(a + b) b + (c + b)b + cb ≥ 3b c(c + b) c + (a + c)c + ac ≥ 3c a(a + c)
or , X cyclic
a ≥ 3(a + b + c) − (a + b + c)2 = 6 − 4 = 2 b(a + b)
but the equality case cant occur so the inequality sign becomes strict. ∇ 6. For x, y, z ∈ R+ such that xyz = 1 .Prove that x3 y3 z3 3 + + ≥ (1 + y)(1 + z) (1 + x)(1 + z) (1 + x)(1 + y) 4 (Source: IMO Shortlist 1998) Solution By AM-GM x3 1+y 1+z 3x + + ≥ (1 + y)(1 + z) 8 8 4
=⇒
X cyclic
6. NESSBIT’S INEQUALITY 3 X
x 1 ≥ (1 + y)(1 + z) 4
(2x − 1) ≥
cyclic
3 4
9
Equality for x = y = z = 1 ∇ ˙ 7. For a, b ∈ R+ 0 such that a + b = 1 Prove that a2 b2 + ≥ 13 1+b 1+a Solution By AM-GM, a2 b+1 2a + ≥ b+1 9 3 b2 a+1 2b + ≥ a+1 9 3 1 add these to get the result with equality for a = b = 2 ∇ 6. Nessbit’s Inequality This a very famous,well-known and well discussed inequality. Most problems are probably stronger than this (that is the job of the proposers - if they keep the questions down to elementary inequality then what is the fun?) but nevertheless it is a must to know this beautiful inequalityTheorem 2. For a, b, c ∈ R+ 0 the following inequality holds b c 3 a + + ≥ b+c c+a a+b 2 Proof Note that for any positive real x, y, z we have by AM-GM x y z + + ≥3 y z x Consider the following expressions a b c S= + + b+c c+a a+b b c a M= + + b+c c+a a+b c a b N= + + b+c c+a a+b we have ofcourse: M + N = 3. According to the Lemma, b+c c+a a+b + + ≥3 M +S = b+c c+a a+b a+c b+a c+b N +S = + + ≥3 b+c c+a a+b Therefore, M + N + 2S ≥ 3 or 2S ≥ 3 or, 3 S≥ 2 Practice
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1. AM-GM INEQUALITY
1. Extend the same idea to prove Nessbit’s inequality for 4 non-negative reals a b c d + + + ≥4 b+c c+d d+a a+b 2. For a, b, c ∈ R+ 0 , Prove that -
a2 + 1 b2 + 1 c2 + 1 + + ≥3 b+c c+a a+b (Source : Indian RMO 2006) 3. Prove Nessbits inequality using the Tautogrid Technique (Left to readers) 7. The Reverse Technique In most inequalities AM-GM can be applied after if for a certain inequality we are to prove - x ≥ y, it ultimately comes down to the fact that - a2 ≥ 0 where a2 = x − y so factorizations can help. But for factorizations, one certainly has to be something short of God to be able to prove all inequalities by factorizations into squares but.... algebraic manipulation is certainly a powerful tool. In certain inequalities by a tricky manipulation we oberve that solutions are obtained are more easily. In general it should be understood that ,for a positive real number which is rational , the value increases as we decrease the denominator and the value decreases as we increase the denominator and the converse for a negative rational. As always i shall explain this with an example For a, b, c ∈ R+ 0 , Prove that b3 c3 a+b+c a3 + 2 + 2 ≥ 2 2 +b b +c a + c2 2 2 2 Solution Seeing the terms a + b , ... , instictively a student does the following b3 c3 a3 b3 c3 1 a2 b2 c2 1 a3 + + ≥ + + = + + ≥ (a + b + c) a2 + b2 b2 + c2 a2 + c2 2ab 2bc 2ca 2 b c a 2 a2
Ok, now is that correct? NO .. ok i guess this was anticipated by you.. but then what was the flaw? read the last lines of the paragraph and the fact that you conceived as elementary has been flawed on by the student. Now an inquisitive student wouldnt move on to a certain different try , He would try to correct his flaw to make his idea right. The begining of that important last line said -”Positive” so that is the cause. This means for applying AM-GM for those terms we need to have the fraction negative, so why not express it as some X − a2Y+b2 ? Yes that is the central idea behind this useful technique.Now the solution will be X X X X a3 ab2 ab2 b a+b+c = a− 2 ≥ a− = a− = 2 2 2 a +b a +b 2ab 2 2 cyclic
cyclic
cyclic
cyclic
and Voila! we have a solution and it is correct and it uses AM-GM inequality. ∇ Here the main idea is l t−1 at a k · ab − = kat−1 + lbt−1 k kat−1 + lbt−1 This is only a random form. To justify my statement that we can do several types of problems and to get you used to this technique, The strength and importance of this technique cant be more revealed than the following problems. Problems 1. Let a, b, c ∈ R+ , then prove that we have a3 b3 c3 a+b+c + + ≥ a2 + ab + b2 b2 + bc + c2 c2 + ca + a2 3
7. THE REVERSE TECHNIQUE
11
Solution b3 c3 a3 + + a2 + ab + b2 b2 + bc + c2 c2 + ca + a2 X X ab(a + b) ab(a + b) a+b+c )≥ (a − = (a − 2 )= 2 a + ab + b 3ab 3 LHS =
cyclic
cyclic
with equaliy for a = b = c (by M.Ramchandran) ∇ The application of this technique doesnt come by just seeing the solutions given by the Author.. for really *learning* it, the reader is advised to try the following problems before succumbing to seeing the solutions. 2. Let a, b, c ∈ R+ such that , a + b + c = 3 . Prove that b c 3 a + + ≥ 2 2 2 1+b 1+c 1+a 2 Solution We have; a ab2 =a− 2 1+b 1 + b2 . Summing up cyclically we get that X a X ab2 X ab2 = a + b + c − ≥ a + b + c − 1 + b2 1 + b2 2b cyc cyc cyc 3 3 1 = 3 − (ab + bc + ca) ≥ 3 − = 2 2 2 since from trivial inequality we have that (ab + bc + ca) ≤
1 (a + b + c)2 = 3 3
. Therefore we are done. ∇ Can we extend this problem to four variables? The answer is yes 3. For a, b, c, d ∈ R+ such that a + b + c + d = 4 .Show that we have a b c d + + + ≥2 1 + b2 1 + c2 1 + d2 1 + a2 Solution In the same manner as the previous problem, we have that a ab ≥a− 1 + b2 2 Summing up we haveX a 1 1 ≥ a + b + c + d − (ab + bc + cd + ad) = 4 − (a + c)(b + d) 2 1 + b 2 2 cyc ≥4−
1 · 2
a+b+c+d 2
Equality holds for a = b = c = d = 1 ∇
2 =4−
1 ·4=2 2
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1. AM-GM INEQUALITY
4. Let a, b, c ∈ R+ satisfy a + b + c = 3. Prove that a2 b2 c2 + + ≥ a + 2b2 b + 2c2 c + 2a2 (by Pham Kim Hung) Solution Obviously 2ab2 2ab2 2 2 2 a2 =a− ≥a− √ = a − · a3 b3 3 2 2 a + 2b a + 2b 3 3 a · b4 So we have X cyc
2 2 2 2 a2 2 2 2 3 b3 + b3 c3 + c3 a3 ≥ 3 − a a + 2b2 3
Hence it suffices to show that
2
2
2
2
2
2
a3 b3 + b3 c3 + c3 a3 ≤ 3 But, 2
2
2
1 [ab + ab + 1 + bc + bc + 1 + ca + ca + 1] 3 2 2 = 1 + (ab + bc + ca) ≤ 1 + · 3 = 1 + 2 = 3 3 3 2
2
2
a3 b3 + b3 c3 + c3 a3 ≤
because (a + b + c)2 ≥ 3(ab + bc + ca) ⇒ ab + bc + ca ≤ 3 . Hence we finish our proof here. Note that this is also true for ab + bc + ca = 3. However, here I pose a challenge for the readers. ∇ 5. For a, b, c ∈ R+ such that a + b + c = 3,Prove that a2 b2 c2 + + ≥1 a + 2b3 b + 2c3 c + 2a3 Solution Obviously
Hence we have
a2 2ab3 ab3 =a− ≥a−2· √ 3 3 3 a + 2b a + 2b 3 a · b2 √ 2 3 =a− b a2 3 2 X a √ √ 2 √ 3 2 + c 3 b2 + a 3 c2 a ≥ a + b + c − b a + 2b3 3 cyc
2 [b(a + a + 1) + c(b + b + 1) + a(c + c + 1)] 9 2 2 = 3 − {2(ab + bc + ca) + 3} ≥ 3 − {6 + 3} = 1 9 9 The last inequality is true by AM-GM, and since we have 1 ab + bc + ca ≤ (a + b + c)2 = 3. 3 ≥a+b+c−
Equality occurs if and only if a = b = c = 1. ∇ 6. For a, b, c ∈ R+ such that a + b + c = 3, Prove that b+1 c+a a+1 + + ≥3 b2 + 1 c2 + 1 a2 + 1
7. THE REVERSE TECHNIQUE
13
Solution Since a+1 b2 (a + 1) b2 (a + 1) ab b = a + 1 − ≥ a + 1 − =a+1− − 2 2 b +1 b +1 2b 2 2 . Therefore we have X a+1 a+b+c 1 ≥3+ − (ab + bc + ca) 2+1 b 2 2 cyc ≥3+
3 3 − =3 2 2
Equality for a = b = c = 1 Hence proved. ∇ 7. For a, b, c ∈ R+ such that a2 + b2 + c2 = 3, prove that 1 1 1 + + ≥1 a3 + 2 b3 + 2 c3 + 2 (Pham Kim Hung) Solution Note that X cyc
≥
X1 1 a3 = 1− 3 a3 + 2 2 a +2 cyc 3 1 X a3 3 1 − · = − =1 2 2 cyc 3a 2 2 ∇
8. For a, b, c > 0. Prove that a3 b3 c3 1 2 + + ≥ a + b2 + c2 a+b b+c c+a 2 Solution X a3 X a2 b = a2 − a+b a+b X a2 b X 1 a2 b = a2 + b2 + c2 − ≥ a2 + b2 + c2 − ·√ a+b 2 ab X1 √ Xa = a2 + b2 + c2 − (a ab) ≥ a2 + b2 + c2 − (a + b) 2 4 1 1 = a2 + b2 + c2 − (a2 + b2 + c2 + ab + bc + ca) ≥ a2 + b2 + c2 − (a2 + b2 + c2 ) 4 2 1 2 2 2 = (a + b + c ) 2 Therefore we are done. Equality occurs if and only if a = b = c. 9. Let a, b, c ∈ R+ that sum up to 3. Prove that we always have 1 1 1 + + ≥1 2 2 1 + 2b c 1 + 2c a 1 + 2a2 b Solution Note that 1 2b2 c 2 b2 c √ = 1 − ≥ 1 − 1 + 2b2 c 1 + 2b2 c 3 3 b4 c2 2√ 2 3 =1− b2 c ≥ 1 − (2b + c) 3 9
14
1. AM-GM INEQUALITY
Therefore we have that X cyc
" # 1 2 X ≥3− (2b + c) 1 + 2b2 c 9 cyc
2 · 3(a + b + c) = 3 − 2 = 1 9 Therefore we are done. Equality occurs if and only if a = b = c = 1. =3−
∇ Practice Problems 10. For a, b, c, d ∈ R+ such that a + b + c + d = 1.Prove that b2 c2 d2 1 a2 + + + ≥ a+b b+c c+d d+a 2 11. Given a, b, c, d ∈ R+ , show that we have a3
b4 c4 d4 a+b+c+d a4 + 3 + 3 + 3 ≥ 3 3 3 3 + 2b b + 2c c + 2d d + 2a 3
12. For a, b, c ∈ R+ such that a + b + c = 3; show that we have ab bc ca 3 + + ≤ b3 + 1 c3 + 1 a3 + 1 2
(by Gibbenergy) 13. For given four positives a, b, c, d with sum 4; show that a b c d + + + ≥1 2 2 2 1 + b c 1 + c a 1 + d a 1 + a2 b (by Pham Kim Hung) 15.Let a, b, c, d > 0 satisfy a + b + c + d = 4; show that 1 + ab 1 + bc 1 + cd 1 + ad + + + ≥4 2 2 2 2 2 2 1+b c 1+c d 1+d a 1 + a2 b2
(by Pham Kim Hung) 16. For all a, b, c, d ∈ R+ satisfying a + b + c + d = 4, Prove that we have a+1 b+1 c+1 d+1 + + + ≥4 b2 + 1 c2 + 1 d2 + 1 a2 + 1
17. For a, b, c, d ∈ R+ with sum 4. Prove that 1 1 1 1 + + + ≥2 a2 + 1 b2 + 1 c2 + 1 d2 + 1 ∇
8. THE WEIGHTED AM-GM INEQUALITY
15
8. The Weighted AM-GM Inequality Theorem 3. Suppose that a1 , a2 , a3 , ...., an are positive real numbers.If n nonegative real numbers x1 , x2 , x3 .., xn have sum 1 then, x1 a1 + x2 a2 + .... + xn an ≥ ax1 1 ax2 2 ...axnn In a way similar to the second proof provided for AM-GM. We have to prove that if x, y ≥ 0, x + y = 1 and a, b > 0 ax + by ≥ ax by m Consider rational numbers x, y then take a limit. Certainly if x, y, are rational numbers then - x = m+n ,y = n , m, n ∈ N, the problem is true according to AM-GM inequality m+n m
n
ma + nb ≥ (m + n)a m+n b m+n =⇒ ax + by ≥ ax by If x, y are real numbers,there exist two sequences of rational numbers (rn )n≥0 and (sn )n≥0 for which rn → x, sn → y, rn + sn = 1.Certainly arn + bsn ≥ arn bsn or arn + b(1 − rn ) ≥ arn b1−rn Taking the limit when n → ∞, we have ax + by ≥ ax by ∇ Problems 1. Let a, b, c be the sidelengts of a triangle.Prove that (a + b − c)a (b + c − a)b (c + a − b)c ≤ aa bb cc Solution Applying the weighted AM-GM inequality, we conclude that 1 " a b c # a+b+c a+b−c b+c−a c+b−a · · a b c 1 a+b−c b+c−a c+a−b ≤ a· +b· +c· =1 a+b+c a b c Or equivalenty, (a + b − c)a (b + c − a)b (c + a − b)c ≤ aa bb cc Equality occurs for a = b = c ∇ 2. Let a, b, c ∈ R+ such that abc = 1.Prove that ab+c · bc+a · ca+b ≤ 1 (Source : INMO 2001) Solution 1 Write the LHS as - (ab)c (bc)a (ca)b .So we have to prove that (ab)c (bc)a (ca)b ≤ 1 By Weighted AM-GM inequality we have 1 (ab)c (bc)a (ca)b a+b+c ≤
1 3 (a.bc + b.ca + c.ab) = ≤1 a+b+c a+b+c (By Ramchandran)
Solution 2 WLOG:c ≥ b ≥ a abc = 1, so c ≥ 1,so ab ≤ 1 and a ≤ 1 Now: ab+c · bc+a · ca+b = (abc)a + bac−a bc−b = ac−a bc−b = (ab)c−b ab−a
16
1. AM-GM INEQUALITY c−b b−a
≤1
1
=1 Therefore ab+c · bc+a · ca+b ≤ 1 (by mathlinks user: rem) Solution 3 Note that 1 ≤1 aa bb cc the last inequality is true because for a, b, c individually greater or lesser than 1 we have ab+c · bc+a · ca+b =
aa ≥ a, bb ≥ b, cc ≥ c (by mathlinks user:Maharjun) ∇ 3. Let a, b, c ∈ R+ 0 , Prove that 1
1
4(a + b + c) ≥ 3(a + (ab) 2 + (abc) 3 ) Solution √ √ √ a a a 3 3a + ( + b) + ( + 2b) + ( + b + 4c) ≥ 3a + ab + 2 ab + 3 abc 4 2 4 (by Aravind Srinivas) 9. Method Of Balancing Co-efficients by AM-GM In most inequalities we have to group terms suitably so that classical inequalities like AM-GM , C-S etc.. can be aplied to get the result.This is not as easy as it looks, it requires proper terms and proper grouping. We usually need some additional variables to solve the equations for finding out the original variables.This is the Method of Balancing co-efficients. In this chapter we shall see the method using AM-GM Inequality. Let me demonstrate it with a simple example 1. If x, y, z ∈ R+ such that xy + yz + zx = 1 ,then find the minimum of the following expressionk(x2 + y 2 ) + z 2 Solution Lets experiment with some values of k shall we? Let k = 10, so we are now required to find the minimum of this non-symmetric expression10(x2 + y 2 ) + z 2 How do we apply a classical basic inequality like AM-GM for this? Well it does seem horrendously difficult, so lets take a sneak-peek at the magical solution? By AM-GM we have the following inequalities, 2x2 + 2y 2 ≥ 4xy 1 8x2 + z 2 ≥ 4yz 2 1 8y 2 + z 2 ≥ 4zx 2 and summing up these we have, 10(x2 + y 2 ) + z 2 ≥ 4(xy + yz + zx) = 4
9. METHOD OF BALANCING CO-EFFICIENTS BY AM-GM
17
Equality holds for x=y 4x = z 4y = z 1 =⇒ x = y = 3 4 z= 3 Hurray ! We did it!!(don’t get envious i like you was flabbergasted at this cause it aint mine) how can a guy arrive at this?? why not pair up 1 and 9 , 5 and 5 or something? All answers shall be revealed in the following lines. back to our general problem, Lets choose some l ≤ k , then apply AM-GM this way, lx2 + ly 2 ≥ 2lxy p 1 (k − l)y 2 + z 2 ≥ yx 2(k − l) 2 p 1 2 (k − l)x + z 2 ≥ xz 2(k − l) 2 summing up we get this p k(x2 + y 2 ) + z 2 ≥ 2lxy + (yz + zx) 2(k−) now we have a condition given - xy + yz + zx = 1 so for obtaining a numerical value we have to have the co-efficients in the RHS the same and without a variable hopefully. so intuitively lets just equate themp 2l = 2(k − l) and solving this we obtain that √ −1 + 1 + 8k l= 4 and ofcourse substituting this in the equation we get the minimal value we are looking for to be √ −1 + 1 + 8k 2 so we observe that there is a unique pair of integers - l, k − l that show us the way by AM-GM. Now that is the reason we were seeing the use of 8,2 and not 3,7 etc.. sure enough you can check the credibility of the pairings now! Note ∇ 2. Let x, y, z, t be real numbers satisfying xy + yz + zt + tx = 1. Find the minimum of the expression 5x2 + 4y 2 + 5z 2 + t2 Solution Here k = 5, so we chooose l ≤ 5,
√ lx2 + 2y 2 ≥ 2 2lxy √ 2y 2 + lz 2 ≥ 2 2lyz p 1 (5 − l)z 2 + t2 ≥ 2(5 − l)tx 2 Summing up these results, We conclude that p √ 5x2 + 4y 2 + 5z 2 + t2 ≥ 2 2l(xy + tz) + 2(5 − l)(zt + tx) The condition xy + yz + zt + tx = 1 suggests us to choose a number l(0 ≤ l ≤ 5) such that p √ 2 2l = 2(5 − l) √ . A simple calculation yields l = 1, thus the minimum of 5x2 + 4y 2 + 5z 2 is 2 2
18
1. AM-GM INEQUALITY
Note Using the same method, solve the following problem : Let x, y, z, t be arbirary real numbers. Prove that 1 2kl 2 2 2 2 2 · (xy + yz + zt + tx) x + ky + z + lt ≥ k+l 3. Let x, y, z be positive real numbers with sum 3. Find the minimum of the expression x2 + y 2 + z 3 (Pham Kim Hung) Solution Let a and b be teo positive real numbers.. Then,by AM-GM inequality we have x2 + a2 ≥ 2ax y 2 + a2 ≥ 2ay z 3 + b2 + b2 ≥ 3zb2 Combining these we have x2 + y 2 + z 3 + 2(a2 + b3 ) ≥ 2a(x + y) + 3b2 z with equality for x = y = a and z = b.In this case, we could have 2a + b = x + y + z = 3(?). Moreover, in order for 2a(x + y) + 3b2 z to be represented as x + y + z, we must have 2a = 3b2 (??).Accordin to (?) and (??) we can find out that, √ √ −1 + 37 3−b 19 − 37 b= ,a = = 6 2 12 Therfore the minimum of x2 + y 2 + z 3 is 6a − 2(a2 + b3 ) where a, b are as determined.The proof is completed. ∇ 4. For x, y, z ∈ R+ such that xy + yz + zx = 1 , Prove that - 15x2 + 7y 2 + 3z 2 ≥ 6 (Own Inequality) Solution Rewrite the Inequality as 7 5x2 + y 2 + z 2 ≥ 2 3 By AM-GM we have , x2 + y 2 ≥ 2xy 1 4x2 + z 2 ≥ 2xz 4 4 2 3 2 y + z ≥ 2xz 3 4 now adding these we get the desired. 10. Quasiliearisation This is a very intrigueing idea due to Russian problem proposer - Fedor Petrov We know by AM-GM that , 2ab ≤ a2 + b2 Introduce a parameter and get the following: b2 t Then read the last inequality from the other point: for any positive a, b there exist positive t such that 2ab ≤ ta2 +
2ab = ta2 +
b b2 (t = ) t a
Or, we may write: 2ab = min(ta2 +
b2 ) t
11. EQUIVALENT SUMMATION TECHNIQUE
19
How may this observation help? Put a=
X X (x2i ), b = (yi2 )
Then for appropriate t we have: X √ √ b X 2 yi2 2 a b = ta + = (txi + ) ≥ (2xi yi ) t t So, we get the Cauchy-Schwarz Inequality. A lot of other inequalities also may be proved by this idea Problem Prove that for any four nonnegative reals a, b, c, d the following inequality holds1
1
1
(ab) 3 + (cd) 3 ≤ ((a + c + b)(a + c + d)) 3
(Source:Proposed at 239 Lyceum Traditional Olympiad) (Author : Fedor Petrov) We have
1 xy And for any positive A and B there exist appropriate x and y ,for which equality holds 1
3(AB) 3 ≤ Ax + By + 1
x=
1
(AB) 3 (AB) 3 ,y = A B
Let A = (a + c + b), B = (a + c + d) in terms of the problem. For some positive x, y we have 1 1 1 3(AB) 3 = Ax + By + = (a + c + b)x + (a + c + d)y + = xy xy 1 1 1 1 1 1 + = a(x+y)+bx+ + c(x+y)+dy+ ≥ 3(ab) 3 +3(cd) 3 (a+c+b)x+(a+c+d)y+ x(x + y) y(x + y) x(x + y) y(x + y) By AM-GM and we are done! 11. Equivalent Summation Technique This is an interesting technique that helps us to solve elegantly many problems. It involves finding a suitable equivalent summation and using it prove the inequality given. Let me demonstrate it through the following already discussed example 1. Prove that for a, b, c ∈ R+ a3 b3 c3 a+b+c + + ≥ a2 + ab + b2 b2 + bc + c2 c2 + ca + a2 3 Solution 1 X X a3 − b3 a3 b3 1 X a 3 + b3 a+b+c = a − b =⇒ = = ≥ 2 2 2 2 2 2 2 2 a + ab + b a + ab + b a + ab + b 2 cyc a + ab + b 3 cyc cyc But there exists another nice solution using the reverse technique. The solution runs as follows: X X X ab(a + b) a3 ab(a + b) = a− 2 ≥a+b+c− 2 2 2 a + ab + b a + ab + b 3ab cyc cyc cyc Since we have a2 + ab + b2 ≥ 3ab from AM-GM. Therefore we have that X a3 2 a+b+c ≥ a + b + c − (a + b + c) = 2 2 a + ab + b 3 3 cyc Hence proved. Equality occurs if and only if a = b = c.
20
1. AM-GM INEQUALITY
Solution 2 By the same idea, LHS =
1 X a3 + b3 2 a2 + b2 + ab
. But ab ≤
a2 + b2 2
and 2(a3 + b3 ) ≥ (a + b)(a2 + b2 ) Therefore LHS ≥
a+b+c 3 ∇
2. For a, b, c ∈ R+ with sum 1,prove that √
ab bc ca 1 +√ +√ ≤√ ab + bc bc + ca ca + ab 2 (MOSP 2007 3.2)
Solution √
bc ca 1 ab +√ +√ ≤√ ab + bc bc + ca ca + ab 2 X
√ X 2ab ab √ √ √ ≤ ab + bc ab + bc
just need to prove X
√
ab a+b+c √ ≤ 2 ab + bc
note that X
√
X ab bc √ = √ √ ab + bc ab + bc
this is true because : X
√
√ √ X (ab − bc)( ab − bc) X √ √ ab − bc √ = = ( ab − bc) = 0 ab − bc ab + bc
so equivalent to X
√
ab + bc √ ≤a+b+c ab + bc
and this is equivalent to √
X
√ √ ab √ ( a − b)2 ≥ 0 √ √ √ ( c + a)( c + b)
and hence proved. (by mathlinks user : kuing) ∇
12. THE G FUNCTION
21
12. The G function This is a beautiful idea for which credit goes to inequality solver - Pham Kim Hung (hungkhtn). Definition 3. For a, b, c ∈ R+ , we define, G(a, b, c) =
a b c + + −3 b c a
It is trivial to oberve by AM-GM that always G(a, b, c) ≥ 0 Some nice properties have been found and some tough inequalities have been solved by this idea. Pham Kim Hung’s Nice Factorisation
G(a, b, c) =
a b c + + −3= b c a
a b b c b (a − b)2 (b − c)(a − c) + −2 + + − −1 = + b a c a a ab ca
this factorization plays an important role in many proofs Note The inequality - G(a, b, c) ≥ 0 is a cyclic inequality and thus no pair-wise order can be assumed. I shall present some properties of this important function. Properties 1. For a, b, c, k ∈ R+ we have G(a, b, c) ≥ G(a + k, b + k, c + k) Proof WLOG : c = min{a, b, c} We have , G(a, b, c) =
(a − b)2 (b − c)(a − c) + ab ca
So it is enough to prove that (a − b)2 (b − c)(a − c) (a − b)2 (b − c)(a − c) + ≥ + ab ca (a + k)(b + k) (c + k)(a + k) This is true as k > 0 and by assumption - (a − c)(b − c) ≥ 0. Hence proved with equality for a = b = c ∇ 2. For a, b, c, k ∈ R+ we have G(a, b, c) ≥ G(a + b, b + c, c + a) Proof We only have to show that (a − b)2 (b − c)(a − c) (a − b)2 (b − c)(a − c) + ≥ + ab ca (a + c)(b + c) (a + b)(a + c) It is trivial to see that the inequality is true. Note The following problem (equivalent to the property discusses above) was asked in the Mathlinks contest 2003 a b c a+b b+c c+a + + ≥ + + b c a b+c c+a a+b ∇ The following properties(same conditions as above)can be solved using the same method 3. For a, b, c, k ∈ R+ we have -
G (a − b)2 , (b − c)2 , (c − a)2 ≥ 2 (by Darij Grinberg)
4. For a, b, c, k ∈ R+ we have -
G(a2 , b2 , c2 ) ≥ 4G(a, b, c)
22
1. AM-GM INEQUALITY
(by M.Ramchandran) 5. Let a, b, c, k ∈ R+ , then If and only if the sum of any two of {a, b, c} is less than 2 G(ab, bc, ac) ≥ G(a, b, c)
(by M.Ramchandran) 6. For a, b, c, k ∈ R+ and let a ≥ b ≥ c then, G(
a2 b2 c2 , , ) ≥ G(a, b, c) bc ca ab (by M.Ramchandran)
7. For a, b, c, k ∈ R+ and let k ≥ max{a2 , b2 , c2 } then, G(a, b, c) ≥ G(a2 + k, b2 + k, c2 + k) (Pham Kim Hung) 8. For a, b, c, k ∈ R+ we have -
G(a3 , b3 , c3 ) ≥ 3G(a2 , b2 , c2 ) (by M.Ramchandran)
9. For a, b, c, k ∈ R+ we have -
G(a2 , b2 , c2 ) ≥ G(a, c, b)
(by M.Ramchandran) If more properties are invited to be shared with the author by e-mail. 13. Problem Set This section consists of problems a step more difficult then the problems already discussed. 1. Let a, b, c ∈ R+ such that, a + b + c = 3.Prove that √ √ √ a + b + c ≥ ab + bc + ca (Source : Russian MO 2004) Solution By AM-GM,
√ √ a + a ≥ 3a √ √ b2 + b + b ≥ 3b √ √ c2 + c + c ≥ 3c Thus,by adding the above and using a + b + c = 3 , √ √ √ a2 + b2 + c2 + 2( a + b + c) ≥ 3(a + b + c) = (a + b + c)2 √ √ √ =⇒ 2( a + b + c) ≥ 2(ab + bc + ca) √ √ √ =⇒ a + b + c ≥ ab + bc + ca With equality for a = b = c = 1 ∇ 2.Let x, y, z ∈ R+ .Prove that x y z 2(x + y + z) 1+ 1+ 1+ ≥2+ 1 y z x (xyz) 3 a2 +
13. PROBLEM SET
23
(Source : APMO 1998) Solution After expanding, the inequality reduces to x y z x+y+z + + ≥ 1 y z x (xyz) 3 For the proof of this inequality see Lemma . Equality for x = y = z 3.Let a, b, c be positive real numbers.Prove that 1 1 1 1 + 3 + 3 ≤ 3 3 3 3 a + abc + b b + abc + c c + abc + a abc (USA MO 1998) Solution Using the lemma : a3 + b3 ≥ ab(a + b) abc abc c ≤ = a3 + b3 + abc ab(a + b) + abc a+b+c Construction two more similar inequalities and adding we get the desired result with equality for a = b = c ∇ 4. If x1 , x2 , ...xn ∈ R+ such that 1 1 1 + + ... + =1 1 + x1 1 + x2 1 + xn , Prove that x1 x2 x3 ....xn ≥ (n − 1)n Solution The condition is equivalent to 1 1 1 xn + + ... + = 1 + x1 1 + x2 1 + xn−1 1 + xn Using AM-GM inequality for all the terms in the LHS we get xn n−1 ≥ 1 1 + xn ((1 + x1 )(1 + x2 )(1 + x3 )...(1 + xn )) n−1 Similarly constructing n more inequalities and multiplying all we get the desired. Equality for xi = n − 1 for i ∈ {1, 2, 3, ...n} ∇ 5. Suppose that x, y, z ∈ R+ and x5 + y 5 + z 5 = 3.Prove that x4 y4 z4 + 3 + 3 ≥3 3 y z x Solution Notice that (x5 + y 5 + z 5 )2 = x10 + 2x5 y 5 + y 10 + 2y 5 z 5 + z 10 + 2z 5 x5 = 9 This suggests the use of AM-GM in this way 10 ·
100 x4 + 6x5 y 5 + 3x10 ≥ 19x 19 3 y
Adding all the cyclic results we get, 100 100 100 x4 y4 z4 5 5 5 2 19 + y 19 + z 19 + + ) + 3(x + y + z ) ≥ 19 x y3 z3 x3 Therfore it is enough if we prove 10(
100
(x 19 + y
100 19
+z
100 19
≥ x5 + y 5 + z 5
which is true by AM-GM because X X X 100 3 + 19 = (1 + 19x 19 ) ≥ 20 x5 cyclic
cyclic
cyclic
24
1. AM-GM INEQUALITY
Equality for x = y = z = 1 ∇ 6. Suppose that x, y, z ∈ R+ and x4 + y 4 + z 4 = 3.Prove that 25
x 6 +y
25 6
+z
25 6
≥3 (Own Inequality)
Solution The inequality is equivalent to 3 + 24
25
X
x 6 ≥ 25x4
cyclic
which is true by AM-GM : X
(1 + 24
cyclic
X
25
X
x 6 ) ≥ 25
cyclic
x4
cyclic
Equality for x = y = z = 1 ∇ 7. Let a, b, c ∈ R+ such that abc = 1.Prove that r r r a+b b+c c+a + + ≥3 a+1 b+1 c+1 (Source : Mathlinks Contest) Solution After applying AM-GM to the left hand side we get (a + b)(b + c)(c + a) ≥ (a + 1)(b + 1)(c + 1) and since abc = 1 it is equivalent to ab(a + b) + bc(b + c) + ca(c + a) ≥ a + b + c + ab + bc + ca By AM-GM, 2LHS +
X
X
ab ≥
cyclic
X X 1 1 a2 b + a2 b + a2 c + a2 c + bc ≥ 5 [a5 · (abc) 3 ] 5 = 5 a
cyclic
X
2LHS +
cyclic
a=
cyclic
cyclic
X
X 1 a2 b + a2 b + b2 a + b2 a + c ≥ 4 [(ab)5 · .abc] 5
cyclic
cyclic
Therefore, 4LHS +
X cyclic
ab +
X
a≥5
cyclic
X
a+5
cyclic
=⇒ 4LHS ≥ 4
X
ab
cyclic
X
a+
cyclic
X
ab
cyclic
Hence Proved. Equality for a = b = c = 1 ∇ 8. Let a, b, c, d ∈ R+ 0 such that a + b + c + d = 4. Prove that a2 + b2 + c2 + d2 − 4 ≥ 4(a − 1)(b − 1)(c − 1)(d − 1) (Pham Kim Hung) Solution By AM-GM , p a2 + b2 + c2 + d2 − 4 = (a − 1)2 + (b − 1)2 + (c − 1)2 + (d − 1)2 ≥ 4 |(a − 1)(b − 1)(c − 1)(d − 1)| If the RHS of the question is negative, then the question is meaningless . So we only have to consider the case when - a ≥ b ≥ 1 ≥ c ≥ d Since a + b ≤ 4 and c, d ≤ 1 and by AM-GM, (1 − c)(1 − d) ≤ 1
13. PROBLEM SET
(a − 1)(b − 1) ≤
25
1 · (a + b − 2)2 ≤ 1 4
Therefore (a − 1)(b − 1)(1 − c)(1 − d) ≤ 1 and the conclusion follows. Equality for a = b = c = d = 1, a = b = 2, c = d = 0 and cylic permutations. ∇ 9. Let a, b, c ∈ R+ and a + b + c = 3 . Prove that 0 p p p a 1 + b3 + b 1 + c3 + c 1 + a3 ≤ 5 (Pham Kim Hung) Solution We know by AM-GM that, X p X p X 1 a 1 + b3 = a (1 + b)(1 − b + b2 ) ≤ · a(2 + b2 ) 2 cyclic
cyclic
cyclic
Thus we are left to prove that ab2 + bc2 + ca2 ≤ 4 WLOG : Let b be the middle number in {a, b, c} So we have , a(b − a)(b − c) ≤ 0 =⇒ ab2 + a2 c ≤ abc + a2 b Thus is it is enough if we prove that abc + a2 b + bc2 ≤ 4 ⇔ b(a2 + ac + c2 ) ≤ 4 By AM-GM inequality, b(a2 + ac + c2 ) ≤ b · (a + c)2 = 4 · b ·
a+c a+c · ≤4· 2 2
a+b+c 3
3 =4
This finished our proof with equality for a = 1, b = 2, c = 0 and cyclic permutations. ∇ 10. Let a, b, c ∈ R+ .Prove that a3 b3 c3 a2 b2 c2 + 2+ 2 ≥ + + 2 b c a b c a (Source : JBMO Shortlist 2002) Solution 1 Note that, a3 a2 ≥ +a−b 2 b b or, a3 ≥ a2 b + ab2 − b3 which is true by Lemma . Add all cyclic results to get the desired (by mathlinks user : limes123) Solution 2 By AM-GM, we know that X 2 X a3 2a +a ≥ 2 b b cyc cyc We shall prove , X 2a2 cyc
b
−a−b−c≥
X a2 cyc
or, X a2 cyc
b
≥a+b+c
b
26
1. AM-GM INEQUALITY
It is true by AM-GM, X a2 b
X
≥
+b
2a
cyc
Hence Proved. (by Johan Gunardi) Solution 3 Lemma For a, b ∈ R+ , a3 + b3 ≥ ab(a + b) We have, 3 X 3 X a3 X ab (a + b) X a2 a b3 c3 a + b3 + 2 + 2 + (a + b + c) = +b = ≥ = +a b2 c a b2 b2 b2 b =⇒ ⇒
b3 c3 a2 b2 c2 a3 + 2+ 2 ≥ + + 2 b c a b c a (by mathlinks user : trungk42sp)
Solution 4 Note by AM-GM that, a3 b3 c3 14 2 + 3 2 + 2 2 ≥ b c a
r 19
a38 a2 = b19 b (by mathlinks administrator : nsato)
∇ 11. For a, b, c ∈ R+ .Prove that -
a6 c4 a4 b6 c6 b4 + + + + ≥ b3 c3 a3 a b c (Vascile Cirtoaje)
Solution Note that by AM-GM we have X b6 X b4 X a6 b6 c6 = ( + + ) ≥ 3 3 b3 c3 c3 a3 a (by mathlinks user : karis) ∇ 12. For a, b, c, d ∈ R+ .Prove that a14 b14 c14 d14 b8 c8 d8 a8 + + + ≥ + + + b7 c7 d7 a7 a b c d (Vascile Cirtoaje) Solution Note that by AM-GM we have X a14 X b14 X b8 c14 d14 7 = (4 + 2 + ) ≥ 7 b7 c7 d7 a7 a (by mathlinks user : karis) ∇ 13. Let a, b, c ∈ R+ such that they are all pairwise distinct . Prove that a + b b + c c + a + + a − b b − c c − a > 1 (Source : Iranian National Olympiad (3rd Round) 2007) Solution Set , a+b =x a−b b+c =y b−c
13. PROBLEM SET
27
a+c =z c−a
=⇒ xy + yz + zx = 1 By AM-GM we have, (x + y + z)2 ≥ 3(xy + yz + zx) = 3 =⇒ |x + y + z| ≥
√
3>
(by mathlinks user : Phm Thnh Quang) ∇ 14.Prove the following inequality for all a, b, c ∈ R+ 0 1
c 3(abc) 3 a b + + + ≥4 b c a a+b+c Solution Lemma For a, b, c ∈ R+ .
a b c a+b+c + + ≥ 1 b c a (abc) 3
Proof By AM-GM we have 1 X a2 3 X a b a 2· + ≥3 =3 1 b c bc (abc) 3 cyclic cyclic cyclic
X
we have to prove that 1
a+b+c
+
1
(abc) 3
3(abc) 3 ≥4 a+b+c
By AM-GM, a+b+c 1 3
+
a+b+c
+
1 3
3(abc) 3(abc) Hence proved with equality for a = b = c
a+b+c 3(abc)
1 3
1
3(abc) 3 + ≥4· a+b+c
s
a+b+c 1
3(abc) 3
≥4
(by mathlinks user : enndb0x) ∇ 15. Let a, b, c ∈ R+ such that abc = 1.Prove that a2 + b2 + c2 + 9(ab + bc + ca) ≥ 10(a + b + c) Solution This beautiful problem has many solutions - mostly involving some high-level methods like uvw method, mixing variables etc..But, the following extra-ordinary generalisation was given by an Indian - Aakansh Gupta X X X a2 + k ab ≥ (k + 1) a∀k∈R cyc
cyc
cyc
Proof We have X
⇒(
X
a ≥ 3 and
cyc 2
a) + ( a2 +
cyc
⇒(
X cyc
a2 + k
X
ab) + (
cyc
X
2
ab) ≥ 3
cyc
X
X
a2 b2 ≥
a2 b2 + k
cyc
⇒
X cyc
a2 + k
X
a+3
X
cyc
cyc
X
ab ≥ 3
cyc
cyc
⇒
X
X
a+
cyc
X
X
a) ≥ (k + 1)
cyc
ab ≥ (k + 1)
ab
cyc
cyc
X
ab
cyc
X cyc
X cyc
a
a + (k + 1)
X cyc
ab
28
1. AM-GM INEQUALITY
Or, X
a 2 b2 + k
cyc
X
a ≥ (k + 1)
cyc
X
ab
cyc
If first one occurs we are done and if the second one occurs then replace a by a1 ; b by get the same expression as the first one and thus we have proved !! ∇ 16.Prove that for a, b, c ∈ R+ r r r 2a 2b 2c (a + b + c) (ab + bc + ca) + + ≥p 2 a+b b+c c+a (a + b2 + c2 ) (a2 b2 + b2 c2 + c2 a2 )
1 b
; c by
1 c
and we
Solution r
2b 4b ≥ b+c 3b + c r 2c 4c ≥ c+a 3c + a r 2a 4a ≥ a+b 3a + b So that, r r a b c 2a 2b 2c + + ≥4 + + a+b b+c c+a 3a + b 3b + c 3c + a From Titu’s Lemma and the following well-known inequality a2 + b2 + c2 ≥ ab + bc + ca r
a b c a2 b2 c2 + + = 2 + 2 + 2 3a + b 3b + c 3c + a 3a + ab 3b + bc 3c + ca 2
2
≥
(a + b + c) (a + b + c) ≥ 2 2 2 3 (a + b + c ) + ab + bc + ca 4 (a2 + b2 + c2 )
Therefore, r r 2 2a 2b 2c (a + b + c) + + ≥ 2 a+b b+c c+a a + b2 + c2 We also have, (just prove analogously) r r r r r r 2a 2b 2c 2ac 2ba 2cb + + = + + a+b b+c c+a ac + bc ba + ca cb + ab r
2
≥
(ab + bc + ca) 2 a b2 + b2 c2 + c2 a2
So that, r
2a + a+b
r
2b + b+c
r
≥p
1 2c ≥ c+a 2
2
2
(a + b + c) (ab + bc + ca) + 2 2 2 2 2 a +b +c a b + b2 c2 + c2 a2
!
(a + b + c) (ab + bc + ca) (a2
+ b2 + c2 ) (a2 b2 + b2 c2 + c2 a2 )
(from AM-GM inequality) The proof is completed. Equality holds if and only if a = b = c (by mathlinks user : leviethai) ∇
13. PROBLEM SET
17. For a, b, c ∈ R+ 0 . Prove that (ab + bc + ca)
1 1 1 + + 2 2 (a + b) (b + c) (c + a)2
29
≥
9 4
(Source : Iran 1996) Solution This is one the most famous, discussed and celebrated inequality of all times. In the year asked it was percieved as a very difficult inequality not solved by any elementary methods but this notion was wronged by this solution given by the Vietnamese Inequality Solver well known for his beautiful solutions Va Quoc Ba Can (mathlinks uers id : canhang2007) Without loss of generality, we may assume that a ≥ b ≥ c. Then, we will show that 1 1 1 2 1 + + ≥ + 2 2 2 (a + b) (a + c) (b + c) 4ab (a + c)(b + c) Indeed, this inequality is equivalent to 1 1 2 1 1 + − ≥ − 2 2 (a + c) (b + c) (a + c)(b + c) 4ab (a + b)2 or
(a − b)2 (a − b)2 ≥ 2 2 (a + c) (b + c) 4ab(a + b)2
This is true because 4ab ≥ 4b2 ≥ (b + c)2 and (a + b)2 ≥ (a + c)2 Now, using the above estimation, it is sufficient to prove that 2 9 1 + ≥ (ab + bc + ca) 4ab (a + c)(b + c) 4 Since ab + bc + ca 1 c(a + b) = + 4ab 4 4ab and
2(ab + bc + ca) 2c2 =2− (a + c)(b + c) (a + c)(b + c)
it is equivalent to 2c2 c(a + b) ≥ 4ab (a + c)(b + c) or (a + b)(b + c)(c + a) ≥ 8abc The last one is true according to the AM-GM Inequality, so our proof is completed It stands out as one of the best solutions for the inequality. ∇ 18. Let a, b, c ∈ R+ such that abc = 1 . Prove that ak bk ck 3 + + ≥ a+b b+c c+a 2 for any positive integer k (Source : China Northern Mathematical Olympiad 2007) Solution Re-write the LHS as ak−1 + bk−1 + ck−1 ≥
X ak−1 b 3 + 2 a+b cyclic
30
1. AM-GM INEQUALITY
by applying AM-GM to the denominators in the RHS we get X 3 1 RHS ≤ ak− 2 b 2 cyclic
Thus, it is enough if we prove that 3
1
3
1
3
1
2(ak−1 + bk−1 + ck−1 ) ≥ 3 + ak− 2 b 2 + bk− 2 c 2 + ck− 2 a 2 This follows directly from AM-GM as ak−1 + bk−1 + ck−1 ≥ 3 · (abc)
k−1 3
=3
And also , 3
1
3
1
3
1
(2k − 3)ak−1 + bk−1 ≥ (2k − 2)ak− 2 b 2 (2k − 3)bk−1 + ck−1 ≥ (2k − 2)bk− 2 c 2 (2k − 3)ck−1 + ak−1 ≥ (2k − 2)ck− 2 a 2 Adding the above we get the desired.Equality for a = b = c = 1 (By Pham Kim Hung) ∇ 19. Prove that for a, b, c ∈ R+ , we have 6 3 ≥ ab + bc + ca a+b+c (Source : Macedonia Team Selection Test 2007) Solution The inequality is equivalent to 3(a + b + c) a+b+c+ ≥6 ab + bc + ca By AM-GM inequality we have, r 3(a + b + c) 3(a + b + c)2 ≥2 ≥6 a+b+c+ ab + bc + ca ab + bc + ca Equality for a = b = c = 1 (By Vo Danh) ∇ 20. Let a, b, c ≥ 0 such that abc ≥ 1 . Prove that 1 1 1 27 a+ b+ c+ ≥ a+1 1+b 1+c 8 1+
(Source : Ukraine Mathematical Festival 2007) Solution By AM-GM we have - , a+1 1 + ≥1 4 1+a and ,
3a 3 3 √ + ≥ · a 4 4 2
Adding the two inequalities we get 1 3 √ ≥ · a 1+a 2 Obtaining similarly all the cyclic inequalities and multiplying them we get 1 1 1 27 √ 27 a+ b+ c+ ≥ · abc ≥ a+1 1+b 1+c 8 8 a+
Equality for a = b = c = 1
13. PROBLEM SET
31
(by Nguyen Dung TN) ∇
