Seemous 2011

Bucharest, March 4 th, 2011

SOUTH EASTERN EUROPEAN MATHEMATICAL OLYMPIAD FOR UNIVERSITY STUDENTS Bucharest, March 4 th, 2011 Solutions and grading schemes Problem 1. For a given integer n decreasing function. Prove that

 

[0, 1] → R be a non≥ 1, let f  : [0,

1

  1

f ( f (x) d x

≤ (n + 1)

0

xn f ( f (x)dx. )dx.

0

Find all non-decreasing continuous functions for which equality holds. [0, 1] we integrate on the interval Solution. For given n and any x, y [0, [0, [0, 1] in terms of  x and in terms of y of  y the obvious inequality

∈

(xn obtaining

n

)(f ((x) − f ( f (y )) ≥ 0, − y )(f 

    −     − 1

0

or

 

1

x f ( f (x) dx

0

)(f ((x) yn )(f 

0

1

n



1

(xn

1

n

y dy

0

− f ( f (y )) dx

    − 1

f ( f (x) dx

0

(1)

dy

≥ 0,

1

n

x dx

0

 

1

f ( f (y) dy +

0

0

y n f ( f (y ) dy

≥0

which which gives gives the desired inequality inequality.. As f  is continuous, when equality holds it forces equality in (1) also, that is f  must be constant. Grading scheme.

  1

(n + 1)

x f ( f  (x) dx =

0

   √  1

n

f 

n+1

0

f  non-decreasing implies f ( f  (x)

3p t dt . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3p

√x) for x ∈ [0, ≤ f ( f  ( [0, 1] n+1

1

. . . . . . . . . . . . . 3p

  1

  1

f (x) dx

0

≤

f (

√x) dx = (n + 1)

n+1

0

  1

0

xn f (x) dx . . . . . . . . . . . . . . . . . . 1p

Equality holds only for constant functions. ..........................3p Alternative solution. (xn y n ) (f (x) f (y)) 0 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3p Integrate the above inequality on [0, 1] [0, 1] to obtain

−

≤       

−

≥

×

1 1

0

((x

0 0

1

n

− y ) (f (x) − f (y)))dxdy =

1

n

y dy

0

  1

n

  1

f (x) dx +

0

−     1

n

x f (x) dx

0

1

n

x dx f (y) dy

0

0

−

y n f (y) dy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3 p

0

  1

1 it obtains the inequality. .....................1p n+1 0 Equality holds only for constants function. ..........................3p Using

xn dx =

Remark 1 The use of the Chebyshev inequality (without proof) ........ 7p Problem 2. Let A = (aij ) be a real n n matrix such that An = 0 and aij a ji 0, for all i, j. Prove that there exist two nonreal numbers among eigenvalues of  A.

×

≤



Solution. Let λ1 , . . . , λn be all eigenvalues of this matrix. From condition

 n

it follows that aii = 0, therefore

λk = 0. The sum



λi λ j equals the sum

i