Advanced book on Mathematics Olympiad

Real Analysis 547

481.Suppose thatx>y. Transform the inequality successively into

mn(x−y)(xm+n−^1 −ym+n−^1 )≥(m+n− 1 )(xm−ym)(xn−yn),

and then

xm+n−^1 −ym+n−^1

(m+n− 1 )(x−y)

≥

xm−ym

m(x−y)

·

xn−yn

n(x−y)

.

The last one can be written as

(x−y)

∫x

y

tm+n−^2 dt≥

∫x

y

tm−^1 dt·

∫x

y

tn−^1 dt.

Here we recognize Chebyshev’s inequality applied to the integrals of the functionsf, g:

[y, x]→R,f(t)=tm−^1 andg(t)=tn−^1.

(Austrian–Polish Competition, 1995)

482.Observe thatfbeing monotonic, it is automatically Riemann integrable. Taking

the mean offon the intervals[ 0 ,α]and[ 1 −α, 1 ]and using the monotonicity of the

function, we obtain

1

1 −α

∫ 1

α

f(x)dx≤

1

α

∫α

0

f(x)dx,

whence

α

∫ 1

α

f(x)dx≤( 1 −α)

∫α

0

f(x)dx.

Adding

∫α

0 f(x)dxto both sides gives

α

∫ 1

0

f(x)dx≤

∫α

0

f(x)dx,

as desired.

(Soviet Union University Student Mathematical Olympiad, 1976)

483.Forx∈[ 0 , 1 ], we havef′(x)≤f′( 1 ), and so

f′( 1 )

f^2 (x)+ 1

≤

f′(x)

f^2 (x)+ 1

.

Integrating, we obtain

f′( 1 )

∫ 1

0

dx

f^2 (x)+ 1

≤

∫ 1

0

f′(x)

f^2 (x)+ 1

=arctanf( 1 )−arctanf( 0 )=arctanf( 1 ).