Advanced book on Mathematics Olympiad

548 Real Analysis

Becausef′( 1 )>0 and arctany≤yfory≥0, we further obtain

∫ 1

0

dx

f^2 (x)+ 1

≤

arctanf( 1 )

f′( 1 )

≤

f( 1 )

f′( 1 )

,

proving the inequality. In order for equality to hold we must have arctanf( 1 )=f( 1 ),
which happens only whenf( 1 )=0. Then

∫ 1

0

dx
f^2 (x)+ 1 =0. But this cannot be true since
the function that is integrated is strictly positive. It follows that the inequality is strict.
This completes the solution.
(Romanian Mathematical Olympiad, 1978, proposed by R. Gologan)

484.The Leibniz–Newton fundamental theorem of calculus gives

f(x)=

∫x

a

f′(t)dt.

Squaring both sides and applying the Cauchy–Schwarz inequality, we obtain

f(x)^2 =

(∫b

a

f′(t)dt

) 2

≤(b−a)

∫b

a

f′(t)^2 dt.

The right-hand side is a constant, while the left-hand side depends onx. Integrating the
inequality with respect toxyields
∫b

a

f(x)^2 dx≤(b−a)^2

∫b

a

f′(t)^2 dt.

Substitutetbyxto obtain the inequality as written in the statement of the problem.

485.This is an example of a problem in which it is important to know how to organize
the data. We start by lettingAbe the subset of[ 0 , 1 ]on whichfis nonnegative, andB
its complement. Letm(A), respectively,m(B)be the lengths (measures) of these sets,
andIAandIBthe integrals of|f|onA, respectively,B. Without loss of generality, we
can assumem(A)≥^12 ; otherwise, changefto−f.
We have
∫ 1

0

∫ 1

0

|f(x)+f(y)|dxdy

=

∫

A

∫

A

(f (x)+f (y))dxdy+

∫

B

∫

B

(|f(x)|+|f(y)|)dxdy

+ 2

∫

A

∫

B

|f(x)+f(y)|dxdy.

Let us first try a raw estimate by neglecting the last term. In this case we would have to
prove