Advanced book on Mathematics Olympiad

Real Analysis 545

We have

0 ≤

∫ 1

0

(f (x)−g(x))^2 dx=

∫ 1

0

f (x)(f (x)−g(x))dx−

∫ 1

0

g(x)(f (x)−g(x))dx

=

∫ 1

0

f^2 (x)−f (x)g(x)dx=

∫ 1

0

f^2 (x)dx− 6

∫ 1

0

xf (x)dx+ 2

∫ 1

0

f(x)dx

=

∫ 1

0

f^2 (x)dx− 4.

The inequality is proved.
(Romanian Mathematical Olympiad, 2004, proposed by I. Ra ̧sa)

476.We change this into a minimum problem, and then relate the latter to an inequality
of the formx≥0. Completing the square, we see that

x(f (x))^2 −x^2 f(x)=

√

xf (x))^2 − 2

√

xf (x)

x

(^32)
2

=

(

√

xf (x)−

x

(^32)
2

) 2

−

x^3

4

.

Hence, indeed,

J(f)−I(f)=

∫ 1

0

(

√

xf (x)−

x

(^32)
2

) 2

dx−

∫ 1

0

x^3

4

dx≥−

1

16

.

It follows thatI(f)−J(f)≤ 161 for allf. The equality holds, for example, for
f:[ 0 , 1 ]→R,f(x)=x 2. We conclude that

max

f∈C^0 ([ 0 , 1 ])

(I (f )−J(f))=

1

16

.

(49th W.L. Putnam Mathematical Competition, 2006, proposed by T. Andreescu)

477.We can write the inequality as
∑

i,j

xixj(ai+aj−2 min(ai,aj))≤ 0.

Note that

∑

i,j

xixjai=xj

∑n

i= 1

aixi= 0 ,

and the same stays true if we exchangeiwithj. So it remains to prove that
∑

i,j

xixjmin(ai,aj)≥ 0.