Advanced book on Mathematics Olympiad

158 3 Real Analysis

As an instructive example we present in detail the proof of another famous inequality.

Chebyshev’s inequality.Letfandgbe two increasing functions onR. Then for any
real numbersa<b,

(b−a)

∫b

a

f (x)g(x)dx≥

(∫b

a

f(x)dx

)(∫b

a

g(x)dx

)

.

Proof.Becausefandgare both increasing,

(f (x)−f (y))(g(x)−g(y))≥ 0.

Integrating this inequality over[a, b]×[a, b], we obtain

∫b

a

∫b

a

(f (x)−f (y))(g(x)−g(y))dxdy≥ 0.

Expanding, we obtain

∫b

a

∫b

a

f (x)g(x)dxdy+

∫b

a

∫b

a

f (y)g(y)dxdy−

∫b

a

∫b

a

f (x)g(y)dxdy

−

∫b

a

∫b

a

f (y)g(x)dxdy≥ 0.

By eventually renaming the integration variables, we see that this is equivalent to

(b−a)

∫b

a

f (x)g(x)dx−

(∫b

a

f(x)dx

)

·

(∫b

a

g(x)dx

)

≥ 0 ,

and the inequality is proved. 

478.Letf:[ 0 , 1 ]→Rbe a continuous function. Prove that

(∫ 1

0

f(t)dt

) 2

≤

∫ 1

0

f^2 (t)dt.

479.Find the maximal value of the ratio

(∫ 3

0

f(x)dx

) 3 /∫ 3

0

f^3 (x)dx,

asfranges over all positive continuous functions on[ 0 , 1 ].