Advanced book on Mathematics Olympiad

146 3 Real Analysis

436.Prove that for any natural numbern≥2 and any|x|≤1,

( 1 +x)n+( 1 −x)n≤ 2 n.

437.Prove that for any positive real numbersa, b, cthe following inequality holds

a+b+c

3

−^3

√

abc≤max{(

√

a−

√

b)^2 ,(

√

b−

√

c)^2 ,(

√

c−

√

a)^2 }.

438.Letfbe a real-valued continuous function onRsatisfying

f(x)≤

1

2 h

∫x+h

x−h

f(y)dy, for allx∈Randh> 0.

Prove that (a) the maximum offon any closed interval is assumed at one of the

endpoints, and (b) the functionfis convex.

An important property of convex (respectively, concave) functions is known as Jen-
sen’s inequality.

Jensen’s inequality.For a convex functionfletx 1 ,x 2 ,...,xnbe points in its domain
and letλ 1 ,λ 2 ,...,λnbe positive numbers withλ 1 +λ 2 +···+λn= 1. Then

f(λ 1 x 1 +λ 2 x 2 +···+λnxn)≤λ 1 f(x 1 )+λ 2 f(x 2 )+···+λnf(xn).

Iffis nowhere linear and thexi’s are not all equal, then the inequality is strict. The
inequality is reversed for a concave function.

Proof.The proof is by induction onn. The base case is the definition of convexity. Let
us assume that the inequality is true for anyn−1 pointsxiand anyn−1 weightsλi.
Considernpoints and weights, and letλ=λ 1 +···+λn− 1. Note thatλ+λn=1 and
λ 1
λ+

λ 2

λ+···+

λn− 1

λ =1. Using the base case and the inductive hypothesis we can write

f(λ 1 x 1 +···+λn− 1 xn− 1 +λnxn)=f

(

λ

(

λ 1

λ

x 1 +···+

λn− 1

λ

xn− 1

)

+λnxn

)

≤λf

(

λ 1

λ

x 1 +···+

λn− 1

λ

xn− 1

)

+λnf(xn)

≤λ

(

λ 1

λ

f(x 1 )+···+

λn− 1

λ

f(xn− 1 )

)

+λnf(xn)

=λ 1 f(x 1 )+···+λn− 1 f(xn− 1 )+λnf(xn),

as desired. For the case of concave functions, reverse the inequalities. 

As an application, we prove the following.