Advanced book on Mathematics Olympiad

528 Real Analysis

434.We assume thatα ≤β ≤γ, the other cases being similar. The expression is a
convex function in each of the variables, so it attains its maximum for somex, y, z=a
orb.
Now let us fix three numbersx, y, z∈[a, b], withx≤y≤z. We have

E(x, y, z)−E(x, z, y)=(γ−α)((z−x)^2 −(y−z)^2 )≥ 0 ,

and henceE(x, y, z)≥E(x, z, y). Similarly,E(x, y, z)≥E(y, x, z)andE(z, y, x)≥
E(y, z, x). So it suffices to consider the casesx=a,z=borx=bandz=a. For
these cases we have

E(a, a, b)=E(b, b, a)=(β+γ )(b−a)^2

and

E(a, b, b)=E(b, a, a)=(α+γ )(b−a)^2.

We deduce that the maximum of the expression under discussion is(β+γ )(b−a)^2 ,
which is attained forx=y=a,z=band forx=y=b,z=a.
(Revista Matematica din Timi ̧soara ̆ (Timi ̧soara Mathematics Gazette), proposed by
D. Andrica and I. Ra ̧sa)

435.The left-hand side of the inequality under discussion is a convex function in each
xi. Hence in order to maximize this expression we must choose some of thexi’s equal to
aand the others equal tob. For such a choice, denote byuthe sum of theti’s for which
xi =aand byvthe sum of theti’s for whichxi=b. It remains to prove the simpler
inequality

(ua+bv)

(u

a

+

v

b

)

≤

(a+b)^2

4 ab

(u+b)^2.

This is equivalent to

4 (ua+vb)(ub+va)≤(ua+vb+ub+va)^2 ,

which is the AM–GM inequality applied toua+vbandub+va.
(L.V. Kantorovich)

436.Expanding with Newton’s binomial formula, we obtain

( 1 +x)n+( 1 −x)n=

∑n^2 

k= 0

(

n

2 k

)

x^2 k.