Advanced book on Mathematics Olympiad

522 Real Analysis

number ofxi’s that are less thanaand the number ofxi’s that are greater thana. The
global minimum is attained where the derivative changes sign. Fornodd, this happens
precisely atxn/ 2 + 1 .Ifnis even, the minimum is achieved at any point of the interval
[xn/ 2 ,xn/ 2 + 1 ]at which the first derivative is zero and the function is constant.
So the answer to the problem isa=xn/ 2 + 1 ifnis odd, andais any number in the
interval[xn/ 2 ,xn/ 2 + 1 ]ifnis even.

Remark.The required numberxis called the median ofx 1 ,x 2 ,...,xn. In general, if the
numbersx∈Roccur with probability distributiondμ(x)then their medianaminimizes

E(|x−a|)=

∫∞

−∞

|x−a|dμ(x).

The median is any number such that
∫a

−∞

dμ(x)=P(x≤a)≥

1

2

and

∫∞

a

dμ(x)=P(x≥a)≥

1

2

.

In the particular case of our problem, the numbersx 1 ,x 2 ,...,xnoccur with equal prob-
ability, so the median lies in the middle.

426.The functionf(t)= tcis convex, whileg(t)=xt is convex and increasing.
Therefore,h(t)=g(f (t))=xt
c
is convex. We thus have

xa

c

+xb

c

=h(a)+h(b)≥ 2 h

(

a+b

2

)

= 2 x(

a+ 2 b)^2 c

≥ 2 x(ab)

c/ 2

.

This completes the solution.
(P. Alexandrescu)

427.We can assume that the triangle is inscribed in a circle of diameter 1, so thata=sinA,
b=sinB,c=sinC,A≥B≥C. The sine function is concave on the interval[ 0 ,π],
and sinceBis betweenAandC, and all three angles lie in this interval, we have

sinB−sinC

B−C

≥

sinA−sinC

A−C

.

Multiplying out, we obtain

(A−C)(sinB−sinC)≥(B−C)(sinA−sinC),