Real Analysis 529The coefficients in the expansion are positive, so the expression is a convex function inx
(being a sum of power functions that are convex). Its maximum is attained when|x|=1,
in which case the value of the expression is 2n. This proves the inequality.
(C. Nast ̆ ̆asescu, C. Ni ̧ta, M. Brandiburu, D. Joi ̧ta, ̆ Exerci ̧tii ̧si Probleme de Algebra ̆
(Exercises and Problems in Algebra), Editura Didactica ̧ ̆si Pedagogica, Bucharest, 1983) ̆
437.Without loss of generality, we may assume thatbis the number in the middle. The
inequality takes the form
a+b+c− 33√
abc≤ 3 (a+c− 2√
ac).For fixedaandc, definef:[a, c]→R,f(b)= 3 (a+c− 2
√
ac)−a−b−c+ 3√
abc.
This function is concave becausef′′(b)=−^23 (ac)^1 /^3 b−^5 /^3 <0, so it attains its minimum
at one of the endpoints of the interval[a, c]. Thus the minimum is attained forb=aor
b=c. Let us try the caseb=a. We may rescale the variables so thata=b=1. The
inequality becomes
2 c+ 3 c^1 /^3 + 1
6
≥c^1 /^2 ,and this is just an instance of the generalized AM–GM inequality. The casea =cis
similar.
(USA Team Selection Test for the International Mathematical Olympiad, 2002, pro-
posed by T. Andreescu)
438.For (a) we apply Sturm’s principle. Givenx∈(a, b)chooseh>0 such thata<
x−h<x+h<b. The mean value theorem implies thatf(x)≤maxx−h≤y≤x+yf(y),
with equality only whenfis constant on[x−h, x+h]. Hencef(x)is less than or equal
to the maximum offon[a, b], with equality if and only iffis constant on[a, b].We
know that the maximum offis attained on[a, b]. It can be attained atxonly iffis
constant on[a, b]. This proves that the maximum is attained at one of the endpoints of
the interval.
To prove (b) we define the linear function
L(x)=(x−a)f (b)+(b−x)f (a)
b−a.
It is straightforward to verify thatLitself satisfies the mean value inequality from the
statement with equality, and so does−L. Therefore, the functionG(x)=f(x)−L(x)
satisfies the mean value inequality, too. It follows thatGtakes its maximum value ata
or atb. A calculation shows thatG(a)=G(b)=0. Therefore,G(x)≤0 forx∈[a, b].
This is equivalent to
f(x)≤(x−a)f (b)+(b−x)f (a)
b−a