IMO 1978 SL 11

Origin: SWE

Problem

A function f : I \toR, defined on an interval I, is called concave if f(\thetax + (1 −\theta)y) \geq\thetaf(x) + (1 −\theta)f(y) for all x, y \inI and 0 \leq\theta \leq1. Assume that the functions f1, . . . , fn, having all nonnegative values, are concave. Prove that the function (f1f2 . . . fn)1/n is concave.

Solution

Set F(x) = f1(x)f2(x) \cdot \cdot \cdot fn(x): we must prove concavity of F 1/n. By the assumption,

F(\thetax + (1 −\theta)y) \geq n

i=1 [\thetafi(x) + (1 −\theta)f(y)]

n  k=0 \thetak(1 −\theta)n−k  fi1(x) . . . fik(x)fik+1(y)fin(y), where the second sum goes through all
n k  k-subsets {i1, . . . , ik} of {1, . . . , n}. The inequality between the arithmetic and geometric means now gives us  fi1(x)fi2(x) \cdot \cdot \cdot fik(x)fik+1(y)fin(y) \geq n k  F(x)k/nF(y)(n−k)/n. Inserting this in the above inequality and using the binomial formula, we finally obtain F(\thetax + (1 −\theta)y) \geq n  k=0 \thetak(1 −\theta)n−k n k  F(x)k/nF(y)(n−k)/n

 \thetaF(x)1/n + (1 −\theta)F(y)1/n n , which proves the assertion.