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Problem 925

AIME late

Algebra Difficulty 5.8 Prove it

29th USAMO 2000 Problem A1 Show that there is no real-valued function f on the reals such that ( f(x) + f(y) )/2 ≥ f( (x+y)/2 ) + |x - y| for all x, y.

This one wants a proof. Work it on paper, read the official solution, then mark yourself honestly — the ladder only means something if the record is true.

Official solution

Put x = a + b, y = a - b with b > 0. Then we have f(a) ≤ 1/2 f(a+b) + 1/2 f(a-b) - 2b. Also f(a + b/2) ≤ 1/2 f(a) + 1/2 f(a+b) - b, f(a - b/2) ≤ 1/2 f(a-b) + 1/2 f(a) - b, and f(a) ≤ 1/2 f(a - b/2) + 1/2 f(a + b/2) - b ≤ 1/4 f(a-b) + 1/2 f(a) + 1/4 f(a+b) - 2b. Hence f(a) ≤ 1/2 f(a-b) + 1/2 f(a+b) - 4b. But a and b are arbitrary (apart from b > 0) so this argument can now be repeated to show that f(a) ≤ 1/2 f(a-b) + 1/2 f(a+b) + 2 n b for any positive integer n. Contradiction. 29th USAMO 2000 © John Scholes [email protected] 11 May 2002

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