Find all real-valued functions f(x) on the reals such that f(xy)/2 + f(xz)/2 - f(x) f(yz) ≥ 1/4 for all x, y, z.
Answer
f(x) = 1/2 for all x
Solution
Put x = y = z = 0, then (1/2 - f(0))2 ≤ 0, so f(0) = 1/2. Put z = 0, then f(xy) ≥ f(x) for all x,y. Taking x = 1 we get f(x) ≥ f(1) for all x. Taking y = 1/x we get f(1) ≥ f(x) for all x except possibly x = 0, so f(x) = f(1) for all x except possibly x = 0. But putting x = y = z = 1 we get (1/2 - f(1))2 ≤ 0, so f(1) = 1/2. Hence f(x) = 1/2 for all x.
Thanks to Suat Namli
© John Scholes
jscholes@kalva.demon.co.uk
13 February 2004
Last corrected/updated 13 Feb 04