Russian 2000 Problem 9 R is the reals. Find all functions f: R → R which satisfy f(x+y) + f(y+z) + f(z+x) ≥ 3f(x+2y+3z) for all x, y, z.
f constant. Solution Put x = a, y = z = 0, then 2f(a) + f(0) ≥ 3f(a), so f(0) ≥ f(a). Put x = a/2, y = a/2, z = -a/2. Then f(a) + f(0) + f(0) ≥ 3f(0), so f(a) ≥ f(0). Hence f(a) = f(0) for all a. But any constant function obviously satisfies the given relation. Russian 2000 © John Scholes jscholes@kalva.demon.co.uk 20 December 2003