IMO 2003 SL A5

Origin: KOR | Category: Algebra

Problem

Let R+ be the set of all positive real numbers. Find all functions f : R+ \toR+ that satisfy the following conditions: (i) f(xyz) + f(x) + f(y) + f(z) = f(\sqrtxy)f(\sqrtyz)f(\sqrtzx) for all x, y, z \in R+. (ii) f(x) < f(y) for all 1 \leqx < y.

Solution

Placing x = y = z = 1 in (i) leads to 4f(1) = f(1)3, so by the condition f(1) > 0 we get f(1) = 2. Also putting x = ts, y = t s, z = s t in (i) gives f(t)f(s) = f(ts) + f(t/s). (1) In particular, for s = 1 the last equality yields f(t) = f(1/t); hence f(t) \geqf(1) = 2 for each t. It follows that there exists g(t) \geq1 such that f(t) = g(t) + g(t). Now it follows by induction from (1) that g(tn) = g(t)n for every integer n, and therefore g(tq) = g(t)q for every rational q. Consequently, if t > 1 is fixed, we have f(tq) = aq + a−q, where a = g(t). But since the set of aq (q \inQ) is dense in R+ and f is monotone on (0, 1] and [1, \infty), it follows that f(tr) = ar + a−r for every real r. Therefore, if k is such that tk = a, we have f(x) = xk + x−k for every x \inR.