22nd Putnam 1961 Problem A2 For which real numbers α, β can we find a constant k such that x α y β < k(x + y) for all positive x, y? Solution
: α, β ≥ 0 and α + b = 1. If α  k(x + y) for any k. Similarly for β  1, then taking x = y sufficiently large gives x α y β > k(x + y) for any k. On the other hand, if α + β  k(x + y) for any k. Suppose, on the other hand that α, b ≥ 0 and α + β = 1. Take x ≥ y. Then x α y β ≤ x α+β = x, so the result holds with k = 1. 22nd Putnam 1961 © John Scholes jscholes@kalva.demon.co.uk 15 Feb 2002