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