The positive reals x, y satisfy x2 + y3 ≥ x3 + y4. Show that x3 + y3 ≤ 2.
Solution
If x, y both ≤ 1, then the inequality is obvious. If x, y both > 1, then they do not satisfy the condition. So either x ≥ 1 ≥ y or y ≥ 1 ≥ x. We show first that x3 + y3 ≤ x2 + y2. In the first case we have x2 + y2(1-y) ≥ x2 + y3(1-y) ≥ x3 + y4 - y4 = x3, so x2 + y2 ≥ x3 + y3, as required. In the second case we have x2 ≥ x3 + y3(y-1) ≥ x3 + y2(y-1), so again x2 + y2 ≥ x3 + y3.
Now by Cauchy-Scwartz, x2 + y2 = x3/2x1/2 + y3/2y1/2 ≤ √(x3+y3) √(x+y). But x2 + y2 ≥ x3 + y3, so x2 + y2 ≤ x + y. But 2xy ≤ (x2+y2), so (x + y)2 ≤ & 2(x2 + y2) and we have just shown that 2(x2 + y2) ≤ 2(x + y), so x + y ≤ 2. Thus we have x3 + y3 ≤ x2 + y2 ≤ x + y ≤ 2.
Thanks to Bekjan Jumabaev
© John Scholes
jscholes@kalva.demon.co.uk
16 February 2004
Last corrected/updated 16 Feb 04