w, x, y, z are non-negative reals such that wx + xy + yz + zw = 1. Show that w3/(x + y + z) + x3/(w + y + z) + y3/(w + x + z) + z3/(w + x + y) ≥ 1/3.
Solution
Solution by Demetres Christofides
Put s = w + x + y + z. Put A = w3/(s-w) + x3/(s-x) + y3/(s-y) + z3/(s-z), B = w2 + x2 + y2 + z2, C = w(s-w) + x(s-x) + y(s-y) + z(s-z) = 2 + 2wy + 2xz. By Cauchy-Schwartz, we have AC >= B2.
We have (w-x)2 + (x-y)2 + (y-z)2 + (z-w)2 ≥ 0, so B ≥ (wx + xy + yz + zw) = 1. Also (w - y)2 + (x - z)2 ≥ 0, so B ≥ 2wy + 2xz. So if 2wy + 2xz ≤ 1, then A ≥ 1/C ≥ 1/3. If 2wy + 2xz > 1, then C > 3, so (C-2)/C > 1 - 2/C > 1/3. Hence AC ≥ B2 ≥ B ≥ (C-2), so A > 1/3.
© John Scholes
jscholes@kalva.demon.co.uk
21 Nov 2002