x, y, z are positive real numbers with product 1. Show that xy/(x5 + xy + y5) + yz/(y5 + yz + z5) + zx/(z5 + zx + x5) ≤ 1. When does equality occur?
Solution
Answer: x = y = z.
x5 + y5 = (x + y)(x4 - x3y + x2y2 - xy3 + y4). Now x4 - x3y - xy3 + y4 = (x - y)(x3 - y3) ≥ 0, with equality iff x = y. Hence x5 + y5 ≥ (x + y)x2y2 with equality iff x = y.
Hence xy/(x5 + xy + y5) ≤ 1/( xy(x+y) + 1) = z/( xyz(x+y) + z) = z/(x + y + z), with equality iff x = y. Similarly for the other terms. Adding, we get the required inequality, with equality iff x = y = z.
© John Scholes
jscholes@kalva.demon.co.uk
16 Dec 2002