Let a, b and c be the lengths of the sides of a triangle. Prove that
a2b(a - b) + b2c(b - c) + c2a(c - a) ≥ 0.
Determine when equality occurs.
Solution
Put a = y + z, b = z + x, c = x + y. Then the triangle condition becomes simply x, y, z > 0. The inequality becomes (after some manipulation):
xy3 + yz3 + zx3 ≥ xyz(x + y + z).
Applying Cauchy's inequality we get (xy3 + yz3 + zx3)(z + x + y) ≥ xyz(y + z + x)2 with equality iff xy3/z = yz3/x = zx3/y. So the inequality holds with equality iff x = y = z. Thus the original inequality holds with equality iff the triangle is equilateral.
Solutions are also available in Murray S Klamkin, International Mathematical Olympiads 1978-1985, MAA 1986, and in István Reiman, International Mathematical Olympiad 1959-1999, ISBN 189-8855-48-X.
© John Scholes
jscholes@kalva.demon.co.uk
14 Oct 1998