IMO 1983 SL 9

Origin: USA

Problem

If a, b, and c are sides of a triangle, prove that a2b(a −b) + b2c(b −c) + c2a(c −a) \geq0. Determine when there is equality.

Solution

For any triangle of sides a, b, c there exist 3 nonnegative numbers x, y, z such that a = y +z, b = z +x, c = x+y (these numbers correspond to the division of the sides of a triangle by the point of contact of the incircle). The inequality becomes

(y +z)2(z +x)(y −x)+ (z +x)2(x+y)(z −y)+ (x+y)2(y +z)(x−z) \geq0. Expanding, we get xy3 + yz3 + zx3 \geqxyz(x + y + z). This follows from Cauchy’s inequality (xy3+yz3+zx3)(z+x+y) \geq
\sqrtxyz(x + y + z) 2 with equality if and only if xy3/z = yz3/x = zx3/y, or equivalently x = y = z, i.e., a = b = c.