IMO 1984 SL 5

Origin: FRG

Problem

Let x, y, z be nonnegative real numbers with x+y +z = 1. Show that 0 \leqxy + yz + zx −2xyz \leq7 27.

Solution

Let f(x, y, z) = xy + yz + zx −2xyz. The first inequality follows imme- diately by adding xy \geqxyz, yz \geqxyz, and zx \geqxyz (in fact, a stronger inequality xy + yz + zx −9xyz \geq0 holds). Assume w.l.o.g. that z is the smallest of x, y, z. Since xy \leq(x + y)2/4 = (1 −z)2/4 and z \leq1/2, we have xy + yz + zx −2xyz = (x + y)z + xy(1 −2z) \leq(1 −z)z + (1 −z)2(1 −2z) = 7 27 −(1 −2z)(1 −3z)2 \leq7 27.