IMO 1967 LL HUN23

Origin: HUN

Problem

Prove that for an arbitrary pair of vectors f and g in the plane, the inequality af 2 + bfg + cg2 \geq0 holds if and only if the following conditions are fulfilled: a \geq0, c \geq0, 4ac \geqb2.

Solution

Suppose that a \geq0, c \geq0, 4ac \geqb2. If a = 0, then b = 0, and the inequality reduces to the obvious cg2 \geq0. Also, if a > 0, then af 2 + bfg + cg2 = a  f + b 2ag 2

• 4ac −b2 4a g2 \geq0. Suppose now that af 2+bfg+cg2 \geq0 holds for an arbitrary pair of vectors f, g. Substituting f by tg (t \inR) we get that (at2 + bt + c)g2 \geq0 holds for any real number t. Therefore a \geq0, c \geq0, 4ac \geqb2.