US-TST 2000 Problem 6

Let $ABC$ be a triangle inscribed in a circle of radius $R$ , and let $P$ be a point in the interior of triangle $ABC$ . Prove that $$\frac {PA}{BC^{2}} + \frac {PB}{CA^{2}} + \frac {PC}{AB^{2}}\ge \frac {1}{R}.$$ Alternative formulation: If $ABC$ is a triangle with sidelengths $BC=a$ , $CA=b$ , $AB=c$ and circumradius $R$ , and $P$ is a point inside the triangle $ABC$ , then prove that

$\frac {PA}{a^{2}} + \frac {PB}{b^{2}} + \frac {PC}{c^{2}}\ge \frac {1}{R}$ .