Let $P$ be a point in the interior of $\triangle ABC$. The lengths of the sides of $\triangle ABC$ is $a,b,c$, and the distance from $P$ to the sides of $\triangle ABC$ is $p,q,r$. Show that the circumradius $R$ of $\triangle ABC$ satisfies \[\displaystyle R\le \frac{a^2+b^2+c^2}{18\sqrt[3]{pqr}}.\] When does equality hold?
Problem
Source: Taiwan 3rd TST, 1st independent study, problem 1
Tags: inequalities, geometry, circumcircle, geometry proposed