Solution:
Let our triangle be $ABC$, with side lengths $a$, $b$, $c$, and let $r$, $R$ be the diameters of the circumscribed and inscribed circles, respectively. We want to show $ab > 4 R r$.

The triangle inequality tells us that $a + b > c$.

Heron's Formula tells us that the area of the triangle is $S = s r$ where $s = \frac{a + b + c}{2}$.

The expression $ab$ also occurs in another expression for the area, $S = \frac{ab \sin(C)}{2}$. Hence,
$$ab = 2S / \sin(C) = 2 r s / (c / 2R) = 2 R r (a + b + c) / c > 2 R r (c + c) / c = 4 R r.$$