Question 25
Let $ABC$ be a triangle with sides $a$, $b$, $c$, circumradius $R$ and inradius $r$.
Prove that
$$
\frac{R}{2r} \ge \left( \frac{64a^2 b^2 c^2}{(4a^2 - (b-c)^2)(4b^2 - (c-a)^2)(4c^2 - (a-b)^2)} \right)^2
$$
Prove that
$$
\frac{R}{2r} \ge \left( \frac{64a^2 b^2 c^2}{(4a^2 - (b-c)^2)(4b^2 - (c-a)^2)(4c^2 - (a-b)^2)} \right)^2
$$