Problem

Source: 2016 KMO Senior #3

Tags: geometry, geometric inequality, Korea



Acute triangle $\triangle ABC$ has area $S$ and perimeter $L$. A point $P$ inside $\triangle ABC$ has $dist(P,BC)=1, dist(P,CA)=1.5, dist(P,AB)=2$. Let $BC \cap AP = D$, $CA \cap BP = E$, $AB \cap CP= F$. Let $T$ be the area of $\triangle DEF$. Prove the following inequality. $$ \left( \frac{AD \cdot BE \cdot CF}{T} \right)^2 > 4L^2 + \left( \frac{AB \cdot BC \cdot CA}{24S} \right)^2 $$