Erdős-Mordell theorem geometry proof 
Using the notation of the Erdős-Mordell theorem, prove that $PA \cdot PB \cdot PC \geq \dfrac{R}{2r}(p_a+p_b)(p_b+p_c)(p_c+p_a)$.

The notation of the Erdős Mordell theorem means that $p_a$ for example is the distance from the point $p$ to the side $a$ and $R$ is the circumradius. I am struggling to see how to use the product $PA \cdot PB \cdot PC$. We also have to relate this somehow to the circumradius. 
EDIT: Sorry there was a typo originally. The $24$ should've been a $2r$.

In order to prove this inequality, it is important that you first prove the identity: $r = 4R\sin\left(\dfrac{A}{2}\right)\sin\left(\dfrac{B}{2}\right)\sin\left(\dfrac{C}{2}\right)$. Thus using Jensen's inequality: $p_c = PA\sin A_1, p_b = PA\sin A_2, A = A_1+A_2\Rightarrow p_c+p_b = PA(\sin A_1+\sin A_2)\leq 2PA\sin\left(\dfrac{A_1+A_2}{2}\right)=2PA\sin\left(\dfrac{A}{2}\right)\Rightarrow \dfrac{(p_a+p_b)(p_b+p_c)(p_c+p_a)}{PA\cdot PB\cdot PC} \leq 8\sin\left(\dfrac{A}{2}\right)\sin\left(\dfrac{B}{2}\right)\sin\left(\dfrac{C}{2}\right)= 8\left(\dfrac{r}{4R}\right)= \dfrac{2r}{R}$. Thus the problem is solved.
