Let $X=\odot (DEF)\cap \overline{AD}$. Observe that $(A,P;X,D)=-1$, thus we simplify the expression as follows: \begin{align*} AD^2=\left(\frac{AE^2}{AX}\right)\cdot AD=\frac{AE^2\cdot DP}{XP}=\frac{AE^2 \cdot DP^2}{PE\cdot PF} \end{align*}Therefore, the inequality is now equivalent to the following: \begin{align*} \frac{AE^2}{PE\cdot PF}\geq \frac{(AB+AC)^2}{BC^2} \end{align*}However, we have \begin{align*} BC^2 \geq (AB+AC)^2\cdot\sin^2{\tfrac{A}{2}} = (AB+AC)^2 \cdot \frac{EF^2}{4AE^2}\geq (AB+AC)^2 \cdot \frac{PE\cdot PF}{AE^2}, \end{align*}as desired. $\blacksquare$