Stewart's yields $PA^2+PB\cdot PC=\frac{AB^2\cdot PC+AC^2\cdot PB}{BC}$. Using Cauchy-Schwarz, \begin{align*} \left(\frac{AB^2}{PC}+\frac{AC^2}{PB} \right)\left( AB^2\cdot PC+AC^2\cdot PB\right)\geq (AB^2+AC^2)^2=BC^4, \end{align*}which rearranges to $\frac{AB^2}{PC}+\frac{AC^2}{PB} \geq \frac{BC^3}{\frac{AB^2\cdot PC+AC^2\cdot PB}{BC}}=\frac{BC^3}{PA^2 + PB \cdot PC}$, as desired.