Question 75
Let $ABC$ be a triangle with $\angle A = 90^\circ$ and let $P$ be a point on the hypotenuse $BC$. Prove that
$$
\frac{AB^2}{PC} + \frac{AC^2}{PB} \geq \frac{BC^3}{PA^2 + PB \cdot PC}
$$
$$
\frac{AB^2}{PC} + \frac{AC^2}{PB} \geq \frac{BC^3}{PA^2 + PB \cdot PC}
$$