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Saudi Arabian Mathematical Competitions

Geometry Difficulty 4.4 AIME Prove it Saudi Arabia

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

Solution

Applying Cauchy-Schwarz inequality we have
$\begin{gathered} \frac{AB^2}{PC} + \frac{AC^2}{PB} = \frac{AB^4}{AB^2 \cdot PC} + \frac{AC^4}{AC^2 \cdot PB} \geq \frac{(AB^2 + AC^2)^2}{AB^2 \cdot PC + AC^2 \cdot PB} \\ = \frac{BC^4}{AB^2 \cdot PC + AC^2 \cdot PB} \end{gathered}$

From Stewart's Theorem we have
$AB^2 \cdot PC + AC^2 \cdot PB = PA^2 \cdot BC + PB \cdot PC \cdot BC$
and we get the inequality.
We have equality if and only if
$\frac{AB^2}{PC \cdot AB^2} = \frac{AC^2}{PB \cdot AC^2},$
that is $PB = PC$ , hence $P$ is the midpoint of segment $BC$ .

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.