AB^2/PC + AC^2/PB >= BC^3/(PA^2 + PB x PC), <BAC=90^o

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AB^2/PC + AC^2/PB >= BC^3/(PA^2 + PB x PC), <BAC=90^o

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#1 h Dec 31, 2021, 8:53 AM

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

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#2 h Feb 20, 2022, 11:55 AM

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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.

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