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Geometry Difficulty 4.4 AIME Prove it Saudi Arabia

Let ABCABC be a triangle with ∠A=90∘\angle A = 90^\circ and let PP be a point on the hypotenuse BCBC. Prove that
AB2PC+AC2PB≥BC3PA2+PB⋅PC \frac{AB^2}{PC} + \frac{AC^2}{PB} \geq \frac{BC^3}{PA^2 + PB \cdot PC}

Solution

Applying Cauchy-Schwarz inequality we have
AB2PC+AC2PB=AB4AB2⋅PC+AC4AC2⋅PB≥(AB2+AC2)2AB2⋅PC+AC2⋅PB=BC4AB2⋅PC+AC2⋅PB \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}
Figure 1
From Stewart's Theorem we have
AB2⋅PC+AC2⋅PB=PA2⋅BC+PB⋅PC⋅BC 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
AB2PC⋅AB2=AC2PB⋅AC2, \frac{AB^2}{PC \cdot AB^2} = \frac{AC^2}{PB \cdot AC^2},
that is PB=PCPB = PC, hence PP is the midpoint of segment BCBC.

Figure 1

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Source: MathNet, licensed CC-BY-4.0. Statement and solution reproduced as published; topic and difficulty added by this site.