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

Source: 2011 Saudi Arabia Pre-TST January p1

December 31, 2021

geometryright trianglegeometric inequality

Problem Statement

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}