Saudi Arabia Mathematical Competitions

Question

Let ABC be a triangle with A = 90^ and let P be a point on the hypotenuse BC. Prove that AB^2PC + AC^2PB BC^3PA^2 + PB PC FIGURE_0

Accepted Answer

Applying Cauchy-Schwarz inequality we have

\frac{A B ^{2}}{P C} + \frac{A C ^{2}}{P B} = \frac{A B ^{4}}{A B ^{2} \cdot P C} + \frac{A C ^{4}}{A C ^{2} \cdot P B} \geq \frac{( A B ^{2} + A C ^{2} ) ^{2}}{A B ^{2} \cdot P C + A C ^{2} \cdot P B} = \frac{B C ^{4}}{A B ^{2} \cdot P C + A C ^{2} \cdot P B}

From Stewart's Theorem we have

A B^{2} \cdot P C + A C^{2} \cdot P B = P A^{2} \cdot B C + P B \cdot P C \cdot B C

and we get the inequality. We have equality if and only if

\frac{A B ^{2}}{P C \cdot A B ^{2}} = \frac{A C ^{2}}{P B \cdot A C ^{2}},

that is

P B = P C

, hence

P

is the midpoint of segment

B C

.