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Austra-Poland triangle inequality

Source: Austra-Poland 2004, single competition, problem 2

February 15, 2005
inequalitiesgeometrytrigonometryangle bisectorgeometry solved

Problem Statement

In a triangle ABCABC let DD be the intersection of the angle bisector of γ\gamma, angle at CC, with the side AB.AB. And let FF be the area of the triangle ABC.ABC. Prove the following inequality: 2⋅ F⋅(1AD−1BD)≤AB.2 \cdot \ F \cdot \left( \frac{1}{AD} -\frac{1}{BD} \right) \leq AB.
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