Problem

Source:

Tags: geometry, geometric inequality, inequalities



Let $M$ be the midpoint of segment $BC$ of $\triangle ABC$. Let $D$ be a point such that $AD=AB$, $AD\perp AB$ and points $C$ and $D$ are on different sides of $AB$. Prove that: $$\sqrt{AB\cdot AC+BC\cdot AM}\geq\frac{\sqrt{2}}{2}CD.$$