Problem

Source: Macedonia JBMO TST 2017, Problem 4

Tags: circumcircle, geometry, Geometric Inequalities



In triangle $ABC$, the points $X$ and $Y$ are chosen on the arc $BC$ of the circumscribed circle of $ABC$ that doesn't contain $A$ so that $\measuredangle BAX = \measuredangle CAY$. Let $M$ be the midpoint of the segment $AX$. Show that $$BM + CM > AY.$$