Point $M$ is the midpoint of side $BC$ of convex quadrilateral $ABCD$. If $\angle{AMD} < 120^{\circ}$. Prove that $$(AB+AM)^2 + (CD+DM)^2 > AD \cdot BC + 2AB \cdot CD.$$
Source: 239 2013 S7
Tags: geometry, Geometric Inequalities
Point $M$ is the midpoint of side $BC$ of convex quadrilateral $ABCD$. If $\angle{AMD} < 120^{\circ}$. Prove that $$(AB+AM)^2 + (CD+DM)^2 > AD \cdot BC + 2AB \cdot CD.$$