We have an isosceles triangle $ABC$ with base $BC$. Through vertex $A$ draw a line $r$ parallel to $BC$. The points $P, Q$ are located on the perpendicular bisectors of $AB$ and $AC$ respectively, such that $PQ\perp BC$. They are points $M$ and $N$ on the line $r$ such that $\angle APM = \angle AQN = 90^o$. Prove that $$\frac{1}{AM} + \frac{1}{AN}\le \frac{2}{ AB}$$