$$PQR$$ is a right triangle with $$\angle QPR = 90^\circ$$. $$I$$ is the incenter of $$\triangle PQR$$. The incircle touches side $$PQ$$ at $$M$$ and side $$PR$$ at $$N$$. A line is drawn through $$I$$ to cut $$PQ$$ at $$A$$ and $$PR$$ at $$B$$. Prove that $$PA \cdot PB \ge 4r^2$$, where $$r$$ is the inradius of $$\triangle PQR$$.