Suppose the points $D, E, F$ lie on sides $BC, CA, AB$, respectively, so that $AD, BE, CF$ are angle bisectors. Define $P_1$, $P_2$, $P_3$ respectively as the intersection point of $AD$ with $EF$, $BE$ with $DF$, $CF$ with $DE$ respectively. Prove that $$\frac{AD}{AP_1}+\frac{BE}{BP_2}+\frac{CF}{CP_3} \ge 6$$
Problem
Source: Indonesia INAMO Shortlist 2009 G8 https://artofproblemsolving.com/community/c1101409_
Tags: geometry, geometric inequality, Angle Bisectors, angle bisector