This is known as Erdős-Mordell Inequality, for the proof you can see Nikolaos Dergiades, Signed Distances and the Erdős-Mordell Inequality, Forum Geometricorum, 4 (2004) 67--68. Claudi Alsina and Roger B. Nelsen, A visual proof of the Erdős-Mordell inequality, Forum Geometricorum, 7 (2007) 99--102.

I will present my favorite solution of this theorem since it was not presented in the above papers. Let $D,E,F$ be the foot of perpendicular from $P$ onto $BC,CA,AB$. Let $H,G$ be the foot of perpendicular from $B,C$ onto $EF$. Clearly, $BC \ge HG = HF + FE + EG$. (1) By easy angle chasing, $\triangle BFH \sim \triangle APE$ and $\triangle CEG \sim \triangle APF$. Therefore, we have $HF = \frac{PE \times BF}{PA}, EG = \frac{PF \times CE}{PA}$ (2) By Ptolemy on $\Box AFBE$, we have $PA \times EF = AF \times PE + AE \times PF$. Therefore, $EF = \frac{AF \times PE + AE \times PF}{PA}$ (3) Using (2), (3) on (1), we have $PA \ge \frac{AB}{BC} \times PE + \frac{AC}{BC} \times PF$. Summing this cyclically and using $AM-GM$ proves the problem. $\blacksquare$

Treegoner posted a stronger inequality in the topic The Erdös-Mordell Inequality (see post #3). No solutions yet.