The Erdős–Mordell inequality

Let $a,b,c$ be the sides, and $A,B,C$ the angles.

Step 1: $EF=PA\sin A$. Since $\angle AFP=\angle AEP=90^\circ$, $E,F$ lie on the circle with diameter $AP$. The chord $EF$ subtends the inscribed angle $A$, so $EF=AP\sin A$.

Step 2: projection on $BC$. Place $B=(0,0)$, $C=(a,0)$, $A$ above the $x$-axis. The vector $\vec{PF}$ points toward line $AB$, perpendicular to it, so it is $|PF|\,(-\sin B,\cos B)$; $\vec{PE}$ is $|PE|\,(\sin C,\cos C)$. Then $\vec{FE}=\vec{PE}-\vec{PF}$ has $x$-component $|PE|\sin C+|PF|\sin B$. Since $EF\ge$ its projection on any line,
$$EF\ge PE\sin C+PF\sin B.$$

Step 3. Combining with Step 1: $PA\sin A\ge PE\sin C+PF\sin B$, i.e.
$$PA\ge\frac ca\,PE+\frac ba\,PF.$$
By cyclic permutation,
$$PB\ge\frac ab\,PF+\frac cb\,PD,\qquad PC\ge\frac bc\,PD+\frac ac\,PE.$$

Step 4: summing.
$$PA+PB+PC\ge PD\left(\frac cb+\frac bc\right)+PE\left(\frac ca+\frac ac\right)+PF\left(\frac ba+\frac ab\right)\ge2(PD+PE+PF),$$
because $x+\frac1x\ge2$. $\blacksquare$

Equality. $a=b=c$ and $P$ the centre.

Lesson. A segment is longer than its projection; choose the projection direction so that the lengths to compare appear as projections of known perpendiculars.