First note that $PD\cdot a+PE\cdot b+PF\cdot c\geq 2S$, so $xc\geq 2S$, where $x=PD+PE+PF$. So $x\geq \frac{2S}{c}$, and equality holds when we bring to $C$ and $PF$ becomes the altitude. Note that the diameter of $\triangle ABC$ is $c$, i.e. the distance between any 2 points in the triangle is at most $c$. Also note that by area chasing, $\frac{PD}{DA}+\frac{PE}{EB}+\frac{PF}{FC}=1$, which means $\frac{x}{c}\leq 1$, so $x\leq c$, with equality when $P$ is on side $AB$. Thus $\frac{2S}{c}\leq PD+PE+PF\leq c$.