Let $P$ be a point inside a triangle $A B C$ and let $a, b, c$ be the side lengths and $p$ the semi-perimeter of the triangle. Find the maximum value of

$$
\min \left(\frac{P A}{p-a}, \frac{P B}{p-b}, \frac{P C}{p-c}\right)
$$

over all possible choices of triangle $A B C$ and point $P$.
If $A B C$ is an equilateral triangle and $P$ its center, then $\frac{P A}{p-a}=\frac{P B}{p-b}=\frac{P C}{p-c}=\frac{2}{\sqrt{3}}$.
We shall prove that $\frac{2}{\sqrt{3}}$ is the required value. Suppose without loss of generality that $\angle A P B \geqslant 120^{\circ}$. Then

$$
A B^{2} \geqslant P A^{2}+P B^{2}+P A \cdot P B \geqslant \frac{3}{4}(P A+P B)^{2}
$$

i.e. $P A+P B \leqslant \frac{2}{\sqrt{3}} A B=\frac{2}{\sqrt{3}}((p-a)+(p-b))$, so at least one of the ratios $\frac{P A}{p-a}$ and $\frac{P B}{p-b}$ does not exceed $\frac{2}{\sqrt{3}}$.