Question:
Let $\Delta ABC$ be a triangle. For any point $P$ inside or on the boundary of triangle, define $d(P)=\min\{\overline{PA},\overline{PB},\overline{PC}\}$. Find the maximum of $d(P)$ (in terms of side length, angle or coordinate of $A$,$B$,$C$).
The question seems harder than it appears unless I miss something. $d$ might not be differentiable to use calculus techniques. I'll be appreciated if anyone can share references, too.
Only see below if you're really curious about motivations of this problem.
Some motivations of this problem: In an exercise (8.1.8) of Dummit and Foote Abstract Algebra textbook, part of the proof of $\mathbb{Z}[\frac{1+\sqrt{D}}{2}]$ is a Euclidean Domain (with corresponding field norm) when $D=-3,-7,-11$ is the following lemma. Every element of $F=\mathbb{Q}[\sqrt{D}]$ differs from an element in $\mathbb{Z}[\frac{1+\sqrt{D}}{2}]$ by an element whose norm is at most $\frac{(1+|D|)^2}{16|D|}$.
We can use $\mathbb{Z}[\frac{1+\sqrt{D}}{2}]=\{a+b\sqrt{D}\vert 2a\in \mathbb{Z},2b\in \mathbb{Z}, a+b\in \mathbb{Z}\}$ to plot elements of $\mathbb{Z}[\frac{1+\sqrt{D}}{2}]$ on complex plane $\mathbb{C}$, field norm of an element that is difference of two elements is just square of their distance. Therefore it reduces to a special case of the question I post.
note: $PA \le PB \le PC$, when $PA$ get max, we always have $PA=PB$ so we only need to find when $PA=PB$ get max.
it is trivial for the none obtuse triangle that when $P$ is on the $O$. as $PA=PB=PC$
for obtuse triangle, suppose $BC>AB>AC$,$M$ is midpoint of $AB, EM \perp AB $ cross $BC$ at $E$, the point $P$ will be on the $E$.