It's sufficient to analize situaltion when $X$ is on the boundary and two of numbers $AX,BX,CX$ are equal or when $XA=XB=XC$. In second situation $max(m(X))=1$. To solution this we must to consider some accidents.

Here's a detailed solution: Denote by $O$ the circumcenter of $ABC$. If $O$ lies inside $ABC$ then taking $X=O$, we have $ m(O)=1>\frac{\sqrt{3}}{3}$, impossible. Thus $O$ lies outside $ABC$, which means that $ABC$ has an angle $>90^{\circ}$. Let this angle be $B$. We also have $BC=1$. Consider the perpendicular bisectors of $AB$ and $BC$. Suppose they intersect $AC$ at $A_{1}$ and $C_{1}$ respectively. Let $P,Q$ be the midpoints of $AB$ and $BC$ respectively. Then, if $X$ lies inside the pentagon $BPA_{1}C_{1}Q$, then $m(X)=BX$. If $X$ lies outside this pentagon, that is, in one of the triangles $APA_{1}$ or $CQC_{1}$, then $m(X)=AX$ (or $CX$, respectively). In both cases - when $m(X)=BX$ and when $m(X)=AX$ (or $CX$) - $m(X)$ is maximal when either $X=A_{1}$ or $X=C_{1}$. We have from Sine Theorem and right triangles: $\frac{\frac{1}{2}c}{BA_{1}}=\cos\frac{\pi}{6}$, or $BA_{1}=\frac{1}{\sqrt{3}}c$. and $ CA_{1}=\frac{\frac{1}{2}a}{\cos C}=\frac{1}{2\cos C}$. 1 $\frac{1}{\sqrt{3}}=\frac{2\sin C}{\sqrt{3}}\geq\frac{1}{2\cos C}$. We get $\sin C=\frac{1}{2}$, and since $C$ is not obtuse, $C=30$ and the inequality is satisfied (actually it becomes equality). 2 $\frac{2\sin C}{\sqrt{3}}<\frac{1}{4\cos C}=\frac{1}{\sqrt{3}}$. We get $\cos C=\frac{\sqrt{3}}{4}$, thus $\sin C=\frac{\sqrt{13}}{4}$. But then the inequality isn't satisfied. So, $A=C=30^{\circ}$, $B=120^{\circ}$.