Proving a point inside a triangle is no further away than the longest side divided by $\sqrt{3}$ Problem:

In a triangle $T$ , all the angles are less than 90 degrees, and the longest side has length $s$. Show that for every point $p$ in $T$ we can pick a corner $h$ in $T$ such that the distance from $p$ to $h$ is less than or equal to $s/\sqrt{3}$.
Source: Problem 2 in http://abelkonkurransen.no/problems/abel_1213_f_prob_en.pdf

Here's my try:
If $p$ is on any of the edges of $T$, it can't be further away than $s/2$. Then I thought that the point furthest away from any $h$ would be a point equidistant to all the vertices. I also know that the equidistant point has to be inside the triangle because no angle is obtuse.
I am assuming an equilateral triangle (I'm not sure if I can). It has no longest side, therefore all must be of length $s$.
Let $h_1,h_2,h_3$ be the vertices, and $z=h_1e=h_2e=h_3e$. To find the equidistant point $e$, I could half all the angles. By trigonometry I will get that, $\cos 30^\circ=\frac{\sqrt{3}}{2}=\frac{s/2}{z} \implies z=\frac{s}{\sqrt{3}}$. Since this is the length from the closest corner to the equidistant point it cannot be further away.

I am not sure if I can rightly assume the triangle is equilateral without loss of generality, probably not. However this is the closest I've gotten to proving this. Could you explain a better approach?
PS. Calculators would not be allowed.

Rephrasing Christian Blatter: for any point interior to an acute triangle, measure the distance to each of the four vertices, and find the smallest value. Now find the point where the this minimum value is largest. Almost obviously, this is the circumcenter. Connect the circumcenter to each of the vertices and drop perpendiculars from the circumcenter to each of the three sides. This divides the original triangle into six right triangles. Any point will be in one of these, and will be closer or equidistant to the vertex than the circumcenter.
