The distances of a point from the vertices of an equilateral triangle fulfill the triangle inequality

Question

Let ABC be an equilateral triangle and let M be a point that does not lie on the triangle's circumscribed circle.

My question is how can I show that the segments AM, BM and CM can be sides of an triangle?

Hint: Show that AM < BM+CM, BM < AM+CM and CM < AM+BM.

I've used Ptolemy's theorem to show that one of the segments is shorter than the other two segments together, but I can't find a way to prove the same thing for the other two segments.

Answer 1

Let $R_{X}^{\alpha}$ be a rotation of the triangle plane around point $X$ of this plane on the angle $\alpha.$

Let $N$ be a point of the triangle plane such that $R_{C}^{60^{\circ}}\left(\Delta BCN\right)=\Delta ACM.$

Thus, $AM=BN$, $MC=MN$ and we see that $\Delta BMN$ is a triangle with sides-lengths $AM$, $BM$ and $CM$.

Answer 2

A proof (almost) without words: by considering the symmetric of $P$ with respect to the sides of $ABC$, a triangle with side lengths $\sqrt{3}\,PA,\sqrt{3}\,PB,\sqrt{3}\,PC$ arises.

This does not happen iff the symmetric of $P$ with respect to the sides are collinear, i.e. iff $P$ lies on the circumcircle of $ABC$, by Simson's theorem. Equivalently, one may show through Ptolemy's theorem that the side lenghts of the pedal triangle of $P$ are proportional to the distances $PA,PB,PC$.