Sum of distances from triangle vertices to interior point is less than perimeter?

Question

Let $M$ be a point in the interior of triangle $ABC$ in the plane. Prove $AM+BM+CM<AB+BC+CA$.

The above question was posed to someone I know who is taking high-school Euclidean geometry. I'm not sure what theorems she can rely on in her proof (though they all follow from Euclid's axioms anyway), but I do know that she does not use trigonometry at all. She turned to me (a mathematician by training) for help; and I can't seem to prove it. So I turn to you all for a proof (using facts from high-school geometry only).

One thing I can prove is that $\sup(AM,BM,CM)<\sup(AB,BC,CA)$. Indeed, say the longest of the interior segments is $\overline{AM}$. Drop an altitude (perpendicular) from $A$ to the point $D\in\overline{BC}$, and consider the side — $\overline{AB}$ or $\overline{AC}$ — such that $\overline{AM}$ lies between $\overline{AD}$ and that side. (If $\overline{AM}\subset\overline{AD}$, consider either $\overline{AB}$ or $\overline{AC}$.) Say it's $\overline{AB}$. Then examining right triangle $ADB$ shows easily that $AM<AB$. However, I can't seem to prove that each of the three sides can be used in turn for one of the interior segments in that proof — which would suffice for the problem above.

Another idea I had was to prove that angle $AMB$ is strictly larger than angle $ACB$ (and likewise for the other two angles) and to use that to prove the claim. But I can't seem to do either: neither to prove the inequality of angle measures, nor, assuming that inequality, to prove the claim sought.

Any help would be much appreciated — again, using high-school geometry only. I suspect there's a simple proof I'm not seeing.

Answer 1

The method is to prove $AC+BC>AM+BM$:

Extend $AM$, let $ME=MB \implies \angle MBE=\angle MEB$, $AM$ cross $BC$ at $F$ (because $M$ is inside of $\triangle ABC$).

If $E$ is on $MF$, then $AF\ge AE=AM+ME$. $BC>CF \implies BC+AC> FC+AC>AF \ge AE=AM+BM$,

If $E$ is on the extension $MF$, $\angle MBE> \angle CBE, \angle BEC>\angle MEB =\angle MBE > \angle CBE$, so in $\triangle CEB$, $BC>CE, \implies BC+AC>AC+CE>AE=AM+BM$.

So we've proven $AC+BC>AM+BM$. For the same reason, $BC+AB>AM+MC$, $AB+AC>BM+MC$.

Finally, we have $AC+BC+AB>AM+BM+CM$.

edit1: more simple way:

$AC +CF> AF =AM +MF, MF + BF > BM \\ \implies AC+CF+BF+MF > AM+BM+MF \\ \implies AC+BC> AM+BM$

Answer 2

The distance from a fixed point is a convex function, and the sum of convex functions gives a convex function. Any triangle is a convex set, and convex functions on convex sets attain their maxima on the boundary. Through this simple principle, if $R$ ranges over a triangle $ABC$ we have

$$ RA+RB+RC \leq AB+AC+BC-\min(AB,AC,BC).$$