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

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:

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