A point is taken inside the triangle. prove that the sum of the segments from this point to the vertices

A point is taken inside the triangle. prove that the sum of the segments from this point to the vertices is less than the perimeter of the triangle.

Let the point M be taken inside the triangle ABC, and it is required to prove that (AM + BM + CM) <(AB + BC + AC).

Take the inequality of a triangle, in which any of its sides is less than the sum of the two sides: AB <(BC + AC).

The inequality is used: if the point M is inside the triangle ABC, then it is true: MA + MC <AB + BC. Let’s write for each pair of sides:

MA + MC <AB + BC; MB + MA <BC + AC; MB + MC <AB + AC, we add three inequalities, we get:

MA + MC + MB + MA + MC + MB <AB + BC + BC + AC + AB + AC;

2 * (MA + MB + MC) <2 * (AB + BC + AC):

MA + MB + MC <AB + BC + AC.