We know that,

In a triangle,

The sum of the length of either two sides of the triangle is always greater than the third side

In the given question, we have to show that:

AB + BC + CA > 2 AM

Therefore,

In ABM, we have:

AB + CM > AM (i)

Same is the case with ACM, in which:

AC + CM > AM (ii)

Adding (i) and (ii), we get

AB + BM + MC + AC > AM + AM

AB + BC + AC > 2 AM