Question

AM is a median of a triangle ABC. Is AB + BC + CA > 2 AM? (Consider the sides of triangles ΔABM and ΔAMC.)

Answer 1

Ina triangle, the sum of the lengths of either two sides is always greater than the third side.
In $ΔABM$,
$AB + BM > AM$ (i)
Similarly, in $ΔACM$,
$AC + CM > AM$ (ii)
Adding equation (i) and (ii),
$AB + BM + MC + AC > AM + AM$
$AB + BC + AC > 2\;AM$

Yes , the given expression is true.