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

In a 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.