Question

AM is a median of a triangle ABC. Is AB + BC + CA > 2 AM ?

(Consider the sides of triangles ∆ ABM and ∆ AMC.)

Answer 1

Consider the triangle ABM

In ∆ ABM

AB + BM > AM — 1 (Sum of the length of any 2 sides of a triangle is greater that the 3^rd side.)

In ∆ ACM

CM + CA > AM — 2 (Sum of the length of any 2 sides of a triangle is greater than the 3^rd side.)

Add 1 and 2

(AB + BM) + (CM + CA) > AM + AM

AB + (BM + CM) + CA > 2AM

AB + BC + CA > 2AM (yes, it is proved.)