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

Since, the sum of lengths of any two sides in a triangle should be greater than the length
of third side.
Therefore, In ∆ABM, AB + BM > AM … (i)
In ∆AMC, AC + MC > AM … (ii)
Adding eq. (i) and (ii),
AB + BM + AC + MC > AM + AM
⇒ AB + AC + (BM + MC) > 2AM
⇒ AB + AC + BC > 2AM
Hence, it is true.

Class 7 Maths Chapter 6 Exercise 6.4

for more answers vist to:
https://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-6/