Question

Am is a median of a ∆ABC prove that AB+BC+CA >2 AM

Answer 1

Proof

In triangle ABM

AB + BM > 2AM ( 1)

( Sum of two side is greater than third side )

In triangle ACM

AC + CM > AM ( 2 )

( Sum of two side is greater than third side )

From ( 1 ) & ( 2 )

AB + BC + CM + CA > AM + AM

AB + BC + CA > 2AM

Answer 2

$\underline{\mathfrak{\huge{Question:}}}$

AM is the median of /_\ ABC. Prove that AB + BC + CA > 2AM

$\underline{\mathfrak{\huge{Answer:}}}$

You can refer to the attachment for the figure of the question.

In /_\ ABM :-

AB + BM > AM ( Why ? Because the sum of the two sides of a triangle is always greater than the third side, no matter, what type of triangle it is ) ...(1)

In /_\ ACM :-

AC + CM > AM ( Why ? Because the sum of the two sides of a triangle is always greater than the third side, no matter, what type of triangle it is ) ...(2)

Do : (1) + (2) ;-

=》 AB + BM + AC + CM > AM + AM

Now, we know from the figure that :-

BM + CM = BC

=》 AB + BM + CM + AC > AM + AM

=》 AB + BC + AC > 2AM

$\boxed{\tt{Hence\:Proved!!}}$