Question

AM is a median of ABC. Prove that (AB + BC +CA) > 2AM. Hint (ABBM) AM On ABM CAC +MC) > AM in L. ACM) Now, add the two inequalities,​

Answer 1

We know,

In a triangle sum of two sides is greater than the third side

Hence,In triangle ABM

AB + BM > AM ____1

Also in triangle ACM

AC + MC > AM ____2

Adding equation 1 and 2 we get,

AB + BM + AC + MC > AM + AM

AB + AC + (BM + MC) > 2AM

AB + AC + BC > 2AM

Answer 2

Given that,

AM is the median of ∆ABC.

In other words, AM divides ABC into two triangles ABM and ACM.

In any triangle, the sum of two sides is greater than the third side.

In ∆ABM,

AB + BM > AM (1)

In ∆ACM,

AC + CM > AM (2)

Adding equations (1) and (2),

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

⟶ AB + BC + AC > 2AM

Hence, proved.