Question

AM is a median of A ABC. Prove that (AB + BC + CA) > 2AM.
Now, add the two inequalities.
(In ∆ ABM)
(In ∆ACM)
Hint. (AB + BM) > AM
(AC +MC) > AM​

Answer 1

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

$\longrightarrow$ AB + BC + AC > 2AM

Hence, proved.