Question

AM is a median of a triangle ABC.
IS AB+ BC + CA > 2 AM?
(Consider the sides of triangles $\Delta ABM$ and $\Delta AMC$

Answer 1

As we know that the sum of lengths of any two sides in a triangle should be greater than the length of third side.

Therefore,

In $△ABM$

$AB+BM>AM.....\left(i\right)$

In $△AMC$

$AC+MC>AM.....\left(2\right)$

Adding ${eq}^n\left(1\right)\&\left(2\right)$, we have

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

$\Rightarrow AB+(BM+MC)+AC>2AM$

$\Rightarrow AB+BC+AC>2AB$

Hence $AB+BC+AC>2AB$