Q.3 AM is a median of a \(\triangle \) ABC.
Is AB + BC + CA > 2 AM?
(Consider the sides of triangles \(\triangle \)ABM and \(\triangle \)AMC.)

We know that,
The sum of the length of any two sides is always greater than the third side.
Now In \(\triangle \)ABM,
Here, AB + BM > AM … [equation i]
Then, In \(\triangle \)ACM
Here, AC + CM > AM … [equation ii]
By adding equation [i] and [ii] we get,
AB + BM + AC + CM > AM + AM
From the figure we have, BC = BM + CM
AB + BC + AC > 2 AM
Hence, the given expression is true.