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Question

AM is a median of a triangle ABC.
IS AB+ BC + CA > 2 AM?
(Consider the sides of triangles ΔABM and ΔAMC

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Solution

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.....(i)
In AMC
AC+MC>AM.....(2)
Adding eqn(1)&(2), we have
(AB+BM)+(AC+MC)>AM+AM
AB+(BM+MC)+AC>2AM
AB+BC+AC>2AB
Hence AB+BC+AC>2AB

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