AM is a median of ∆ABC. Prove that (AB + BC + CA) > 2AM.

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Solution

Sum of any two sides of a triangle is greater than the third side.

In $△$ AMB:
AB + BM >AM........(i)

In $△$ AMC:
AC + CM >AM.........(ii)

Adding the above two equation:
AB + BM + AC + CM >AM + AM
AB + BC + AC > 2AM

Hence, proved.

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