AM is a median of a triangle $A B C$ .
Is $A B + B C + C A > 2 A M$ ?
(Consider the sides of triangles $Δ A B M$ and $Δ A B M$ .)

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Solution

Triangle ABM
$A B + B M > A M$ ………. $(1)$ (Sum of $2$ sides $>$ $3^{r d}$ side)
In triangle ACM
$A C + C M > A M$ ………. $(2)$
By adding $1$ & $2$
$(A B + B M) + (C M + A C) > 2 A M$
$A B + A C + (B M + C M) > 2 A M$
$A B + A C + B C > 2 A M$
$∴$ Hence proved.

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