Question

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

Answer 1

Since, from the property of triangle inequality, the sum of the lengths of any two sides of a triangle must be greater than or equal to the length of the third side.
In $\Delta ABM$, we have $(AB+BM)>AM$ .......(i)
In $\Delta ACM$, we have $(AC+MC)>AM$ .......(ii)
Now , add the above two inequalities, we get
$AB+BM+AC+MC>AM+AM$
$\therefore AB+BC+AC>2AM$ [Since, $BM+MC=BC$]