Prove that the perimeter of a triangle is greater than the sum of the three medians.

Explanation

Given: In `/_\A B C` in which `A D ,B E` and `C F` are its medians.
To Prove: `A B+B C+A C > A D+B E+C F`
Proof: We know that the sum of two sides of a triangle is always greater than the twice the median drawn on third side.
`AB+AC>2AD` …(1)
`AB+BC>2BE` …(2)
`BC+AC>2CF` ...(3)
Adding equation (1), (2) and (3),we get:
`(AB+AC)+(AB+BC)+(BC+AC)>2AD+2BE+2CF`
`2(AB+BC+AC)>2(AD+BE+CF)`
`AB+BC+AC> AD+BE+CF`