Prove that the perimeter of a triangle is greater than the sum of its three medians. In the` triangle ABC`, D,E and F are the midpoints of sides BC,CA and AB respectively.

Explanation

"We know that the sum of two sides of a triangle is greater than twice the median bisecting the third side.
Hence In `triangle ABD`, AD is a median
`AB+AC>2(AD)`_____1
Similarly, we get`BC+AC>2(CF)`______2
`BC+AB>2(BE)`____3
On adding 1,2 and 3 equation, we get
`(AB+AC)+(BC+AC)(BC+AB)>2AD+2CF+2BE`
`2(AB+BC+AC)>2(AD+BE+CF)`

`thereforeAB+BC+AC>AD+BE+CF`
Hence, The perimeter of a triangle is greater than sum of its three medians."