Let ABC be the triangle where D, E and F are the mid-points of BC, CA and AB respectively

As, we know that the sum of two sides of the triangle is greater than twice the median bisecting the third side

∴ AB + AC > 2AD

Similarly, BC + AC > 2CF

Also, BC + AB > 2BE

Now, by adding all these we get:

(AB + BC) + (BC + AC) + (BC + AB) > 2AD + 2CD + 2BE

2 (AB + BC + AC) > 2(AD + BE + CF)

∴ AB + BC + AC > AD + BE + CF

Hence, the perimeter of the triangle is greater than the sum of its medians