Question

prove that the sum of the side of a triangle is greater than the sum of its three median

Answer 1

Figure is in the attachment

Let AD,BE & CF be the three medians of a ∆ABC.

WE KNOW THAT THE SUM OF ANY TWO SIDES OF A TRIANGLE IS GREATER THAN TWICE THE MEDIAN DRAWN TO THE THIRD SIDE.

AB+AC>2AD;. AB+BC>2BE & BC+AC>2CF.

Adding We get,
2(AB+BC+AC) >2(AD+BE+CF)

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

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