Prove that three times the perimeter of any triangle is greater than two times the sum of medians.

I think your question is wrong because we know that sum of any two side of a triangle is greater than twice the median with respect to third side,

From here we can say that , if BE and CF are the other median then,

AB+BC>2BE

and AC+BC> 2CF

Now adding all we get,

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

or (AB+BC+CA)>(AD+BE+CF)

Although after this step we can say that if (AB+BC+CA)>(AD+BE+CF) then we can say that 3(AB+BC+CA)>2(AD+BE+CF) will be automatically true.

Hence 3 times perimeter is greater than 2 times sum of medians.

**
**