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

Dear Shreya, 
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.