Correct Option: B

On the basis of given in question , we draw a figure of triangle ABC whose AD, BE and CF are medians

We know that the sum of any two sides of a triangle is greater than twice the median bisecting the third side.
AD is the median on BC.
∴ AB + AC > 2AD .....( 1 )
Similarly,
AB + BC > 2BE .....( 2 )
BC + AC > 2CF .....( 3 )
On adding all three equations, we get
AB + AC + AB + BC + BC + AC > 2AD + 2BE + 2CF
⇒ 2 (AB + BC + AC) > 2 (AD + BE + CF)
⇒ AB+BC+AC > AD + BE + CF