Correct Option: D

Firstly , We draw a figure of triangle ABC whose the medians AD, BE and CF meet at G ,

In any triangle the sum of the squares of any two sides is equal to twice the square of half of the third side together with twice the square of the median bisecting it.
∴ AB² + AC² = 2(AD² + BD²)
⇒ AB² + AC² = 2(AD² + BC²/4)
⇒ 2(AB² + AC²) = 4 AD² + BC²
Similarly,
2(AB² + BC²) = 4 BE² + AC²
2(AC² + BC²) = 4 CF² + AB²
On adding all three, we get
4(AB² + BC² + AC²) = 4(AD² + BE² + CF²) + BC² + AC² + AB²
⇒ 3(AB² + BC² + AC²) = 4(AD² + BE² + CF²)
Again,
AB + AC > 2AD
AB + BC > 2BE
BC + AC > 2CF
On adding , we get
∴ 2(AB + BC + AC) > 2(AD + BE + CF)
⇒ AB + BC + AC > AD + BE + CF