Question

AB+BC+CA > AD+BE+CF, if the medians AD, BEand CF of ∆ABC intersect at G.​

Answer 1

65
Would be the answer

Answer 2

3(AB+BC+CA)<4(AD+BE+CF)

As G is the centroid,hence G divide AD,BE,CF in the ratio 2:1

BG= ⅔ BE

and

CG = ⅔ CF

Again in△BGC,

BG+CG>BC

⟹ ⅔

BE+2CF>3BCor3BC<2BE+2CF−(1)

Similarly,3CA<2CF+2AD−(2)

3AB<2AD+2BE−(3)

By adding eqn(1),(2)&(3)

3BC+3CA+3AB<2BE+2CF+2CF+2AD+2AD+2BE

⟹3(AB+BC+CA)<4(AD+BE+CF)