Question

In a $△ABC$, medians $AD,BE$ & $CF$ intersect each other at $G$. Prove that.

$4(AD+BE+CF)>3(AB+BC+CA)$

Answer 1

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

$BG=\frac{2}{3}BEandCG=\frac{2}{3}CF$

Again in$△BGC,$

$BG+CG>BC$

$⟹\frac{2}{3}BE+2CF>3BCor3BC<2BE+2CF-\left(1\right)$

Similarly,$3CA<2CF+2AD-\left(2\right)$

$3AB<2AD+2BE-\left(3\right)$

By adding $eqn\left(1\right),\left(2\right)\&\left(3\right)$

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

$⟹3(AB+BC+CA)<4(AD+BE+CF)$