2021/011) Given positive a,b,c show that $a^2(b+c) + b^2(a+c) + c^2(a+b) >= 6abc$

 We have 

LHS $= abc(\frac{a}{c} + \frac{a}{b}) +  abc(\frac{b}{c} + \frac{b}{a})+ abc(\frac{c}{b} + \frac{c}{a})$

$= abc(\frac{a}{c} + \frac{a}{b} + \frac{b}{c} + \frac{b}{a} + \frac{c}{b} + \frac{c}{a})$

$= abc(\frac{a}{c} + \frac{c}{a} + \frac{b}{c} + \frac{c}{b} + \frac{a}{b} + \frac{b}{a})$

as $\frac{a}{c} + \frac{c}{a} \ge 2 $ 

$\frac{b}{c} + \frac{c}{b} \ge 2 $

$\frac{a}{b} + \frac{b}{a} \ge 2 $

The above 3 relations follow from am gm inequality

putting the value we get

LHS $>= abc(2+2+2)$ or LHS $>= 6abc$