Question

Prove that $a^4+b^4+c^4\ge a^2{b}^2+b^2{c}^2+c^2{a}^2$.

Answer 1

We know A M GM inequality

$a^4+b^4\ge 2a^2{b}^2$

$b^4+c^4\ge 2b^2{c}^2$

$c^4+a^4\ge 2a^2{c}^2$

Add all and divided by 2 we get

$\therefore a^4+b^4+c^4\ge a^2{b}^2+b^2{c}^2+a^2{c}^2$

Hence prove