Question

Prove that $a^4+b^4+c^4\ge abc(a+b+c)$.

Answer 1

We know AM, 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

$\therefore a^4+b^4+c^4\ge a^2{b}^2+b^2{c}^2+a^2{c}^2$.........(1)

We know AM, GM inequality

$a^2{b}^2+b^2{c}^2\ge 2b^{2}ac$

$b^2{c}^2+c^2{a}^2\ge 2c^{2}ab$

$c^2{a}^2+a^2{b}^2\ge 2a^{2}bc$

Add all and divided by 2 we get

$a^2{b}^2+b^2{c}^2+c^2{a}^2\ge b^{2}ac+c^{2}ab+a^{2}bc$

$a^2{b}^2+b^2{c}^2+c^2{a}^2\ge abc(b+c+a)$.........(2)

As per (1) and (2) we get

$a^4+b^4+c^4\ge abc(a+b+c)$