Question

Prove that $a^4+b^4+c^4>abc(a+b+c)$, where $a,b,c$ are different positive real numbers.

Answer 1

From the AM and GM inequality, we have
$a^4+b^4\ge 2a^2{b}^2$
$b^4+c^4\ge 2b^2{c}^2$
$c^4+a^4\ge 2a^2{c}^2$
adding above inequalities and dividing by $2$, we get
$a^4+b^4+c^4\ge a^2{b}^2+b^2{c}^2+c^2{a}^2......1$
now we repeat the process of $a^2{b}^2,b^2{c}^2$ and $c^2{a}^2$ to get as below
$a^2{b}^2+b^2{c}^2\ge 2b^{2}ac$
$b^2{c}^2+c^2{a}^2\ge 2c^{2}ab$
$c^2{a}^2+a^2{b}^2\ge 2a^{2}bc$
adding the above and dividing by $2$ we get
$a^2{b}^2+b^2{c}^2+c^2{a}^2\ge (b^{2}ac+c^{2}ab+a^{2}bc)\text{or}abc(b+c+a).....2$
from $\left(1\right)$ and $\left(2\right)$ it follows
$a^4+b^4+c^4\ge abc(a+b+c)$