# An Application of Schur's Inequality II

Dan Sitaru has kindly communicated a problem he invented and solved in collaboration with Leo Giugiuc. Theirs is Proof 1. Proof 2 is by Imad Zak.

### Proof 1

We use Schur's inequality twice:

With $r=1\;$ in the form $\sum x^3+3xyz\ge \sum xy(x+y)\;$ and,

with $r=2\;$ in the form $\sum x^4+xyz\sum x\geq \sum xy(x^2+y^2).\;$

Since $xyz=1$ the two can be rewritten as

$\displaystyle \sum x^3+3\ge \sum \left(\frac{x+y}{z}\right),\\ \displaystyle \sum x^4+\sum x\geq \sum \left(\frac{x^2+y^2}{z}\right).$

$\displaystyle\sum (x^4+x^3+x)+3\ge \sum \left(\frac{x^2+y^2}{z}\right)+\frac{x}{z}+\frac{z}{x}+\frac{y}{x}+\frac{x}{y}+\frac{y}{z}+\frac{z}{y}.$

But $\displaystyle\sum \left(\frac{x}{z}+\frac{z}{x}\right)\ge 2+2+2=6.\;$ It follows that

$\displaystyle\sum (x^4+x^3+x)\ge \sum \left(\frac{x^2+y^2}{z}\right)+6-3$

or

$\displaystyle\sum (x^4+y^3+z)\geq \sum \left(\frac{x^2+y^2}{z}\right)+3.$

### Proof 2

$\displaystyle\sum x^4+xyz\sum x=\sum x^4+\sum x=\sum (x^4+x)\ge\sum xy(x^2+y^2)=\sum\frac{x^2+y^2}{z}.$

Further, by the AM-GM inequality,

$\sum x^3\ge 3xyz=3.$

Adding this to the previous inequality yields the required result. Equlity holds fot $x=y=z=1.$