# A Cyclic Inequality in Three Variables VIII

### Solution

Dividing by $xyz\,$ throughout, we get an equivalent inequality:

$\displaystyle 4\sum_{cycl}\left(\frac{x}{y}+\frac{y}{x}\right)+ 4\sum_{cycl}\frac{1}{\displaystyle\frac{x}{y}+\frac{y}{x}+2} \ge 27.$

We'll show that

$\displaystyle 4\left(\frac{x}{y}+\frac{y}{x}\right)+4\frac{1}{\displaystyle\frac{x}{y}+\frac{y}{x}+2} \ge 9.$

Denote $\displaystyle\alpha=\frac{x}{y}+\frac{y}{x}\ge 2\,$ (by the AM-GM inequality.) In terms of $\alpha\,$ the latter inequality becomes $\displaystyle 4\alpha+\frac{4}{\alpha +2}\ge 9,\,$ which reduces to

$4\alpha^2-\alpha+14=(\alpha-2)(4\alpha+7)\ge 0,$

which is true because $\alpha\ge 2.$

### Acknowledgment

Dan Sitaru has kindly posted the above problem (from the Romanian Mathematical Magazine) at the CutTheKnotMath facebook page, along with a solution by Mihalcea Andrei Stefan.