# Dan Sitaru's Cyclic Inequality in Three Variables VII

### Solution

We have a sequence of equivalent inequalities:

\displaystyle\begin{align} &\frac{5x+3y+z}{5z+3y+x}+\frac{5y+3z+x}{5x+3z+y}+\frac{5z+3x+y}{5y+3x+z}\ge 3\\ &\frac{5x+3y+z}{5z+3y+x}+1+\frac{5y+3z+x}{5x+3z+y}+1+\frac{5z+3x+y}{5y+3x+z}+1\ge 6\\ &\frac{6(x+y+z)}{5z+3y+x}+\frac{6(x+y+z)}{5x+3z+y}+\frac{6(x+y+z)}{5y+3x+z}\ge 6\\ &\frac{x+y+z}{5z+3y+x}+\frac{x+y+z}{5x+3z+y}+\frac{x+y+z}{5y+3x+z}\ge 1. \end{align}

Define $a=5z+3y+x,$ $b=5x+3z+y,$ $c=5y+3x+z.$ $a+b+c=9(x+y+z).$ The last inequality can be rewritten as

$\displaystyle \frac{a+b+c}{a}+\frac{a+b+c}{b}+\frac{a+b+c}{c}\ge 9.$

This is indeed true:

\displaystyle\begin{align} \sum_{cycl}\frac{a+b+c}{a}&=\sum_{cycl}\left(1+\frac{b}{a}+\frac{c}{a}\right)\\ &=3+\sum_{cycl}\left(\frac{b}{a}+\frac{a}{b}\right)\ge 3+6=9, \end{align}

due to the AM-GM inequality $\displaystyle \frac{u}{v}+\frac{v}{u}\ge 2.$

Equality is attained for $a=b=c,$ i.e., $x=y=z.$

### Acknowledgment

Dan Sitaru has kindly posted the problem at the CutTheKnotMath facebook page, with a solution by Rustem Zeynalov. The problem was previously published at the Romanian Mathematical Magazine.