# A Cyclic Inequality in Three Variables XXVI

### Solution 1

For $a \ge 0,$ we $2(a^6+1)\ge (a^4+1)(a^2+1)$ which follows from $(a^2-1)^2(a^2+1)\ge 0.$ Thus, using the AM-GM and AM-HM inequalities,

\displaystyle \begin{align} \sum_{cycl}\frac{(x+y)^4+1}{(x+y)^6+1}\;&\le\sum_{cycl}\frac{2}{(x+y)^2+1}\\ &\le \sum_{cycl}\frac{(1}{x+y}\le\frac{1}{4}\sum_{cycl}\left(\frac{1}{x}+\frac{1}{y}\right)\\ &=\frac{1}{2}\sum_{cycl}\frac{1}{x}. \end{align}

### Solution 2

\displaystyle \begin{align} \frac{1}{2}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)&\;\le\frac{1}{4}\sum_{cycl}\left(\frac{1}{x}+\frac{1}{y}\right)\\ &\overset{AM-HM}{\ge}\sum_{cycl}\frac{1}{x+y}=\sum_{cycl}\frac{2}{2(x+y)}\\ &\overset{AM-GM}{\ge}\sum_{cycl}\frac{2}{(x+y)^2+1}\\ &\quad =~ 2\sum_{cycl}\frac{(x+y)^4+1}{((x+y)^2+1)((x+y)^4+1)}\\ &\overset{\small{Chebyshev}}{\ge}\sum_{cycl}\frac{(x+y)^4+1}{(x+y)^6+1}. \end{align}

### Acknowledgment

Dan Sitaru has kindly posted the problem of his from the Romanian Mathematical Magazine at the CutTheKnotMath facebook page and later commented with several solutions. Solution 1 is by Diego Alvariz; Solution 2 is by Yadamsuren Myagmarsuren. More solutions are available at the problem link.