# A Cyclic Inequality by Seyran Ibrahimov

### Solution 1

From $x^3+y^3+z^3=3,$ with the AM-GM inequality, we obtain

$3=x^3+y^3+z^3\ge 3\sqrt[3]{x^3y^3z^3}=3xyz,$ such that $xyz\le 1.$

Similarly, $y^4+y^2z^2+z^4\ge 3\sqrt[3]{y^6z^6}=3y^2z^2,$ $z^4+z^2x^2+x^4\ge 3z^2x^2,$ $x^4+x^2y^2+y^4\ge 3x^2y^2.$ It follows that

\displaystyle \begin{align} \sum_{cycl}\frac{x}{y^4+y^2z^2+z^4}&\le \sum_{cycl}\frac{x}{3y^2z^2}=\sum_{cycl}\frac{x^3}{3x^2y^2z^2}\\ &=\frac{1}{3x^2y^2z^2}(x^3+y^3+z^3)=\frac{1}{x^2y^2z^2}=\frac{1}{(xyz)^2}. \end{align}

### Solution 2

Here's another way of using the AM-GM inequality: $y^4+z^4+y^2z^2\ge 2\sqrt{y^4z^4}+y^2z^2=3y^2z^2,$ so that, as in the first solution,

\displaystyle \begin{align} \sum_{cycl}\frac{x}{y^4+y^2z^2+z^4}&\le \sum_{cycl}\frac{x}{3y^2z^2}=\sum_{cycl}\frac{x^3}{3x^2y^2z^2}\\ &=\frac{1}{3x^2y^2z^2}(x^3+y^3+z^3)=\frac{1}{x^2y^2z^2}=\frac{1}{(xyz)^2}. \end{align}

### Solution 3

$\displaystyle \frac{x^3y^2z^2}{y^4+y^2z^2+z^4} =\frac{x^3}{\frac{y^2}{z^2}+\frac{z^2}{y^2}+1} \leq\frac{x^3}{3} ~\text{(AM-GM)}.$

Adding the other two cyclical terms,

$\displaystyle\sum_{cycl}\frac{x^3y^2z^2}{y^4+y^2z^2+z^4}\leq\frac{1}{3}\sum_{cycl}x^3=1.$

Dividing both sides by $(xyz)^2$, we get the inequality to be proved.

### Acknowledgment

Dan Sitaru has kindly posted at CutTheKnotMath facebook page this problem by Seyran Ibrahimov, along with a solution of his own. Solution 2 is by Pankaj Agarwal; Solution 3 is by Amit Itagi.