# An Inequality with Two Cyclic Sums

### Solution

Let $\sqrt[3]{a}=x,\,$ $\sqrt[3]{b}=y,\,$ $\sqrt[3]{c}=z.\,$ Then still $xyz=1,\,$ $x,y,z\gt 0.\,$

By the Cauchy-Schwarz inequality,

$(1+y^2+z)(x^2+1+z)\ge (x+y+z)^2\ge 3^2=9,$

implying $\displaystyle \frac{9}{1+y^2+z}\le x^2+1+z.\,$ Similarly, $\displaystyle \frac{9}{1+z^2+x}\le y^2+1+x\,$ and $\displaystyle \frac{9}{1+x^2+y}\le z^2+1+y.\,$ Adding the three gives,

\displaystyle \begin{align} 9\sum_{cycl}\frac{1}{1+y^2+x}&\le 3+\sum_{cycl}x+\sum_{cycl}x^2\\ &\le\sum_{cycl}x^3+\sum_{cycl}x+\sum_{cycl}x^2\\ &=\sum_{cycl}a+\sum_{cycl}\sqrt[3]{a}+\sum_{cycl}a^2\\ &=\sum_{cycl}(a+\sqrt[3]{b^2}+c). \end{align}

and this is the required inequality.

### Acknowledgment

This problem from the Romanian Mathematical Magazine has been kindly posted at the CutTheKnotMath facebook page by Dan Sitaru. The above solution is by Nguyen Tien Lam.