An Inequality with Two Cyclic Sums

Problem

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.

Inequalities with the Product of Variables as a Constraint

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An Inequality with Two Cyclic Sums $\left(\displaystyle \sum_{cycl}(a+\sqrt[3]{a}+\sqrt[3]{a^2})\ge 9\sum_{cycl}\frac{1}{1+\sqrt[3]{b^2}+\sqrt[3]{c}}\right)$
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An Inequality with Cycling Sums $\left(\displaystyle\sum_{cycl}(x^4+y^3+z)\ge \sum_{cycl}\frac{x^2+y^2}{z}+3\right)$
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Leo Giugiuc's Cyclic Inequality with a Constraint $\left(\displaystyle a^3+b^3+c^3+\frac{16}{(a+b)(b+c)(c+a)}\ge 5\right)$

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