# A Cyclic Inequality in Three Variables XV

### Solution

Consider function $\displaystyle f:\,(0,\infty)\to \mathbb{R};$ defined as

\displaystyle\begin{align} f(x)&=\frac{x^2+x^6}{1+x^6}-x=\frac{x^2+x^6-x-x^7}{1+x^6}\\ &=\frac{x^6(1-x)-x(1-x)}{1+x^6}=\frac{x(1-x)(x^5-1)}{x^6+1}\\ &=\frac{-x(x-1)(x^5-1)}{x^6+1}\\ &=\frac{-x(x-1)^2(x^4+x^3+x^2+x+1)}{x^6+1}\\ &\leq 0. \end{align}

Now,

\displaystyle\begin{align} f\left(\sqrt{\frac{a}{b}}\right)&=\frac{\displaystyle\frac{a}{b}+\frac{a^3}{b^3}}{\displaystyle 1+\frac{a^3}{b^3}}-\sqrt{\frac{a}{b}}\\ &=\frac{a(a^2+b^2)}{a^3+b^3}-\sqrt{\frac{a}{b}}. \end{align}

It follows that $\displaystyle f\left(\sqrt{\frac{a}{b}}\right)\le 0\,$ is equivalent to

$\displaystyle\frac{a(a^2+b^2)}{a^3+b^3}-\sqrt{\frac{a}{b}}\le 0,$

i.e., $\displaystyle\frac{a(a^2+b^2)}{a^3+b^3}\le\sqrt{\frac{a}{b}}.$ The required inequality is nothing but

$\displaystyle f\left(\sqrt{\frac{a}{b}}\right)+f\left(\sqrt{\frac{b}{c}}\right)+f\left(\sqrt{\frac{c}{a}}\right)\le 0.$

### Challenge

Prove that, for $x,y\gt 0,$

$\displaystyle\frac{1+x^2}{1+x^3}+\frac{1+y^2}{1+y^3}+\frac{xy(1+x^2y^2)}{1+x^3y^3}\le 3.$

Visual support:

and a contour plot:

### Acknowledgment

This problem with the solution has been kindly communicated to me by Dan Sitaru. wolframalpha was instrumental in obtaining the 3d plots. N. N. Taleb has kindly supplied enhanced illustration and a printout of Mathemtica's application.