A Cyclic Inequality in Three Variables XV

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

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

Visual support:

and a contour plot:

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.