Given $x,\,y,\,z$ such that $x,\,y,\,z\in [\,1,\,8\,]$$.$ Prove$:$ $$\frac{x}{y}+ \frac{y}{z}+ \frac{z}{x}\geqq \frac{2\,x}{y+ z}+ \frac{2\,y}{z+ x}+ \frac{2\,z}{x+ y}$$ By computer$,$ I have found the generalization of the above inequality $($If you are interested in$)$$:$ Given $x,\,y,\,z$ such that $x,\,y,\,z\in [\,1,\,\text{k}\,]$$.$ $$\frac{x}{y}+ \frac{y}{z}+ \frac{z}{x}\geqq \frac{2\,x}{y+ z}+ \frac{2\,y}{z+ x}+ \frac{2\,z}{x+ y}$$ is true for $\text{k}> 8$ and its largest value is the largest root of irreducible equation$:$ $$4\,\text{k}^{\,17}- 52\,\text{k}^{\,16}+ 192\,\text{k}^{\,15}- 187\,\,\text{k}^{\,14}- 506\,\text{k}^{\,13}- 567\,\,\text{k}^{\,12}- 3460\,\,\text{k}^{\,11}- 1552\,\text{k}^{\,10}+ 4468\,\text{k}^{\,9}+ 1430\,\text{k}^{\,8}- 2812\,\text{k}^{\,7}- 562\,\text{k}^{\,6}+ 728\,\text{k}^{\,5}- 28\,\text{k}^{\,4}- 156\,\text{k}^{\,3}- 19\,\text{k}^{\,2}+ 6\,\text{k}+ 1= 0$$ My hard work is substituting $a= \frac{x}{y},\,b= \frac{x}{z}$ $($Suppose $x= \max\{\,x,\,y,\,z\,\}$$)$$.$ We will have$:$ $$f(\,a,\,b\,)= a+ \frac{b}{a}+ \frac{1}{b}- \frac{2\,ab}{a+ b}- \frac{2\,b}{ab+ a}- \frac{2\,a}{ab+ b}\geqq 0$$ then$:$ $$f(\,a,\,8\,)= a+ \frac{8}{a}+ \frac{1}{b}- \frac{16\,a}{a+ 8}- \frac{16}{9\,a}- \frac{2\,a}{8\,a+ 8}$$ We have$:$ $$f(\,a,\,8\,)\geqq f(\,1,\,8\,)$$ So I tried$:$ $f(\,a,\,b\,)- f(\,a,\,8\,)= (\,8- a\,)\,g(\,a,\,b\,)$$,$ then I continued $g(\,a,\,b\,)- g(\,a,\,8\,)$$,$$...$$,$ but $n(\,1,\,8\,)\leqq 0$$.$ So I can$'$t continue doing it$!$ Can you try to solve my own$?$ Thanks for your interest$!$

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    $\begingroup$ These inequalities have no suprema, which would reduce your equation. You are proving the tangent lie group SU(2), which are reducing will heirarch these rationals under an infinum integer resolved by k. $\endgroup$ – Cppg Mar 23 at 20:44

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