Conjecture about : $\sum_{cyc}\frac{214x^4}{133x^3 + 81y^3}\ge\frac{x+y+z}{12}+\sum_{cyc}\frac{13x^4}{8x^3+5y^3}-0.25(x-\frac{2yx}{2y+x})\ge x+y+z$

Question

Hi it's a conjecture refinement in the simple case of Prove that $\sum_{\mathrm{cyc}} \frac{214x^4}{133x^3 + 81y^3} \ge x + y + z$ for $x, y, z > 0$ .

Conjecture :

Let $x,y,z>0$ such that $x\geq y \ge z$ then it seems we have :

$$\sum_{cyc}\left(\frac{214x^4}{133x^3 + 81y^3}\right)\geq \frac{x+y+z}{12}+\sum_{cyc}\left(\frac{13x^4}{8x^3+5y^3}-\frac{1}{4}\left(x-\frac{2yx}{2y+x}\right)\right)\geq x+y+z$$

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My attempt :

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I can show the RHS because the function $f(x)=\frac{13}{8+5x^{3}}-0.25\left(1-\frac{2x}{2x+1}\right)-\frac{214}{133+81x^{3}}$ is concave on $(0,\infty)$ also we can use Jensen's inequality .

For the LHS I have tried Buffalo's way without success .

Edit 08/04/2022 :

Using again Jensen inequality I can show :

Let :

$$h\left(x\right)=\frac{2}{1+x^{3}}+\frac{x^{3}+x}{\frac{4}{3}+x^{3}}+\frac{\frac{1}{3}x}{x+2}$$

And :

$$g\left(x\right)=\frac{13}{8+5x^{3}}$$

We have :

$$g''(x)-h''(x)>0$$

So we get a lower bound for the inequality due to user HN_NH (see Olympiad Inequality $\sum\limits_{cyc} \frac{x^4}{8x^3+5y^3} \geqslant \frac{x+y+z}{13}$ )

Next it seems we have for $x,y,z>0$ and $x\geq y\geq z$ :

$$\frac{x}{x+y+z}h\left(\frac{y}{x}\right)+\frac{y}{x+y+z}h\left(\frac{z}{y}\right)+\frac{z}{x+y+z}h\left(\frac{x}{z}\right)-1+g\left(1\right)-h\left(1\right)\ge0$$

I think it's not sufficient to show the conjecture anyway it provides a half solution for HN_NH's inequality .

My question :

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How to deal with it or in other word how to show the inequality/Conjecture ?

Thanks a lot .