It's a conjecture wich (if true) refine this problem Olympiad Inequality $\sum\limits_{cyc} \frac{x^4}{8x^3+5y^3} \geqslant \frac{x+y+z}{13}$ :
First define for $x>0$ these two funtions :
$$f\left(x\right)=\left(\frac{17}{8}\right)^{-2-x^{3}},g\left(x\right)=\frac{13}{5x^{3}+8}$$
Then it seems we have for $x,y,z>0$ reals numbers :
$$ 0\leq\frac{x}{x+y+z}\left(f\left(\frac{y}{x}\right)-g\left(\frac{y}{x}\right)\right)+\frac{y}{x+y+z}\left(f\left(\frac{z}{y}\right)-g\left(\frac{z}{y}\right)\right)+\frac{z}{x+y+z}\left(f\left(\frac{x}{z}\right)-g\left(\frac{x}{z}\right)\right)+g(1)-f\left(1\right)$$
I have tried Bernouli's inequality without a success . On the other hand it seems also interesting to note that it's much harder to use Buffalo's way .
In the link I provide a proof of the HN_NH inequality and also there is a nice proof due to user Michael Rozenberg .
Some progress :
Using Bernoulli's inequality (not to show it) it seems we have for $x,y,z\in(0,1]$ :
$$\left(\frac{17}{8}\right)^{-2-x^{3}}-\frac{x\left(\frac{17}{8}\right)^{-3}}{x+y+z}\le\frac{\left(\frac{17}{8}\right)^{-2+4}1}{1+\frac{11}{8}\left(x^{3}+4\right)}-\frac{x}{x+y+z}\frac{\left(\frac{17}{8}\right)^{-2+4}1}{1+\frac{11}{8}\left(1+4\right)}$$
It seems also that the inequality is saved using this upper bound .
How to (dis)prove it properly ?
