<s>[INST] Prove, that for every real numbers $ x \ge y \ge z > 0 $, and $x+y+z=\frac{9}{2}, xyz=1$, the following inequality takes place Prove, that for every real numbers $ x \ge y \ge z > 0  $, and $x+y+z=\frac{9}{2}, xyz=1$, the following inequality takes place: 
$$ \frac{x}{y^3(1+y^2x)} + \frac{y}{z^3(1+z^2y) } + \frac{z}{x^3(1+x^2z)} > \frac{1}{3}(xy+zx+yz) $$
I've tried using the fact that $(xy+yz+zx)^2 \ge xyz(x+y+z) $ or $xy+yz+zx \le \frac{(x+y+z)^2}{3} $
I've also arrived to the fact that the inequality is equivalent to 
$$ \sum_{cyc}{\frac{(xz)^{7/3}}{y^{5/3}(z+y)} > \frac{1}{3}(xy+yz+zx)} $$ which is homogenous. 
I can't seem to find a nice way of using the given conditions for the sum and their order, thank you.
 [/INST] With the two equations $$x+y+z=\frac{9}{2}$$ and $$xyz=1$$ we can express the variables $$x,y$$ by $z$ for instance and your inequality problem reduces to a one variable problem
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