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.

Note: I have found a solution.
Shall we observe that each term of the LHS sum is of the form $\frac{x^4z^4}{y+z}$, the inequality is equivalent to
$$\sum_{cyc}{\frac{x^4z^4}{y+z}} > \frac{1}{3}{(xy+yz+zx)} $$
But from Titu's Lemma, we have
$$  \sum_{cyc}{\frac{x^4z^4}{y+z}} =  \sum_{cyc}{\frac{(x^2z^2)^2}{y+z}}  \ge \frac{({x^2z^2+y^2x^2+z^2y^2})^2}{2(x+y+z)} \ge^{(Quadratic Mean\ge AM)} \frac{(xy+yz+xz)^4}{18(x+y+z)}  $$ 
Hence it suffices to prove $$\frac{(xy+yz+yz)^4}{18(x+y+z)} > \frac{1}{3}{(xy+yz+zx)} $$which is equivalent to $$(xy+yz+xz)^3 > 6(x+y+z)=27 $$
or, equivalently, $$ xy+yz+xz > 3$$ which is true by AM-GM inequality: 
$$xy+yz+xz \ge 3(x^2y^2z^2)^{\frac{1}{3}}=3$$
With the equality case being impossible, since it would imply $x=y=z$, implying both $x=y=z=1$ ( from $xyz=1$) and $x=y=z=\frac{3}{2}$ (from $x+y+z=\frac{9}{2}$), we have only the strict version taking place:  $ xy+yz+xz > 3$
Q. E. D.
