# 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.

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
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: $$xy+yz+xz > 3$$