an interesting inequality with Muirhead If $x,y,z>0$ I have to prove that
$\sum\limits_{cyc}^{} \frac { x(x^3 yz+x^2-x y^3 z-yz) }{(1+x y^2)(1+xyz)} \ge 0$ holds. My approach is that from Muirhead's inequality the inequality is true since $(4,1,1)≻(2,3,1)$ and $(2,1)≻(1,1)$. Am I right?

We need to prove that
$$\sum_{cyc}\frac{x(x^2(1+xyz)-yz(1+xy^2)}{(1+xy^2)(1+xyz)}\geq0$$ or
$$\sum_{cyc}\frac{x^3}{1+xy^2}\geq\frac{3xyz}{1+xyz}.$$
Now, by C-S
$$\sum_{cyc}\frac{x^3}{1+xy^2}=\sum_{cyc}\frac{x^4}{x+x^2y^2}\geq\frac{(x^2+y^2+z^2)^2}{\sum\limits_{cyc}(x+x^2y^2)}.$$
Id est, it's enough to prove that 
$$(1+xyz)(x^2+y^2+z^2)\geq3xyz\sum_{cyc}(x+x^2y^2),$$ which is true because
$$(x^2+y^2+z^2)^2\geq3xyz(x+y+z)$$ and
$$xyz(x^2+y^2+z^2)^2\geq3xyz(x^2y^2+x^2z^2+y^2z^2).$$
Indeed, we can say that both last inequalities are true by Muirhead. 
