Inequality with x,y,z fractions $\frac{x}{y}+\frac{y}{z+x}+\frac{z}{x}\ge 2$ If $x,y,z>0$, show:
$$\frac{x}{y}+\frac{y}{z+x}+\frac{z}{x}\ge 2$$
I expand and to prove
$$x^3 - 2 x^2 y + x^2 z + x y^2 - x y z + y z^2\ge 0$$
I don't know how to do this.

This might be pretty cumbersome to prove, if you don't see the trick is to sum $1$ in both sides:
$$\frac{x}{y}+\frac{y}{z+x}+\frac{z}{x} + 1\geq 3$$
And this follows directly from AM-GM:
$$\frac{x}{y}+\frac{y}{z+x}+\left(\frac{z}{x} + 1\right) = \frac{x}{y}+\frac{y}{z+x}+\frac{z+x}{x} \geq 3\sqrt[3]{\frac{x}{y}\cdot \frac{y}{z+x}\cdot \frac{z+x}{x}} = 3$$
with equality when $x=y$ and $z = 0$.
