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.

By C-S and AM-GM $$\frac{x}{y}+\frac{y}{x+z}+\frac{z}{x}=\frac{x^2}{xy}+\frac{y^2}{xy+yz}+\frac{z^2}{xz}\geq\frac{(x+y+z)^2}{xy+xy+yz+xz}=$$
$$=\frac{x^2+y^2+z^2+2xy+2xz+2yz}{2xy+xz+yz}\geq \frac{4xy+2xz+2yz}{2xy+xz+yz}=2.$$
Also, we can end your work.
Indeed, we need to prove that:
$$yz^2+(x^2-xy)z+x(x-y)^2\geq0,$$ which is true for $x\geq y$.
But for $x<y$ it's enough to prove that:
$$x^2(x-y)^2-4xy(x-y)^2\leq0$$ or
$$x(x-y)^2(x-4y)\leq0,$$ which is obvious.
