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.
 A: 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 $z = 0$.
A: 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.
A: If $x, y, z > 0$, then we see that
$$
\begin{align}
\frac{x}{y} + \frac{ y }{z+x } + \frac{z }{x } - 2 &= \frac{ x^2(z+x) + xy^2 + yz(z+x) - 2xy(z+x)  }{ xy(z+x) } \\
&= \frac{ yz^2 + x(x - y)z + x\left(x^2 - y^2 \right) }{ xy(z+x) }.
\end{align}
$$
Now consider the numerator, which is a quadratic in $z$; it has the discriminant 
$$ 
\begin{align}
x^2(x- y)^2 - 4 xy \left(x^2 - y^2 \right) &= x^4 - 2x^3y + x^2y^2 - 4x^3y + 4xy^3 \\
&= x^4 - 6x^3y + x^2y^2 + 4xy^3 \\
&= 
\end{align}
$$
The idea is to see if the discriminant is negative for all values of $x, y > 0$, or not.
A: I found
$$x^3 - 2 x^2 y + x^2 z + x y^2 - x y z + y z^2 = xy(x-y)^2+x^2(z+x-y)^2+yz^2(y+z) \geqslant 0$$
