Inequality with condition $x+y+z=xy+yz+zx$ I'm trying to prove the following inequality:
For $x,y,z\in\mathbb{R}$ with $x+y+z=xy+yz+zx$, prove that
$$
\frac{x}{x^2+1}+\frac{y}{y^2+1}+\frac{z}{z^2+1}\ge-\frac{1}{2}
$$
My approach:
After slight manipulation the inequality is equivalent to:
$$
\sum_{cyc}\frac{(x+1)^2}{x^2+1}\ge 2
$$
Now, applying CS is legitimate and it reduces the inequality to proving:
$$
s^2-10s-3\le0
$$
with $s=x+y+z=xy+yz+zx$, but I'm not quite sure if this is still true. Could anybody give me a hint in the right direction? Any help is highly appreciated.
 A: Here is a way to use CS to solve the inequality.  First, we re-write what we want to prove as
$$\frac{(x+1)^2}{x^2+1}+\frac{(y+1)^2}{y^2+1} \ge 2-\frac{(z+1)^2}{z^2+1} = \frac{(z-1)^2}{z^2+1}$$
Now the constraint gives $z = \dfrac{x+y-xy}{x+y-1}$.  Using this, we need to only show
$$\frac{(x+1)^2}{x^2+1}+\frac{(y+1)^2}{y^2+1} \ge \frac{(xy-1)^2}{(x+y-xy)^2+(x+y-1)^2}$$
Using CS on the LHS, we have
$$\left(\frac{(x+1)^2}{x^2+1}+\frac{(y+1)^2}{y^2+1}\right)\left((x^2+1)(y-1)^2+(y^2+1)(x-1)^2 \right) \ge \left((x+1)(y-1)+(y+1)(x-1) \right)^2=4(xy-1)^2$$
Thus it is sufficient to show
$$4(x+y-xy)^2+4(x+y-1)^2 \ge (x^2+1)(y-1)^2+(y^2+1)(x-1)^2$$
which reduces to showing that the following quadratic (in say $x$) is non-negative:
$$(y^2-3y+3)x^2 - (3 y^2-8 y+3)x + (3y^2-3y+1) \ge 0$$
which is easy to show as its discriminant, $-3(y^2-1)^2$ is never positive.
A: It's just $$(x-yz)^2+(y-zx)^2+(z-xy)^2+(x+y+z+1)^2+(x+y+z+xyz)^2\geq0.$$
Done!
A: Hint: $(x+y+z)^2=x^2+y^2+z^2+2(xy+yz+xz)$ 
Hint 2: this is a quadratic equation which places constraints on $x+y+z$ and $x^2+y^2+z^2$
