Inequality with condition $x+y+z=xy+yz+zx$

Question

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.

Answer 1

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.

Answer 2

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 $x^2+y^2+z^2$

Answer 3

It's just $$(x-yz)^2+(y-zx)^2+(z-xy)^2+(x+y+z+1)^2+(x+y+z+xyz)^2\geq0.$$ Done!