Iranian Mathematical Olympiad

Let's define $f(t)=\frac{t}{\sqrt{t^4+t^2+1}}$, that is, $f(t)=-f(-t)$, $f(\frac{1}{t})=f(t)$, furthermore, $|f(t)|\le \frac{1}{\sqrt{3}}$. Then, if we change $(x,y,z)$ with $(\frac{1}{x},\frac{1}{y},\frac{1}{z})$, nothing has been changed. Moreover, one can find that, $z=\frac{x+y-xy}{x+y-1}$, so if $x,y>0$ and $z<0$ then $$\frac{x}{\sqrt{x^4+x^2+1}}+\frac{y}{\sqrt{y^4+y^2+1}}+\frac{z}{\sqrt{z^4+z^2+1}}\ge \frac{-1}{\sqrt{3}}$$ Moreover, if $x,y,z>0$, we are done. Assume now, $x\le y\le 0\le z$, Without loss of generality, we can assume $xy\ge 1$, otherwise, replace $(x,y,z)$ with $(\frac{1}{x},\frac{1}{y},\frac{1}{z})$. If $z=0$, then $0>x+y=xy>0$, a contradiction, hence $z>0$. Now, we can write $$z=\frac{x+y-xy}{x+y-1}=\frac{|x|+|y|+xy}{|x|+|y|+1}\ge 1$$ Hence, $$z-|x|=z+x=\frac{x^2+y}{x+y-1}=\frac{|y|-x^2}{|x+y|+1}$$ Then, $1\le xy\le x^2$ implies that $|x|\ge 1$. Moreover, $x\le y\le 0$ ensures that $|y|\le |x|\le x^2$, hence, $z-|x|\le 0$. Thus, $1\le z\le |x|$. Finally, it is clear that $$f(t)=\frac{1}{\sqrt{t^2+t^{-2}+1}}$$ The function, $t^2+t^{-2}$ is increasing for $|t|\ge 1$, hence $f(t)$ is decreasing on $|t|\ge 1$, therefore $f(|x|)\le f(z)$. That is $$-f(x)=f(-x)=f(|x|)\le f(z)$$ Hence $f(x)+f(z)\ge 0$ and $f(y)\ge -\frac{1}{\sqrt{3}}$, thus $$f(x)+f(y)+f(z)\ge \frac{-1}{\sqrt{3}}.$$