46. Given that $x, y, z$ are real numbers, and satisfy $x+y+z=xy+yz+zx$, find the minimum value of $\frac{x}{x^{2}+1}+\frac{y}{y^{2}+1}+$ $\frac{z}{z^{2}+1}$. (2008 Brazilian Mathematical Olympiad Problem)
46. Let $x=1, y=z=-1$. Then $\frac{x}{x^{2}+1}+\frac{y}{y^{2}+1}+\frac{z}{z^{2}+1}=-\frac{1}{2}$. Conjecture that the minimum value of $\frac{x}{x^{2}+1}+$ $\frac{y}{y^{2}+1}+\frac{z}{z^{2}+1}$ is $-\frac{1}{2}$.

It is sufficient to prove
$$\frac{x}{x^{2}+1}+\frac{y}{y^{2}+1}+\frac{z}{z^{2}+1} \geqslant-\frac{1}{2} \Leftrightarrow \frac{x}{x^{2}+1}+\frac{y}{y^{2}+1} \geqslant \frac{(z-1)^{2}}{z^{2}+1}$$

Notice that
$$z(x+y-1)=x+y-x y$$

If $x+y-1=0$, then $x+y=x y=1$, which is a contradiction. Hence, $x+y-1 \neq 0$. Therefore, $z=\frac{x+y-x y}{x+y-1}$.
Substituting into equation (1) gives
$$\frac{x}{x^{2}+1}+\frac{y}{y^{2}+1} \geqslant \frac{(x+y-1)^{2}}{(x+y-1)^{2}+(x+y-x y)^{2}}$$

By the Cauchy-Schwarz inequality, we have
$$\frac{x}{x^{2}+1}+\frac{y}{y^{2}+1} \geqslant \frac{[(1+x)(1-y)+(1+y)(1-x)]^{2}}{\left(1+x^{2}\right)(1-y)^{2}+\left(1+y^{2}\right)(1-x)^{2}}$$

Thus, it is sufficient to prove
$$\begin{array}{l}
4(x+y-1)^{2}+4(x+y-x y)^{2} \geqslant\left(1+x^{2}\right)(1-y)^{2}+\left(1+y^{2}\right)(1-x)^{2} \Leftrightarrow \\
f(x)=\left(y^{2}-3 y+3\right) x^{2}-\left(3 y^{2}-8 y+3\right) x+3 y^{2}-3 y+1 \geqslant 0 \\
\quad \text { By } \Delta=\left(3 y^{2}-8 y+3\right)^{2}-4\left(y^{2}-3 y+3\right)\left(3 y^{2}-3 y+1\right)=-3\left(y^{2}-1\right)^{2} \leqslant
\end{array}$$
0, hence $f(x) \geqslant 0$ always holds. Therefore, the conjecture is true, i.e., $\frac{x}{x^{2}+1}+\frac{y}{y^{2}+1}+\frac{z}{z^{2}+1} \geqslant-\frac{1}{2}$.