7. Given $x+y+z=xy+yz+zx$, find the minimum value of $\frac{x}{x^{2}+1}$ $+\frac{y}{y^{2}+1}+\frac{z}{z^{2}+1}$.
7. Solution: Take $x=-1, y=-1, z=1$,
the required expression is $-\frac{1}{2}$.
Next, we prove $\frac{x}{x^{2}+1}+\frac{y}{y^{2}+1}+\frac{z}{z^{2}+1} \geqslant-\frac{1}{2}$,
Note that if $x \in \mathbf{R}$, then $\frac{x}{1+x^{2}} \in\left[-\frac{1}{2}, \frac{1}{2}\right]$,
If $x, y, z>0$, then $\frac{x}{x^{2}+1}+\frac{y}{y^{2}+1}+\frac{z}{z^{2}+1} \geqslant$ $-\frac{1}{2}$ holds;

If $x, y, z$ have only one not greater than zero, assume $x \leqslant 0$ $\leqslant y \leqslant z$, then $\frac{y}{y^{2}+1}+\frac{z}{z^{2}+1} \geqslant 0, \frac{x}{x^{2}+1} \geqslant-\frac{1}{2}$, holds;

If $x, y, z$ have only one greater than zero, assume $x \leqslant y \leqslant 0$ $1$, then the condition can be changed to $\frac{1}{x} \times \frac{1}{y}+\frac{1}{y} \times \frac{1}{z}$ $+\frac{1}{z} \times \frac{1}{y}=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}$, and the required $\frac{x}{x^{2}+1}+$ $\frac{y}{y^{2}+1}+\frac{z}{z^{2}+1}=\frac{\frac{1}{x}}{1+\left(\frac{1}{x}\right)^{2}}+\frac{\frac{1}{y}}{1+\left(\frac{1}{y}\right)^{2}}+$ $\frac{\frac{1}{z}}{1+\left(\frac{1}{z}\right)^{2}}$

Still denote $\frac{1}{x}, \frac{1}{y}, \frac{1}{z}$ as $x, y, z$, then $x<0, y$ $<0,0<z<1$, the above proof conclusion still holds,
Assume $\frac{x}{x^{2}+1}+\frac{z}{z^{2}+1} \geqslant 0$, then $\frac{x}{x^{2}+1}+$
$$
\begin{array}{l}
\frac{y}{y^{2}+1}+\frac{z}{z^{2}+1} \geqslant\left(\frac{x}{x^{2}+1}+\frac{z}{z^{2}+1}\right)+\frac{y}{y^{2}+1} \geqslant \\
\frac{y}{y^{2}+1} \geqslant-\frac{1}{2}
\end{array}
$$

Equality holds if and only if $\frac{x}{x^{2}+1}+\frac{z}{z^{2}+1}=0$, $y=-1$, i.e., $x=y=-1, z=1$.