Given real numbers $a, b, c$ satisfy $a+b+c=ab+bc+ca$, try to find
$$
\frac{a}{a^{2}+1}+\frac{b}{b^{2}+1}+\frac{c}{c^{2}+1}
$$

the minimum value.
II. Solution: The minimum value is $-\frac{1}{2}$.
Let $f(a, b, c)=\frac{a}{a^{2}+1}+\frac{b}{b^{2}+1}+\frac{c}{c^{2}+1}=\sum_{c y c} \frac{a}{a^{2}+1}$
If $a=1$, then $b c=1, f \geq \frac{1}{2}-\frac{1}{2}-\frac{1}{2}=-\frac{1}{2}$, when $a=1, b=c=-1$ then $f$ reaches $-\frac{1}{2}$;
If $a=-1$, then $1+b c=2(b+c), f=-\frac{1}{2}+\frac{2(b+c)^{2}}{\left(1+b^{2}\right)\left(1+c^{2}\right)} \geq-\frac{1}{2}$, equality holds when $b+c=0, b c=-1$, i.e., one of $b, c$ is 1, the other is -1.
The discussion for $b=-1$ and $c=-1$ is similar.

Next, we prove that when $a, b, c$ are not -1, $f(a, b, c)=\sum_{c y c} \frac{a}{a^{2}+1} \geq-\frac{1}{2}$. We make the following substitution:
$a=\frac{1-x}{1+x}, b=\frac{1-y}{1+y}, c=\frac{1-z}{1+z}$, substituting into the condition and simplifying, the condition becomes $x+y+z=3 x y z$, at this point $\frac{a}{a^{2}+1}=\frac{1-x^{2}}{2\left(1+x^{2}\right)}=\frac{1}{1+x^{2}}-\frac{1}{2}$, hence $\sum_{c y c} \frac{a}{a^{2}+1}=\sum_{c y c} \frac{1}{1+x^{2}}-\frac{3}{2}$. We need to prove: $\sum_{c y c} \frac{1}{1+x^{2}} \geq 1$.

By Cauchy's inequality, we have $\sum_{c y c} \frac{1}{1+x^{2}} \sum_{c y c}\left(1+x^{2}\right)(y z)^{2} \geq\left(\sum_{c y c} y z\right)^{2}$, thus we only need to prove
$\left(\sum_{c y c} y z\right)^{2} \geq \sum_{c y c}\left(1+x^{2}\right)(y z)^{2}$, and
$$
\left(\sum_{c y c} y z\right)^{2}-\sum_{c y c}\left(1+x^{2}\right)(y z)^{2}=\sum_{c y c} y^{2} z^{2}+2 x y z \sum_{c y c} x-\sum_{c y c} y^{2} z^{2}-3 x^{2} y^{2} z^{2}=3 x^{2} y^{2} z^{2} \geq 0
$$

Equality holds when at least one of $x, y, z$ is 0, without loss of generality, let $x=0$, then $a=1$, thus $b c=1, f \geq \frac{1}{2}-\frac{1}{2}-\frac{1}{2}=-\frac{1}{2}$,

When $a=1, b=c=-1$ then $f$ reaches $-\frac{1}{2}$.
In summary, the minimum value of the original expression is $-\frac{1}{2}$, which is achieved when and only when two of $a, b, c$ are 1 and the other is -1.