II. (40 points) Let positive real numbers $a, b, c$ satisfy $a+b+c=ab+bc+ca$.
If $\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c} \geqslant k$ always holds, find the maximum value of $k$.

保留了源文本的换行和格式。
II. The maximum value of $k$ is $\frac{4}{3}$.
Let $c=\frac{1}{2}$. Then $a+b+\frac{1}{2}=a b+\frac{1}{2} b+\frac{1}{2} a \Rightarrow a=\frac{b+1}{2 b-1}$. Let $b \rightarrow \frac{1}{2}\left(b>\frac{1}{2}\right)$. Then $a \rightarrow+\infty$. Hence $\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c} \rightarrow \frac{4}{3}$.
Therefore, $k \leqslant \frac{4}{3}$.
Below is the proof:
$$
\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c} \geqslant \frac{4}{3} \text {. }
$$

Without loss of generality, assume $c$ is the smallest. Then $00$.
And $-(2 c-1)^{2}<0$, so inequality (1) holds.
In conclusion, the maximum value of $k$ is $\frac{4}{3}$.