14. Given $a, b, c \in R^{+}$, satisfying $a b c(a+b+c)=1$,
(I) Find the minimum value of $S=(a+c)(b+c)$;
(II) When $S$ takes the minimum value, find the maximum value of $c$.

保留了源文本的换行和格式。
14. Solution: (I) Since $(a+c)(b+c)=a b+a c+b c+c^{2}=a b+(a+b+c) c=a b+\frac{1}{a b}$ $\geq 2 \sqrt{a b \cdot \frac{1}{a b}}=2$, the equality holds when $a b=1$,
When $a=b=1, c=\sqrt{2}-1$, $S$ can achieve the minimum value of 2.
(II) When $S$ takes the minimum value, $a b=1$, thus $c(a+b+c)=1$,
which means $c^{2}+(a+b) c-1=0$. Let $t=a+b$, then $t \geq 2 \sqrt{a b}=2$
Therefore, $c=\frac{-t+\sqrt{t^{2}+4}}{2}$ or $c=\frac{-t-\sqrt{t^{2}+4}}{2}<0$ (discard)
Hence, $c=\frac{-t+\sqrt{t^{2}+4}}{2}=\frac{2}{\sqrt{t^{2}+4}+t}$ is monotonically decreasing in $t \in[2,+\infty)$,
So when $t=2$, $c$ has the maximum value $\sqrt{2}-1$.