15. Let \(a, b \in [0,1]\), find the maximum and minimum values of \(S = \frac{a}{1+b} + \frac{b}{1+a} + (1-a)(1-b)\). (2006 Shanghai High School Mathematics Competition Problem)
$$\begin{array}{l}
\quad \text { 15. } S=\frac{a}{1+b}+\frac{b}{1+a}+(1-a)(1-b)=\frac{1+a+b+a^{2} b^{2}}{(1+a)(1+b)}=1- \\
\frac{a b(1-a b)}{(1+a)(1+b)} \leqslant 1
\end{array}$$

When $a b=0$ or $a b=1$, the equality holds, so the maximum value of $S$ is 1.
$$\begin{array}{l}
\text { Let } T=\frac{a b(1-a b)}{(1+a)(1+b)}, x=\sqrt{a b} . \\
\quad T=\frac{a b(1-a b)}{(1+a)(1+b)} \leqslant \frac{a b(1-a b)}{(1+\sqrt{a b})^{2}}=\frac{a b(1-a b)}{(1+\sqrt{a b})^{2}}= \\
\quad \frac{a b(1-\sqrt{a b})}{1+\sqrt{a b}}=\frac{x^{2}(1-x)}{1+x}
\end{array}$$

We need to prove
$$\begin{array}{c}
\frac{x^{2}(1-x)}{1+x} \leqslant \frac{5 \sqrt{5}-11}{2} \\
\text { Equation } 1 \Leftrightarrow\left(x-\frac{\sqrt{5}-1}{2}\right)^{2}(x+\sqrt{5}-2) \geqslant 0
\end{array}$$

So $T \leqslant \frac{5 \sqrt{5}-11}{2}$. Therefore, $S \geqslant \frac{11-5 \sqrt{5}}{2}$.
When $a=b=\frac{\sqrt{5}-1}{2}$, the equality holds, so the minimum value of $S$ is $\frac{11-5 \sqrt{5}}{2}$.