11. (20 points) Given the function
$$
f(x)=\left(1-x^{2}\right)\left(x^{2}+b x+c\right)(x \in[-1,1]) \text {. }
$$

Let the maximum value of $\mid f(x)$ | be $M(b, c)$. When $b$ and $c$ vary, find the minimum value of $M(b, c)$.
11. Since for any $x \in[-1,1],|f(x)|$ $\leqslant M(b, c)$, thus, taking $x= \pm \lambda, 0$, we get
$$
\begin{array}{l}
\left\{\begin{array}{l}
|c| \leqslant M(b, c), \\
\left|\left(1-\lambda^{2}\right)\left(\lambda^{2}+b \lambda+c\right)\right| \leqslant M(b, c), \\
\left|\left(1-\lambda^{2}\right)\left(\lambda^{2}-b \lambda+c\right)\right| \leqslant M(b, c)
\end{array}\right. \\
\Rightarrow 2\left|\left(1-\lambda^{2}\right)\left(\lambda^{2}+c\right)\right| \leqslant 2 M(b, c) \\
\Rightarrow\left|\left(1-\lambda^{2}\right)\left(\lambda^{2}+c\right)\right| \leqslant M(b, c) \\
\Rightarrow\left|\left(1-\lambda^{2}\right) \lambda^{2}\right| \\
\quad \leqslant\left|\left(1-\lambda^{2}\right)\left(\lambda^{2}+c\right)\right|+\left(1-\lambda^{2}\right)|c| \\
\quad \leqslant\left(2-\lambda^{2}\right) M(b, c) \\
\Rightarrow M(b, c) \geqslant \frac{\left(1-\lambda^{2}\right) \lambda^{2}}{2-\lambda^{2}} \\
\Rightarrow M(b, c) \geqslant\left(\frac{\left(1-\lambda^{2}\right) \lambda^{2}}{2-\lambda^{2}}\right)_{\max }=3-2 \sqrt{2} .
\end{array}
$$

At this point, we can take
$$
M=3-2 \sqrt{2}, b=0, c=2 \sqrt{2}-3 \text {, }
$$

i.e., $\square$
$$
\begin{array}{l}
\left(1-x^{2}\right)\left(x^{2}+2 \sqrt{2}-3\right) \leqslant 3-2 \sqrt{2} \\
\Leftrightarrow\left(x^{2}-2+\sqrt{2}\right)^{2} \geqslant 0 .
\end{array}
$$

Clearly, the equality in the above expression can be achieved.