11. The quadratic function $f(x)=a x^{2}+b x+c(a, b, c \in \mathbf{R}, a \neq 0)$ satisfies the following conditions:
(1) $f(-1)=0$;
(2) For $x \in \mathbf{R}$, $f(x) \geqslant x$;
(3) For $x \in(0,2)$, $f(x) \leqslant \frac{(x+1)^{2}}{4}$.
If $x_{1}, x_{2}, x_{3} \in(0,2)$, and $\frac{1}{x_{1}}+\frac{1}{x_{2}}+\frac{1}{x_{3}}=3$, find the minimum value of $f\left(x_{1}\right) f\left(x_{2}\right) f\left(x_{3}\right)$.
11. Solution: $f(x) \geqslant x, \therefore f(1) \geqslant 1$, and for $x \in(0,2)$, $f(x) \leqslant \frac{(x+1)^{2}}{4}, \therefore f(1) \leqslant 1, \therefore f(1)=1$. Also, $f(-1)=0, \therefore\left\{\begin{array}{l}a+b+c=1 \\ a-b+c=0\end{array}, b=\frac{1}{2}, a+c=\frac{1}{2}\right.$, considering for $x \in \mathbf{R}$, $f(x) \geqslant x$, we have $a>0$, and for any $x, a x^{2}+(b-1) x+c \geqslant 0$, hence $\Delta=(b-1)^{2}-4 a c \leqslant 0$.
$$
\begin{array}{l}
\text { For any } x, a x^{2}+(b-1) x+c \geqslant 0, \text { hence } \Delta=(b-1)^{2}-4 a c \leqslant 0 . \\
\quad \therefore a c \geqslant \frac{1}{16}, \therefore c \geqslant 0, \sqrt{a c} \leqslant \frac{a+c}{2}=\frac{1}{4} \text {, thus } a c \leqslant \frac{1}{16}, \therefore a c=\frac{1}{16}, \therefore a=c=\frac{1}{4}, f(x)=\frac{(x+1)^{2}}{4}, \\
\because x_{1}+x_{2}+x_{3} \geqslant 3 \sqrt[3]{x_{1} x_{2} x_{3}}, \therefore x_{1} x_{2}+x_{2} x_{3}+x_{3} x_{1} \geqslant 3 \sqrt[3]{\left(x_{1} x_{2} x_{3}\right)^{2}}, \\
\therefore\left(x_{1}+1\right)\left(x_{2}+1\right)\left(x_{3}+1\right) \geqslant\left(\sqrt[3]{x_{1} x_{2} x_{3}}+1\right)^{3}, \frac{\left(x_{1}+1\right)^{2}}{4} \cdot \frac{\left(x_{2}+1\right)^{2}}{4} \cdot \frac{\left(x_{3}+1\right)^{2}}{4} \geqslant \\
{\left[\frac{\left(\sqrt[3]{x_{1} x_{2} x_{3}}+1\right)^{2}}{4}\right]^{3}, \text { and } \sqrt[3]{x_{1} x_{2} x_{3}} \geqslant \frac{3}{\frac{1}{x_{1}}+\frac{1}{x_{2}}+\frac{1}{x_{3}}}=1 \text {, and since } f(x) \text { is increasing on }(-1,+\infty), } \\
\therefore f\left(x_{1}\right) f\left(x_{2}\right) f\left(x_{3}\right) \geqslant 1 \text {, with equality if and only if } x_{1}=x_{2}=x_{3}=1.
\end{array}
$$