15. 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 $f\left(x_{1}\right) f\left(x_{2}\right) f\left(x_{3}\right)$.
15. Since $f(x) \geqslant x$, we have $f(1) \geqslant 1$; also, since $f(x) \leqslant \frac{(x+1)^{2}}{4}$, we have $f(1) \leqslant \frac{(1+1)^{2}}{4}=1$, thus $f(1)=1$. Given $f(-1)=0$, we have
$$
\left\{\begin{array}{l}
a+b+c=1 \\
a-b+c=0
\end{array}\right.
$$

Solving these, we get $b=\frac{1}{2}, a+c=\frac{1}{2}$. Considering $f(x) \geq x$, we have $a>0$, and for any $x$, $a x^{2} + (b-1) x + c \geqslant 0$, thus $\triangle=(b-1)^{2}-4 a c \leqslant 0$, which implies $a c \geqslant \frac{1}{16}$. Therefore, $c>0$. By $\sqrt{a c} \leqslant \frac{a+c}{2}=\frac{1}{4}$, we get $a c \leqslant \frac{1}{16}$, so $a c=\frac{1}{16}$, leading to $a=c=\frac{1}{4}$, i.e., $f(x)=\frac{(x+1)^{2}}{4}$.
Since $x_{1}+x_{2}+\dot{x}_{3} \geqslant 3 \cdot \sqrt[3]{x_{1} x_{2} x_{3}}$,
$$
x_{1} x_{2}+x_{1} x_{3}-x_{2} x_{3} \geqslant 3 \cdot \sqrt[3]{\left(x_{1} x_{2} x_{3}\right)^{2}}.
$$

Thus, $\left(x_{1}+1\right)\left(x_{2}+1\right)\left(x_{3}+1\right) \geq\left(\sqrt[3]{x_{1} x_{2} x_{3}}+1\right)^{3}$, and $\frac{\left(x_{1}+1\right)^{2}}{4} \cdot \frac{\left(x_{2}+1\right)^{2}}{4} \cdot \frac{\left(x_{3}+1\right)^{2}}{4} \geq\left[\frac{\left(\sqrt[3]{x_{1} x_{2} x_{3}}+1\right)^{2}}{4}\right]^{3}$. Also, $\sqrt[3]{x_{1} x_{2} x_{3}} \geqslant \frac{3}{\frac{1}{x_{1}}+\frac{1}{x_{2}}+\frac{1}{x_{3}}}=1$. Since $f(x)$ is monotonically increasing on the interval $[-1,+\infty]$, we have $f\left(x_{1}\right) \cdot f\left(x_{2}\right) \cdot f\left(x_{3}\right) \geqslant 1$, with equality if and only if $x_{1}=x_{2}=x_{3}=1$.