(17) (This sub-question is worth 20 points)
Given the function $f(x)=\frac{a x^{2}}{e^{x}}$, and the line $y=\frac{1}{e} x$ is a tangent line to the curve $y=f(x)$.
(1) Find the value of the real number $a$;
(II) Let $\min \{m, n\}$ denote the minimum of $m$ and $n$. Define the function $g(x)=\min \left\{f(x), x-\frac{1}{x}\right\}(x>0)$. If the function $h(x)=g(x)-c x^{2}$ is an increasing function, find the range of the real number $c$.
Solving:
(1) Differentiating $f(x)$ yields $f^{\prime}(x)=a \cdot \frac{2 x \cdot e^{x}-x^{2} \cdot e^{x}}{\left(e^{x}\right)^{2}}=a \cdot \frac{x(2-x)}{e^{x}}$,
Assuming the line $y=\frac{1}{e} x$ is tangent to the curve $y=f(x)$ at point $P\left(x_{0}, y_{0}\right)$, then
$$
\left\{\begin{array}{c}
\frac{1}{e} x_{0}=\frac{a x_{0}^{2}}{e^{x_{0}}} \\
\frac{1}{e}=a \cdot \frac{x_{0}\left(2-x_{0}\right)}{e^{x_{0}}}
\end{array},\right.
$$

Solving gives $a=x_{0}=1$. Therefore, the value of $a$ is 1.
(II) Let the function $F(x)=f(x)-\left(x-\frac{1}{x}\right)=\frac{x^{2}}{e^{x}}-x+\frac{1}{x}, x>0$, and we will examine the sign of the function $y=F(x)$. Differentiating $y=F(x)$ yields $F^{\prime}(x)=\frac{x(2-x)}{e^{x}}-1-\frac{1}{x^{2}}, x>0$.
When $x \geqslant 2$, $F^{\prime}(x) < 0, F(2)=\frac{4}{e^{2}}-\frac{3}{2} < 0$; $x \in\left(x_{0},+\infty\right), F(x) < 0$; $x \in\left(0, x_{0}\right), F(x) > 0$,
Thus $h(x)=g(x)-c x^{2}=\left\{\begin{array}{c}x-\frac{1}{x}-c x^{2}, 0 < x < x_{0} \\ x-\frac{1}{x}, x > x_{0}\end{array}\right.$
$\therefore h^{\prime}(x)=\left\{\begin{array}{l}1+\frac{1}{x^{2}}-2 c x, 0 < x < x_{0} \\ 1+\frac{1}{x^{2}}, x > x_{0}\end{array}\right.$
Given that the function $h(x)=g(x)-c x^{2}$ is increasing, and the curve $y=h(x)$ is continuous on $(0,+\infty)$, we know that $h^{\prime}(x) \geqslant 0$ holds on $\left(0, x_{0}\right),\left(x_{0},+\infty\right)$.
(1) When $x > x_{0}$, $\frac{x(2-x)}{e^{x}}-2 c x \geqslant 0$ holds on $\left(x_{0},+\infty\right)$, i.e., $2 c \leqslant \frac{2-x}{e^{x}}$ holds on $\left(x_{0},+\infty\right)$. Let $u(x)=\frac{2-x}{e^{x}}, x > x_{0}$, then $u^{\prime}(x)=\frac{x-3}{e^{x}}, x > x_{0}$,
As $x$ changes, $u^{\prime}(x), u(x)$ change as follows:
\begin{tabular}{|c|c|c|c|}
\hline$x$ & $\left(x_{0}, 3\right)$ & 3 & $(3,+\infty)$ \\
\hline$u^{\prime}(x)$ & - & 0 & + \\
\hline$u(x)$ & decreasing function & minimum value & increasing function \\
\hline
\end{tabular}
$$
\therefore u(x)_{\min }=u(x)_{\text {minimum }}=u(3)=-\frac{1}{e^{3}}.
$$

Therefore, “$2 c \leqslant \frac{2-x}{e^{x}}$ holds on $\left(x_{0},+\infty\right)$” only requires $2 c \leqslant u(x)_{\min }=-\frac{1}{e^{3}}$, i.e., $c \leqslant-\frac{1}{2 e^{3}}$.
(2) When $0 < x < x_{0}$, $1+\frac{1}{x^{2}}-2 c x \geqslant 0$ holds on $\left(0, x_{0}\right)$.
Combining (1) and (2), we know that when $c \leqslant-\frac{1}{2 e^{3}}$, the function $h(x)=g(x)-c x^{2}$ is increasing. Therefore, the range of real number $c$ is $\left(-\infty,-\frac{1}{2 e^{3}}\right]$.