(1) 若 $f(x) \leqslant 0$ 恒成立, 求实数 $a$ 的最大值;
(2) 设 $n \in \mathbf{N}^*, n \geqslant 2$, 求证: $\left(1+\frac{1}{3 \sqrt{2}}\right)\left(1+\frac{1}{4 \sqrt{3}}\right)\left(1+\frac{1}{5 \sqrt{4}}\right) \cdots\left[1+\frac{1}{(n+1) \sqrt{n}}\right] < \frac{2}{3} \mathrm{e}^2$.
【答案】 （1） $f(x)=\ln (x+a)-a x$, 定义域 $(-a,+\infty), f^{\prime}(x)=\frac{1}{x+a}-a=\frac{-a x-a^2+1}{x+a}$.

①当 $a=0$ 时, $f^{\prime}(x)=\frac{1}{2}$, 定义域为 $(0,+\infty), f(x)=\ln x$,

②当 $a < 0$ 时, 令 $f^{\prime}(x)=0$, 则 $x=\frac{1-a^2}{a}=\frac{1}{a}-a < -a$,
$f^{\prime}(x)$ 的草图如图（a）所示:

③当 $a > 0$ 时, 令 $f^{\prime}(x)=0$, 则 $x=\frac{1-a^2}{a}=\frac{1}{a}-a > -a$,
$f^{\prime}(x)$ 草图如图 (b) 所示:

（2）在（1）中令 $a=1$, 可得: $\ln (x+1) \leqslant x$,
\begin{aligned} & \text { 从而 } \ln \left(1+\frac{1}{(n+1) \sqrt{n}}\right) < \frac{1}{(n+1) \sqrt{n}}=\frac{n+1-n}{\sqrt{n+1} \cdot \sqrt{n}} \times \frac{1}{\sqrt{n+1}} \\ & =\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n+1} \cdot \sqrt{n}} \times \frac{\sqrt{n+1}+\sqrt{n}}{\sqrt{n+1}}=\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\left(1+\frac{\sqrt{n}}{\sqrt{n+1}}\right) < 2\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right), \\ & \therefore \sum_{i=1}^n \ln \left(1+\frac{1}{(n+1) \sqrt{n}}\right) < 2\left(\frac{1}{1}-\frac{1}{\sqrt{2}}\right)+2\left(\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}\right)+\cdots+2\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right) \\ & =2\left(1-\frac{1}{\sqrt{n+1}}\right) < 2 \\ & \Leftrightarrow\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3 \sqrt{2}}\right)\left(1+\frac{1}{4 \sqrt{3}}\right) \cdots\left(1+\frac{1}{(n+1) \sqrt{n}}\right) < \mathrm{e}^2 \\ & \Leftrightarrow\left(1+\frac{1}{3 \sqrt{2}}\right)\left(1+\frac{1}{4 \sqrt{3}}\right)\left(1+\frac{1}{5 \sqrt{4}}\right) \cdots\left(1+\frac{1}{(n+1) \sqrt{n}}\right) < \frac{2}{3} \mathrm{e}^2 . \end{aligned}