16. (20 points) Given $a>0$, the function
$$
f(x)=\ln x-a(x-1), g(x)=\mathrm{e}^{x} \text {. }
$$
(1) Draw the tangents $l_{1} 、 l_{2}$ to the curves $y=f(x) 、 y=$ $g(x)$ through the origin, respectively. If the slopes of the two tangents are reciprocals of each other, prove: $\frac{\mathrm{e}-1}{\mathrm{e}}<a<\frac{\mathrm{e}^{2}-1}{\mathrm{e}}$;
(2) Let $h(x)=f(x+1)+g(x)$, when $x \geqslant 0$, $h(x) \geqslant 1$ always holds, try to find the range of the real number $a$.
16. (1) Let the tangent line $l_{2}: y=k_{2} x$, with the point of tangency being $\left(x_{2}, y_{2}\right)$.

Then $y_{2}=\mathrm{e}^{x_{2}}, k_{2}=g^{\prime}\left(x_{2}\right)=\mathrm{e}^{x_{2}}=\frac{y_{2}}{x_{2}}$
$\Rightarrow \mathrm{e}^{x_{2}}=\frac{\mathrm{e}^{x_{2}}}{x_{2}} \Rightarrow x_{2}=1, y_{2}=\mathrm{e}$
$\Rightarrow k_{2}=\mathrm{e}$.
By the problem, the slope of the tangent line $l_{1}$ is $k_{1}=\frac{1}{k_{2}}=\frac{1}{\mathrm{e}}$, and the equation is $y=\frac{1}{\mathrm{e}} x$.
Let the point of tangency of $l_{1}$ with the curve $y=f(x)$ be $\left(x_{1}, y_{1}\right)$.
Then $k_{1}=f^{\prime}\left(x_{1}\right)=\frac{1}{x_{1}}-a=\frac{y_{1}}{x_{1}}$
$\Rightarrow y_{1}=\frac{x_{1}}{\mathrm{e}}=1-a x_{1} \Rightarrow a=\frac{1}{x_{1}}-\frac{1}{\mathrm{e}}$.
Also, $y_{1}=\ln x_{1}-a\left(x_{1}-1\right)$, eliminating $y_{1}$, we get $\ln x_{1}-1+\frac{1}{x_{1}}-\frac{1}{\mathrm{e}}=0$.
Let $m(x)=\ln x-1+\frac{1}{x}-\frac{1}{\mathrm{e}}$. Then
$m^{\prime}(x)=\frac{1}{x}-\frac{1}{x^{2}}=\frac{x-1}{x^{2}}$.
Thus, $m(x)$ is monotonically decreasing on the interval $(0,1)$ and monotonically increasing on the interval $(1,+\infty)$.
If $x_{1} \in(0,1)$, by
$m\left(\frac{1}{\mathrm{e}}\right)=-2+\mathrm{e}-\frac{1}{\mathrm{e}}>0, m(1)=-\frac{1}{\mathrm{e}}<0$, which is a contradiction.
In summary, $\frac{\mathrm{e}-1}{\mathrm{e}}<a<2$ when $x \in[0,+\infty)$, and
$$
h^{\prime \prime}(x)=\mathrm{e}^{x}-\frac{1}{(x+1)^{2}}=\frac{(x+1)^{2} \mathrm{e}^{x}-1}{(x+1)^{2}} \geqslant 0,
$$

then $h^{\prime}(x)$ is increasing on the interval $[0,+\infty)$.
Also, $h^{\prime}(0)=2-a<0$, so there exists $x_{0} \in(0,+\infty)$ such that $h^{\prime}\left(x_{0}\right)=0$.

Thus, $h(x)$ is decreasing on the interval $\left(0, x_{0}\right)$ and increasing on the interval $\left(x_{0},+\infty\right)$.

Also, $h\left(x_{0}\right)<h(0)=1$, so $h(x) \geqslant 1$ does not always hold, which does not meet the requirement.
In summary, the range of real number $a$ is $(-\infty, 2]$.