16. Given the functions $f(x)=\frac{\ln x}{x}, g(x)=\mathrm{e}^{x}$.
(1) If the inequality $f(x) \leqslant m x \leqslant g(x)$ holds for all $x$, find the range of the real number $m$;
(2) If $x_{1}>x_{2}>0$, prove that: $\left[x_{1} f\left(x_{1}\right)-x_{2} f\left(x_{2}\right)\right]\left(x_{1}^{2}+x_{2}^{2}\right)>2 x_{2}\left(x_{1}-x_{2}\right)$.
(1) Let $p(x)=\frac{\ln x}{x^{2}}, q(x)=\frac{\mathrm{e}^{x}}{x}$, then $p^{\prime}(x)=\frac{\frac{1}{x} \cdot x^{2}-2 x \ln x}{x^{4}}=\frac{1-2 \ln x}{x^{3}}$, so $p(x)$ is monotonically increasing on $(0, \sqrt{\mathrm{e}})$ and monotonically decreasing on $(\sqrt{\mathrm{e}},+\infty)$, thus $p(x)_{\max }=p(\sqrt{\mathrm{e}})=\frac{1}{2 \mathrm{e}}$; also, $q^{\prime}(x)=\frac{\mathrm{e}^{x}(x-1)}{x^{2}}$, so $q(x)$ is monotonically decreasing on $(0,1)$ and monotonically increasing on $(1,+\infty)$, thus $q(x)_{\text {min }}=q(1)=\mathrm{e}$. In summary, the range of real number $m$ is $\left[\frac{1}{2 \mathrm{e}}, \mathrm{e}\right]$.
(2) $\left(\ln x_{1}-\ln x_{2}\right)\left(x_{1}^{2}+x_{2}^{2}\right)>2 x_{2}\left(x_{1}-x_{2}\right) \Leftrightarrow\left[\left(\frac{x_{1}}{x_{2}}\right)^{2}+1\right] \ln \frac{x_{1}}{x_{2}}>2\left(\frac{x_{1}}{x_{2}}-1\right)$.

Let $t=\frac{x_{1}}{x_{2}}>1, h(t)=\ln t-\frac{2(t-1)}{t^{2}+1}$, then $h^{\prime}(t)=\frac{1}{t}-\frac{2\left(t^{2}+1\right)-4 t(t-1)}{\left(t^{2}+1\right)^{2}}$ $=\frac{(t-1)\left(t^{3}+3 t^{2}+t-1\right)}{t\left(t^{2}+1\right)^{2}}>0$, so $h(t)$ is monotonically increasing $\Rightarrow h(t)>h(1)=0$.

Therefore, $\ln t>\frac{2(t-1)}{t^{2}+1} \Rightarrow\left(t^{2}+1\right) \ln t>2(t-1)$, which means the original inequality holds.