16、设实数$m > 0$，若对任意的$x \in \left( \frac{1}{e^{2}}, + \infty \right)$，不等式$e^{mx} - \frac{\ln x}{m} \geq \frac{1}{m} - \frac{e^{mx}}{mx}$恒成立，则实数$m$的取值范围
解：原式可化为$mxe^{mx} - x\ln x \geq x - e^{mx}$
$\therefore e^{mx}(mx + 1) \geq \ln x(\ln x + 1) = e^{\ln x}(\ln x + 1)$
设$f(x) = e^{x}(x + 1)$，则$f'(x) = e^{x}(x + 2)$
$\therefore x > 0$时$f(x)$单增
$\therefore mx \geq \ln x$
$\therefore m \geq \frac{\ln x}{x}$
设$g(x) = \frac{\ln x}{x}$，则$g'(x) = \frac{1 - \ln x}{x^{2}}$
$\therefore 0 < x < e$时$g(x)$单增，$x > e$时$g(x)$单减
$\therefore g(x)$极大值$= \frac{1}{e}$
$\therefore m \geq \frac{1}{e}$