16、若不等式$ae^{2x} + x + \ln a \geq \ln x$对任意的$x \in (0, +\infty)$恒成立，则实数$a$的取值范围__
解：$ae^{2x} + x + \ln a \geq \ln x$恒成立
$\therefore e^{\ln a}e^{2x} + x + \ln a \geq \ln x$恒成立
$\therefore e^{\ln a + 2x} + \ln a + 2x \geq e^{\ln x} + \ln x$恒成立
设$f(x) = e^x + x$，显然$f(x)$单增
$\therefore \ln a + 2x \geq \ln x$
$\therefore \ln a \geq \ln x - 2x$
设$g(x) = \ln x - 2x$，则$g'(x) = \frac{1}{x} - 2 = \frac{1 - 2x}{x}$
$\therefore 0 < x < \frac{1}{2}$时$g(x)$单增，$x > \frac{1}{2}$时$g(x)$单减
$\therefore g(x)_{极大值} = g\left( \frac{1}{2} \right) = \ln\frac{1}{2} - 1 = \ln\frac{1}{2e}$
$\therefore \ln a \geq \ln\frac{1}{2e}$
$\therefore a \geq \frac{1}{2e}$