16、若正实数$a$，$b$满足$a\left( \ln b - \ln a + a \right) \geq be^{a - 1}$，则$\frac{1}{ab}$的最小值__
解：原式可化为$\ln b - \ln a + a \geq \frac{b}{a}e^{a - 1} = e^{\frac{\ln b}{a}}e^{a - 1}$
$\therefore \ln b - \ln a + a \geq e^{\frac{\ln b}{a} + \ln a + a - 1}$
令$t = \ln b - \ln a + a$
$\therefore t \geq e^{t - 1}$
设$f(x) = e^{x - 1} - t$，则$f'(x) = e^{x - 1} - 1$
$\therefore t > 1$时$f(t)$单增，$t < 1$时$f(t)$单减
$\therefore f(t)$极小值$= f(1) = 0$
$\therefore e^{t - 1} \geq t$
$\therefore e^{t - 1} = t = 1$
$\therefore \ln b - \ln a + a = 1$
$\therefore \ln b = \ln ae^{1 - a}$
$\therefore b = ae^{1 - a}$
$\therefore \frac{1}{ab} = \frac{1}{a^{2}e^{1 - a}} = \frac{e^{a - 1}}{a^{2}}$
设$g(x) = \frac{e^{x - 1}}{x^{2}}$，则$g'(x) = \frac{e^{x - 1}(x - 2)}{x^{3}}$
$\therefore x > 2$时$g(x)$单增，$x < 2$时$g(x)$单减
$\therefore g(x)$极小值$= g(2) = \frac{e}{4}$
$\therefore \frac{1}{ab} \geq \frac{e}{4}$