15、若关于$x$的不等式$e^{x} - \frac{e^{ax}}{x} \leq a - 1 - \frac{\ln x}{x}$恒成立，则实数$a$的取值范围
解：原式可化$xe^{x} - e^{ax} \leq ax - x - \ln x$
$\therefore e^{ax} + ax \geq xe^{x} + (\ln x + x) = e^{\ln x + x} + (\ln x + x)$
设$f(x) = e^{x} + x$，则$f'(x) = e^{x} + 1$
$\therefore f(x)$在$R$上单增
$\therefore ax \geq \ln x + x$
$\therefore a \geq \frac{\ln x + x}{x} = \frac{\ln x}{x} + 1$
设$g(x) = \frac{\ln x}{x} + 1$，则$g'(x) = \frac{1 - \ln x}{x^{2}}$
$\therefore x > e$时，$g(x)$单减，$0 < x < e$时，$g(x)$单增
$\therefore g(x)$极大值$= g(e) = \frac{1}{e} + 1$
$\therefore a \geq \frac{1}{e} + 1$