- 若函数$f(x) = \frac{\left( e^{tx} + 2 \right)tx}{x + 2} - \ln x$，当$x \in \left( 0, + \infty \right)$时，$f(x) > 0$，则实数$t$的取值范围
简析：依题可知$\frac{\left( e^{tx} + 2 \right)tx}{x + 2} > \ln x$恒成立
$\therefore \left( e^{tx} + 2 \right)tx > (x + 2)\ln x$恒成立，只需$e^{tx} > x$且$tx > \ln x$，即$tx > \ln x$恒成立
令$g(x) = (x + 2)\ln x$，则$g'(x) = 1 + \frac{2}{x} + \ln x$
令$h(x) = 1 + \frac{2}{x} + \ln x$，则$h'(x) = \frac{x - 2}{x^{2}}$
$\therefore 0 < x < 2$时$h(x)$单减，$x > 2$时$h(x)$单增
$\therefore h(x)_{\min} = h(2) = 2 + \ln 2 > 0$
$\therefore g'(x) > 0$，$g(x)$单增
$\therefore e^{tx} > x$
$\therefore tx > \ln x$
$\therefore t > \frac{\ln x}{x}$
令$m(x) = \frac{\ln x}{x}$，则$m'(x) = \frac{1 - \ln x}{x^{2}}$
$\therefore 0 < x < e$时$m(x)$单增，$x > e$时$m(x)$单减
$\therefore m(x)_{\max} = m(e) = \frac{1}{e}$
$\therefore t > \frac{1}{e}$