已知实数$a$，$b$，$c$满足$e^{a + c} + be^{- c + 1} \leq a + \ln b + 3$，则$a + b - c$的最小值
解：$\because e^{x} \geq x + 1$（$x = 0$时等号成立）
$\therefore e^{a + c} \geq a + c + 1$（$a + c = 0$时等号成立）
$be^{- c + 1} = e^{\ln b}e^{- c + 1} = e^{\ln b - c + 1} \geq \ln b - c + 2$（$\ln b - c + 1 = 0$时等号成立）
$\therefore e^{a + c} + be^{- c + 1} \geq a + \ln b + 3$
$\because e^{a + c} + be^{- c + 1} \leq a + \ln b + 3$
$\therefore e^{a + c} + be^{- c + 1} = a + \ln b + 3$，此时$a + c = 0$，$\ln b - c + 1 = 0$
$\therefore a + b - c = e^{c - 1} - 2c$
设$f(x) = e^{x - 1} - 2x$，$f'(x) = e^{x - 1} - 2$
$\therefore x > \ln 2 + 1$时$f(x)$单增，$x < \ln 2 + 1$时$f(x)$单减
$\therefore f(x)$极小值$= e^{\ln 2} - 2(\ln 2 + 1) = - 2\ln 2$
$\therefore a + b - c \geq - 2\ln 2$