# 伯努利不等式

${\displaystyle (1+x)^{n}\geq 1+nx}$

${\displaystyle (1+x)^{n}>1+nx\,}$

## 證明和推廣

${\displaystyle (1+x)^{n+1}=(1+x)(1+x)^{n}\geq (1+x)(1+nx)}$
${\displaystyle =1+(n+1)x+nx^{2}\geq 1+(n+1)x}$

${\displaystyle r\leq 0}$${\displaystyle r\geq 1}$，有${\displaystyle (1+x)^{r}\geq 1+rx}$
${\displaystyle 0\leq r\leq 1}$，有${\displaystyle (1+x)^{r}\leq 1+rx}$

${\displaystyle r=0,1}$時，等式顯然成立。

${\displaystyle (-1,\infty )}$上定義${\displaystyle f(x)=(1+x)^{r}-(1+rx)}$，其中${\displaystyle r\neq 0,1}$， 對${\displaystyle x}$求导得${\displaystyle f'(x)=r(1+x)^{r-1}-r}$， 則${\displaystyle f'(x)=0}$當且僅當${\displaystyle x=0}$。分情況討論：

1. ${\displaystyle 0，則對${\displaystyle x>0}$${\displaystyle f'(x)<0}$；對${\displaystyle -1${\displaystyle f'(x)>0}$。因此${\displaystyle f(x)}$${\displaystyle x=0}$時取最大值${\displaystyle 0}$，故得${\displaystyle (1+x)^{r}\leq 1+rx}$
2. ${\displaystyle r<0}$${\displaystyle r>1}$，則對${\displaystyle x>0}$${\displaystyle f'(x)>0}$；對${\displaystyle -1${\displaystyle f'(x)<0}$。因此${\displaystyle f(x)}$${\displaystyle x=0}$時取最小值${\displaystyle 0}$，故得${\displaystyle (1+x)^{r}\geq 1+rx}$

## 相關不等式

${\displaystyle (1+x)^{r}