# 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}