內斯比特不等式

此不等式證明方法很多，例如從平均數不等式我們有：

${\displaystyle {\frac {(a+b)+(a+c)+(b+c)}{3}}\geq {\frac {3}{{\frac {1}{a+b}}+{\frac {1}{a+c}}+{\frac {1}{b+c}}}}}$，

移項得出：

${\displaystyle [(a+b)+(a+c)+(b+c)]\left({\frac {1}{a+b}}+{\frac {1}{a+c}}+{\frac {1}{b+c}}\right)\geq 9}$，

整理左式：

${\displaystyle {\frac {a+b+c}{b+c}}+{\frac {a+b+c}{a+c}}+{\frac {a+b+c}{a+b}}\geq {9 \over 2}}$，

${\displaystyle \left({\frac {a}{b+c}}+1\right)+\left({\frac {b}{a+c}}+1\right)+\left({\frac {c}{a+b}}+1\right)\geq {9 \over 2}}$。

因而不等式得證。