若 $f(x)$ 是區間 $[a,b]$ 上的凹函數，則對任意的 $x_{1},x_{2},...,x_{n} \in [a,b]$，且 $\sum_{i = 1}^{n}\lambda_{i} = 1, \lambda_{i} > 0$，有不等式
$$\sum_{i = 1}^{n}\lambda_{i}f(x_{i}) \geq f\left ( \sum_{i = 1}^{n}\lambda_{i}x_{i} \right )$$
當且僅當 $x_{1} = x_{2} = ... = x_{n}$ 時等號成立。
證明：
證明過程采用數學歸納法。
1）當 $n = 1$ 時，$\lambda_{1} = 1$，則不等式左側為 $f(x_{1})$，不等式右側為 $f(x_{1})$，不等式顯然成立。
2）當 $n = 2$ 時，$\lambda_{1} + \lambda_{2} = 1$，不等式左側為 $\lambda_{1}f(x_{1}) + \lambda_{2}f(x_{2})$，不等式右側為 $f(\lambda_{1}x_{1} + \lambda_{2}x_{2})$，參考博客：函數的凹凸性，可知不等式成立。
3）假設 $n = k$ 時，琴生不等式成立，即
$$\sum_{i = 1}^{k}\lambda_{i}f(x_{i}) \geq f\left ( \sum_{i = 1}^{k}\lambda_{i}x_{i} \right ), \;\;\;\; \sum_{i = 1}^{k}\lambda_{i} = 1$$
則 $n = k + 1$ 時：
$$\sum_{i = 1}^{k+1}\lambda_{i}f(x_{i}) = \lambda_{k+1}f(x_{k+1}) + \sum_{i = 1}^{k}\lambda_{i}f(x_{i}) \\
= \lambda_{k+1}f(x_{k+1}) + C \sum_{i=1}^{k}\frac{\lambda_{i}}{C}f(x_{i})^{k}\lambda_{i} \\
\geq \lambda_{k+1}f(x_{k+1}) + Cf\left ( \sum_{i=1}^{k}\frac{\lambda_{i}}{C}x_{i} \right )$$
其中 $C = \sum_{i = 1}^{k}\lambda_{i}$，所以 $C = 1 - \lambda_{k+1}$。根據凹函數的性質(不懂的話先去閱讀上面的博客)有
$$\lambda_{k+1}f(x_{k+1}) + Cf\left ( \sum_{i=1}^{k}\frac{\lambda_{i}}{C}x_{i} \right )
\geq f\left ( \lambda_{k+1}x_{k+1} + C\sum_{i=1}^{k}\frac{\lambda_{i}}{C}x_{i} \right ) = f\left ( \sum_{i = 1}^{k+1}\lambda_{i}f(x_{i}) \right )$$
所以
$$\sum_{i = 1}^{k + 1}\lambda_{i}f(x_{i}) \geq f\left ( \sum_{i = 1}^{k + 1}\lambda_{i}x_{i} \right )$$
證畢