若 $f(x)$ 是区间 $[a,b]$ 上的凹函数，则对任意的 $x_{1},x_{2},...,x_{n} in [a,b]$，且 $sum_{i = 1}^{n}lambda_{i} = 1, lambda_{i} > 0$，有不等式
$$sum_{i = 1}^{n}lambda_{i}f(x_{i}) geq fleft ( sum_{i = 1}^{n}lambda_{i}x_{i} ight )$$
当且仅当 $x_{1} = x_{2} = ... = x_{n}$ 时等号成立。
证明：
证明过程采用数学归纳法。
1）当 $n = 1$ 时，$lambda_{1} = 1$，则不等式左侧为 $f(x_{1})$，不等式右侧为 $f(x_{1})$，不等式显然成立。
2）当 $n = 2$ 时，$lambda_{1} + lambda_{2} = 1$，不等式左侧为 $lambda_{1}f(x_{1}) + lambda_{2}f(x_{2})$，不等式右侧为 $f(lambda_{1}x_{1} + lambda_{2}x_{2})$，参考博客：函数的凹凸性，可知不等式成立。
3）假设 $n = k$ 时，琴生不等式成立，即
$$sum_{i = 1}^{k}lambda_{i}f(x_{i}) geq fleft ( sum_{i = 1}^{k}lambda_{i}x_{i} ight ), ;;;; sum_{i = 1}^{k}lambda_{i} = 1$$
则 $n = k + 1$ 时：
$$sum_{i = 1}^{k+1}lambda_{i}f(x_{i}) = lambda_{k+1}f(x_{k+1}) + sum_{i = 1}^{k}lambda_{i}f(x_{i}) \
= lambda_{k+1}f(x_{k+1}) + C sum_{i=1}^{k}frac{lambda_{i}}{C}f(x_{i})^{k}lambda_{i} \
geq lambda_{k+1}f(x_{k+1}) + Cfleft ( sum_{i=1}^{k}frac{lambda_{i}}{C}x_{i}
ight )$$
其中 $C = sum_{i = 1}^{k}lambda_{i}$，所以 $C = 1 - lambda_{k+1}$。根据凹函数的性质(不懂的话先去阅读上面的博客)有
$$lambda_{k+1}f(x_{k+1}) + Cfleft ( sum_{i=1}^{k}frac{lambda_{i}}{C}x_{i}
ight )
geq fleft ( lambda_{k+1}x_{k+1} + Csum_{i=1}^{k}frac{lambda_{i}}{C}x_{i}
ight ) = fleft ( sum_{i = 1}^{k+1}lambda_{i}f(x_{i})
ight )$$
所以
$$sum_{i = 1}^{k + 1}lambda_{i}f(x_{i}) geq fleft ( sum_{i = 1}^{k + 1}lambda_{i}x_{i} ight )$$
证毕