數學上的切比雪夫總和不等式或切比雪夫不等式以數學家切比雪夫命名,可用以比較兩組數積的和及兩組數的線性和的積的大小: 关于概率论中相似名稱的不等式,请见「切比雪夫不等式」。 若 a 1 ≥ a 2 ≥ ⋯ ≥ a n {\displaystyle a_{1}\geq a_{2}\geq \cdots \geq a_{n}} 且 b 1 ≥ b 2 ≥ ⋯ ≥ b n {\displaystyle b_{1}\geq b_{2}\geq \cdots \geq b_{n}} ,則: n ∑ k = 1 n a k b k ≥ ( ∑ k = 1 n a k ) ( ∑ k = 1 n b k ) ≥ n ∑ k = 1 n a k b n + 1 − k {\displaystyle n\sum _{k=1}^{n}a_{k}b_{k}\geq \left(\sum _{k=1}^{n}a_{k}\right)\left(\sum _{k=1}^{n}b_{k}\right)\geq n\sum _{k=1}^{n}a_{k}b_{n+1-k}} 。 上式也可以寫作 1 n ∑ k = 1 n a k b k ≥ ( 1 n ∑ k = 1 n a k ) ( 1 n ∑ k = 1 n b k ) ≥ 1 n ∑ k = 1 n a k b n + 1 − k {\displaystyle {\frac {1}{n}}\sum _{k=1}^{n}a_{k}b_{k}\geq \left({\frac {1}{n}}\sum _{k=1}^{n}a_{k}\right)\left({\frac {1}{n}}\sum _{k=1}^{n}b_{k}\right)\geq {\frac {1}{n}}\sum _{k=1}^{n}a_{k}b_{n+1-k}} 。 它是由排序不等式而來。 设 a 1 ≥ a 2 ≥ ⋯ ≥ a n {\displaystyle a_{1}\geq a_{2}\geq \cdots \geq a_{n}} 且 b 1 ≥ b 2 ≥ ⋯ ≥ b n {\displaystyle b_{1}\geq b_{2}\geq \cdots \geq b_{n}} ,由排序不等式可知,最大的和为顺序和: a 1 b 1 + ⋯ + a n b n {\displaystyle a_{1}b_{1}+\cdots +a_{n}b_{n}} 于是: a 1 b 1 + ⋯ + a n b n = a 1 b 1 + a 2 b 2 + ⋯ + a n b n {\displaystyle a_{1}b_{1}+\cdots +a_{n}b_{n}=a_{1}b_{1}+a_{2}b_{2}+\cdots +a_{n}b_{n}} a 1 b 1 + ⋯ + a n b n ≥ a 1 b 2 + a 2 b 3 + ⋯ + a n b 1 {\displaystyle a_{1}b_{1}+\cdots +a_{n}b_{n}\geq a_{1}b_{2}+a_{2}b_{3}+\cdots +a_{n}b_{1}} a 1 b 1 + ⋯ + a n b n ≥ a 1 b 3 + a 2 b 4 + ⋯ + a n b 2 {\displaystyle a_{1}b_{1}+\cdots +a_{n}b_{n}\geq a_{1}b_{3}+a_{2}b_{4}+\cdots +a_{n}b_{2}} ⋮ {\displaystyle \vdots } a 1 b 1 + ⋯ + a n b n ≥ a 1 b n + a 2 b 1 + ⋯ + a n b n − 1 {\displaystyle a_{1}b_{1}+\cdots +a_{n}b_{n}\geq a_{1}b_{n}+a_{2}b_{1}+\cdots +a_{n}b_{n-1}} 将这 n {\displaystyle n} 个不等式分边相加,同时对右边进行因式分解,便得到: n ( a 1 b 1 + ⋯ + a n b n ) ≥ ( a 1 + ⋯ + a n ) ( b 1 + ⋯ + b n ) {\displaystyle n(a_{1}b_{1}+\cdots +a_{n}b_{n})\geq (a_{1}+\cdots +a_{n})(b_{1}+\cdots +b_{n})} 两边都除以 n 2 {\displaystyle n^{2}} ,就得到切比雪夫不等式的第一个不等号: a 1 b 1 + ⋯ + a n b n n ≥ a 1 + ⋯ + a n n ⋅ b 1 + ⋯ + b n n {\displaystyle {\frac {a_{1}b_{1}+\cdots +a_{n}b_{n}}{n}}\geq {\frac {a_{1}+\cdots +a_{n}}{n}}\cdot {\frac {b_{1}+\cdots +b_{n}}{n}}} 同理,右边的不等号可由最小的和为逆序和推得。 切比雪夫不等式的积分形式如下: 若 f {\displaystyle f} 和 g {\displaystyle g} 是区间 [ 0 , 1 ] {\displaystyle [0,1]} 上的可积的实函数,并且两者都是递增或两者都是递减的,则: ∫ f g ≥ ∫ f ∫ g {\displaystyle \int fg\geq \int f\int g} 上式可推广到任意区间。 排序不等式 Mathworld: Chebyshev Sum Inequality (页面存档备份,存于互联网档案馆)