In mathematics, Chebyshev's sum inequality, named after Pafnuty Chebyshev, states that if
and ![{\displaystyle \quad b_{1}\geq b_{2}\geq \cdots \geq b_{n},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b935150896ba3143153bb3aff9680fe4457ff32c)
then
![{\displaystyle {1 \over n}\sum _{k=1}^{n}a_{k}b_{k}\geq \left({1 \over n}\sum _{k=1}^{n}a_{k}\right)\!\!\left({1 \over n}\sum _{k=1}^{n}b_{k}\right)\!.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/18cb6f8b2a870b50e3eebf435015b631bdcc21bd)
Similarly, if
and ![{\displaystyle \quad b_{1}\geq b_{2}\geq \cdots \geq b_{n},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b935150896ba3143153bb3aff9680fe4457ff32c)
then
[1]
Proof
Consider the sum
![{\displaystyle S=\sum _{j=1}^{n}\sum _{k=1}^{n}(a_{j}-a_{k})(b_{j}-b_{k}).}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8ccc5e11d0e1374a0ed66b64c0a7a9ad08867546)
The two sequences are non-increasing, therefore aj − ak and bj − bk have the same sign for any j, k. Hence S ≥ 0.
Opening the brackets, we deduce:
![{\displaystyle 0\leq 2n\sum _{j=1}^{n}a_{j}b_{j}-2\sum _{j=1}^{n}a_{j}\,\sum _{j=1}^{n}b_{j},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/149dd9e37914b12742fd880f76845dfacb002353)
hence
![{\displaystyle {\frac {1}{n}}\sum _{j=1}^{n}a_{j}b_{j}\geq \left({\frac {1}{n}}\sum _{j=1}^{n}a_{j}\right)\!\!\left({\frac {1}{n}}\sum _{j=1}^{n}b_{j}\right)\!.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/da29404006953e3a64a8384643432053da8d0b04)
An alternative proof is simply obtained with the rearrangement inequality, writing that
![{\displaystyle \sum _{i=0}^{n-1}a_{i}\sum _{j=0}^{n-1}b_{j}=\sum _{i=0}^{n-1}\sum _{j=0}^{n-1}a_{i}b_{j}=\sum _{i=0}^{n-1}\sum _{k=0}^{n-1}a_{i}b_{i+k~{\text{mod}}~n}=\sum _{k=0}^{n-1}\sum _{i=0}^{n-1}a_{i}b_{i+k~{\text{mod}}~n}\leq \sum _{k=0}^{n-1}\sum _{i=0}^{n-1}a_{i}b_{i}=n\sum _{i}a_{i}b_{i}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/798c99730453b2096bc3d4c7f79da68c7c4809fa)
Continuous version
There is also a continuous version of Chebyshev's sum inequality:
If f and g are real-valued, integrable functions over [a, b], both non-increasing or both non-decreasing, then
![{\displaystyle {\frac {1}{b-a}}\int _{a}^{b}f(x)g(x)\,dx\geq \!\left({\frac {1}{b-a}}\int _{a}^{b}f(x)\,dx\right)\!\!\left({\frac {1}{b-a}}\int _{a}^{b}g(x)\,dx\right)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/97fd114238df5a948667b6485167a68f10172590)
with the inequality reversed if one is non-increasing and the other is non-decreasing.
See also
Notes
- ^ Hardy, G. H.; Littlewood, J. E.; Pólya, G. (1988). Inequalities. Cambridge Mathematical Library. Cambridge: Cambridge University Press. ISBN 0-521-35880-9. MR 0944909.