Chebyshev, after which the inequality is named In mathematics , Chebyshev's sum inequality , named after Pafnuty Chebyshev , states that if
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{\displaystyle a_{1}\geq a_{2}\geq \cdots \geq a_{n}\quad }
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{\displaystyle \quad b_{1}\geq b_{2}\geq \cdots \geq b_{n},}
then
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{\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)\!.}
In words, if we are given two sequences that are both non-increasing or non-decreasing, then the product of their averages is less than the average of their (termwise) product.
Similarly, if
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{\displaystyle a_{1}\leq a_{2}\leq \cdots \leq a_{n}\quad }
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{\displaystyle \quad b_{1}\geq b_{2}\geq \cdots \geq b_{n},}
then
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{\displaystyle {1 \over n}\sum _{k=1}^{n}a_{k}b_{k}\leq \left({1 \over n}\sum _{k=1}^{n}a_{k}\right)\!\!\left({1 \over n}\sum _{k=1}^{n}b_{k}\right)\!.}
[ 1]
Proof
Consider the sum
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{\displaystyle S=\sum _{j=1}^{n}\sum _{k=1}^{n}(a_{j}-a_{k})(b_{j}-b_{k}).}
The two sequences are non-increasing , therefore a j − a k b j − b k j , k S ≥ 0
Opening the brackets, we deduce:
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{\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},}
hence
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{\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)\!.}
An alternative proof is simply obtained with the rearrangement inequality , writing that
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{\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}.}
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
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{\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)}
with the inequality reversed if one is non-increasing and the other is non-decreasing.
