Chebyshev's sum inequality (original) (raw)

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Mathematical inequality

In mathematics, Chebyshev's sum inequality, named after Pafnuty Chebyshev, states that if

a 1 ≥ a 2 ≥ ⋯ ≥ a n {\displaystyle a_{1}\geq a_{2}\geq \cdots \geq a_{n}\quad } {\displaystyle a_{1}\geq a_{2}\geq \cdots \geq a_{n}\quad } and b 1 ≥ b 2 ≥ ⋯ ≥ b n , {\displaystyle \quad b_{1}\geq b_{2}\geq \cdots \geq b_{n},} {\displaystyle \quad b_{1}\geq b_{2}\geq \cdots \geq b_{n},}

then

1 n ∑ k = 1 n a k b k ≥ ( 1 n ∑ k = 1 n a k ) ( 1 n ∑ k = 1 n b k ) . {\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)\!.} {\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

a 1 ≤ a 2 ≤ ⋯ ≤ a n {\displaystyle a_{1}\leq a_{2}\leq \cdots \leq a_{n}\quad } {\displaystyle a_{1}\leq a_{2}\leq \cdots \leq a_{n}\quad } and b 1 ≥ b 2 ≥ ⋯ ≥ b n , {\displaystyle \quad b_{1}\geq b_{2}\geq \cdots \geq b_{n},} {\displaystyle \quad b_{1}\geq b_{2}\geq \cdots \geq b_{n},}

then

1 n ∑ k = 1 n a k b k ≤ ( 1 n ∑ k = 1 n a k ) ( 1 n ∑ k = 1 n b k ) . {\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)\!.} {\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]

Consider the sum

S = ∑ j = 1 n ∑ k = 1 n ( a j − a k ) ( b j − b k ) . {\displaystyle S=\sum _{j=1}^{n}\sum _{k=1}^{n}(a_{j}-a_{k})(b_{j}-b_{k}).} {\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 and b j − b k have the same sign for any j, k. Hence S ≥ 0.

Opening the brackets, we deduce:

0 ≤ 2 n ∑ j = 1 n a j b j − 2 ∑ j = 1 n a j ∑ j = 1 n b j , {\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},} {\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

1 n ∑ j = 1 n a j b j ≥ ( 1 n ∑ j = 1 n a j ) ( 1 n ∑ j = 1 n b j ) . {\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)\!.} {\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

∑ i = 0 n − 1 a i ∑ j = 0 n − 1 b j = ∑ i = 0 n − 1 ∑ j = 0 n − 1 a i b j = ∑ i = 0 n − 1 ∑ k = 0 n − 1 a i b i + k mod n = ∑ k = 0 n − 1 ∑ i = 0 n − 1 a i b i + k mod n ≤ ∑ k = 0 n − 1 ∑ i = 0 n − 1 a i b i = n ∑ i a i b i . {\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}.} {\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}.}

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

1 b − a ∫ a b f ( x ) g ( x ) d x ≥ ( 1 b − a ∫ a b f ( x ) d x ) ( 1 b − a ∫ a b g ( x ) d x ) {\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)} {\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.

  1. ^ Hardy, G. H.; Littlewood, J. E.; Pólya, G. (1988). Inequalities. Cambridge Mathematical Library. Cambridge: Cambridge University Press. ISBN 0-521-35880-9. MR 0944909.