Chebyshev's sum inequality

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

a_{1}\geq a_{2}\geq \cdots \geq a_{n}

and

b_{1}\geq b_{2}\geq \cdots \geq b_{n},

then

{1 \over n}\sum _{{k=1}}^{n}a_{k}\cdot 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).

Similarly, if

a_{1}\leq a_{2}\leq \cdots \leq a_{n}

and

b_{1}\geq b_{2}\geq \cdots \geq b_{n},

then

{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).

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Proof

Consider the sum

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 $S \geq 0$ .

Opening the brackets, we deduce:

0\leq 2n\sum _{{j=1}}^{n}a_{j}b_{j}-2\sum _{{j=1}}^{n}a_{j}\,\sum _{{k=1}}^{n}b_{k},

whence

\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_{k=1}^n b_k\right).

An alternative proof is simply obtained with the rearrangement inequality, writing that

\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 [0,1], both non-increasing or both non-decreasing, then

\int _{0}^{1}f(x)g(x)\,dx\geq \int _{0}^{1}f(x)\,dx\int _{0}^{1}g(x)\,dx,

with the inequality reversed if one is non-increasing and the other is non-decreasing.

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

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