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}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).

^[1]

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 _{j=1}^{n}b_{j},

hence

{\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

\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

{\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.

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