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

${1 \over n}\sum _{{k=1}}^{n}a_{k}b_{k}\leq \left({1 \over n}\sum _{{k=1}}^{n}a_{k}\right)\left({1 \over n}\sum _{{k=1}}^{n}b_{k}\right).$^[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 ≥ 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.

Continuous version

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

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

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

Notes

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.