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

${1\over n} \sum_{k=1}^n a_kb_k \leq \left({1\over n}\sum_{k=1}^n a_k\right)\left({1\over n}\sum_{k=1}^n b_k\right).$^[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:

$0 \leq 2 n \sum_{j=1}^n a_j b_j - 2 \sum_{j=1}^n a_j \, \sum_{k=1}^n b_k,$

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

$\int_0^1 f(x)g(x) \, dx \geq \int_0^1 f(x) \, dx \int_0^1 g(x) \, dx,$

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.