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# Chebyshev's sum inequality

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### 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 aj − ak and bj − bk have the same sign for any jk. 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.
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