Chebyshev's Sum Inequality

In mathematics, Chebyshev's sum inequality, named after Pafnuty Chebyshev, states that if
:$a_1 \geq a_2 \geq \cdots \geq a_n \quad$ and $\quad b_1 \geq b_2 \geq \cdots \geq b_n,$
then
:$\sum_^n a_k b_k \geq \left(\sum_^n a_k\right)\!\!\left(\sum_^n b_k\right)\!.$

Similarly, if
:$a_1 \leq a_2 \leq \cdots \leq a_n \quad$ and $\quad b_1 \geq b_2 \geq \cdots \geq b_n,$
then
:$\sum_^n a_k b_k \leq \left(\sum_^n a_k\right)\!\!\left(\sum_^n b_k\right)\!.$

Proof

Consider the sum

:$S = \sum_^n \sum_^n (a_j - a_k) (b_j - b_k).$

The two sequences are non-increasing, therefore and have the same sign for any . Hence .

Opening the brackets, we deduce:

:$0 \leq 2 n \sum_^n a_j b_j - 2 \sum_^n a_j \, \sum_^n b_j,$

hence

:$\frac \sum_^n a_j b_j \geq \left( \frac \sum_^n a_j\right)\!\!\left(\frac \sum_^n b_j\right)\!.$

An alternative proof is simply obtained with the rearrangement inequality, writing that
:$\sum_^ a_i \sum_^ b_j = \sum_^ \sum_^ a_i b_j =\sum_^\sum_^ a_i b_ = \sum_^ \sum_^ a_i b_ \leq \sum_^ \sum_^ a_ib_i = n \sum_i a_ib_i.$

Continuous version

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

:$\frac \int_a^b f(x)g(x) \,dx \geq\! \left(\frac \int_a^b f(x) \,dx\right)\!\!\left(\frac\int_a^b g(x) \,dx\right)$

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

See also

* Hardy–Littlewood inequality
* Rearrangement inequality

Notes

{{DEFAULTSORT:Chebyshev's Sum Inequality
Inequalities
Sequences and series