Editing Chebyshev's sum inequality

{{For|the similarly named inequality in [[probability theory]]|Chebyshev's inequality}}

In [[mathematics]], '''Chebyshev's sum inequality''', named after [[Pafnuty Chebyshev]], states that if
:<math>a_1 \geq a_2 \geq \cdots \geq a_n \quad</math> and <math>\quad b_1 \geq b_2 \geq \cdots \geq b_n,</math>
then
:<math>{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)\!.</math>

Similarly, if
:<math>a_1 \leq a_2 \leq \cdots \leq a_n \quad</math> and <math>\quad b_1 \geq b_2 \geq \cdots \geq b_n,</math>
then
:<math>{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)\!.</math><ref>{{cite book|mr=0944909|last=Hardy|first=G. H.|last2=Littlewood|first2=J. E.|last3=Pólya|first3=G.|title=Inequalities|series=Cambridge Mathematical Library|publisher=Cambridge University Press|location=Cambridge|year=1988|isbn=0-521-35880-9}}</ref>

==Proof==
Consider the sum

:<math>S = \sum_{j=1}^n \sum_{k=1}^n (a_j - a_k) (b_j - b_k).</math>

The two [[sequence]]s are [[Sequence#Increasing and decreasing|non-increasing]], therefore {{math|''a''<sub>''j''</sub>&nbsp;−&nbsp;''a''<sub>''k''</sub>}} and {{math|''b''<sub>''j''</sub>&nbsp;−&nbsp;''b''<sub>''k''</sub>}} have the same sign for any {{math|''j'',&nbsp;''k''}}. Hence {{math|''S''&nbsp;&ge;&nbsp;0}}.

Opening the brackets, we deduce:

:<math>0 \leq 2 n \sum_{j=1}^n a_j b_j - 2 \sum_{j=1}^n a_j \, \sum_{j=1}^n b_j,</math>

hence

:<math>\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)\!.</math>

An alternative [[mathematical proof|proof]] is simply obtained with the [[rearrangement inequality]], writing that
:<math>\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_ib_i = n \sum_i a_ib_i.</math>

==Continuous version==
There is also a continuous version of Chebyshev's sum inequality:

If ''f'' and ''g'' are [[real number|real]]-valued, [[integrable function]]s over [''a'', ''b''], both non-increasing or both non-decreasing, then

:<math>\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)</math>

with the [[inequality (mathematics)|inequality]] reversed if one is non-increasing and the other is non-decreasing.

==See also==
* [[Hardy–Littlewood inequality]]
* [[Rearrangement inequality]]

==Notes==
{{Reflist}}

{{DEFAULTSORT:Chebyshev's Sum Inequality}}
[[Category:Inequalities (mathematics)]]
[[Category:Sequences and series]]