Login
Login

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

:a_1 \geq a_2 \geq \cdots \geq a_n \quad and \quad b_1 \geq b_2 \geq \cdots \geq b_n,

then

:{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)\!.

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

:{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)\!.{{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}}

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 {{math|aj − ak}} and {{math|bj − bk}} have the same sign for any {{math|j, k}}. Hence {{math|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_{j=1}^n b_j,

hence

:\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)\!.

An alternative proof is simply obtained with the rearrangement inequality, writing that

:\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.

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{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)

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

See also

Notes

{{Reflist}}

{{DEFAULTSORT:Chebyshev's Sum Inequality}}

Category:Inequalities (mathematics)

Category:Sequences and series