Chebyshev's sum inequality

For the similarly named inequality in probability theory, see Chebyshev's inequality.

In mathematics, Chebyshev's sum inequality, named after Pafnuty Chebyshev, states that if

${\displaystyle a_{1}\geq a_{2}\geq \cdots \geq a_{n}\quad }$ and ${\displaystyle \quad b_{1}\geq b_{2}\geq \cdots \geq b_{n},}$

then

${\displaystyle {1 \over n}\sum _{k=1}^{n}a_{k}b_{k}\geq \left({1 \over n}\sum _{k=1}^{n}a_{k}\right)\!\!\left({1 \over n}\sum _{k=1}^{n}b_{k}\right)\!.}$

In words, if we are given two sequences that are both non-increasing or non-decreasing, then the product of their averages is less than the average of their (termwise) product.

Similarly, if

${\displaystyle a_{1}\leq a_{2}\leq \cdots \leq a_{n}\quad }$ and ${\displaystyle \quad b_{1}\geq b_{2}\geq \cdots \geq b_{n},}$

then

${\displaystyle {1 \over n}\sum _{k=1}^{n}a_{k}b_{k}\leq \left({1 \over n}\sum _{k=1}^{n}a_{k}\right)\!\!\left({1 \over n}\sum _{k=1}^{n}b_{k}\right)\!.}$^[1]

Proof

Consider the sum

${\displaystyle S=\sum _{j=1}^{n}\sum _{k=1}^{n}(a_{j}-a_{k})(b_{j}-b_{k}).}$

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:

${\displaystyle 0\leq 2n\sum _{j=1}^{n}a_{j}b_{j}-2\sum _{j=1}^{n}a_{j}\,\sum _{j=1}^{n}b_{j},}$

hence

${\displaystyle {\frac {1}{n}}\sum _{j=1}^{n}a_{j}b_{j}\geq \left({\frac {1}{n}}\sum _{j=1}^{n}a_{j}\right)\!\!\left({\frac {1}{n}}\sum _{j=1}^{n}b_{j}\right)\!.}$

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

${\displaystyle \sum _{i=0}^{n-1}a_{i}\sum _{j=0}^{n-1}b_{j}=\sum _{i=0}^{n-1}\sum _{j=0}^{n-1}a_{i}b_{j}=\sum _{i=0}^{n-1}\sum _{k=0}^{n-1}a_{i}b_{i+k~{\text{mod}}~n}=\sum _{k=0}^{n-1}\sum _{i=0}^{n-1}a_{i}b_{i+k~{\text{mod}}~n}\leq \sum _{k=0}^{n-1}\sum _{i=0}^{n-1}a_{i}b_{i}=n\sum _{i}a_{i}b_{i}.}$

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

${\displaystyle {\frac {1}{b-a}}\int _{a}^{b}f(x)g(x)\,dx\geq \!\left({\frac {1}{b-a}}\int _{a}^{b}f(x)\,dx\right)\!\!\left({\frac {1}{b-a}}\int _{a}^{b}g(x)\,dx\right)}$

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

See also

• Hardy–Littlewood inequality
• Rearrangement inequality