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

$a_{1}\geq a_{2}\geq \cdots \geq a_{n}$

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

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

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 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:

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

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.