Chebyshev's sum inequality

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

$a_1\ge a_2\ge \cdots \ge a_n$ and $b_1\ge b_2\ge \cdots \ge b_n,$

then

$\frac{1}{n}\sum _{k=1}^n{a}_k{b}_k\ge \left(\frac{1}{n}\sum _{k=1}^n{a}_k\right)\left(\frac{1}{n}\sum _{k=1}^n{b}_k\right).$

Similarly, if

$a_1\le a_2\le \cdots \le a_n$ and $b_1\ge b_2\ge \cdots \ge b_n,$

then

$\frac{1}{n}\sum _{k=1}^n{a}_k{b}_k\le \left(\frac{1}{n}\sum _{k=1}^n{a}_k\right)\left(\frac{1}{n}\sum _{k=1}^n{b}_k\right).$

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 Template:Math and Template:Math have the same sign for any Template:Math. Hence Template:Math.

Opening the brackets, we deduce:

$0\le 2n\sum _{j=1}^n{a}_j{b}_j-2\sum _{j=1}^n{a}_j\sum _{j=1}^n{b}_j,$

hence

$\frac{1}{n}\sum _{j=1}^n{a}_j{b}_j\ge \left(\frac{1}{n}\sum _{j=1}^n{a}_j\right)\left(\frac{1}{n}\sum _{j=1}^n{b}_j\right).$

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

$\sum _{i=0}^{n-1}{a}_i\sum _{j=0}^{n-1}{b}_j=\sum _{i=0}^{n-1}\sum _{j=0}^{n-1}{a}_i{b}_j=\sum _{i=0}^{n-1}\sum _{k=0}^{n-1}{a}_i{b}_{i+k\text{mod}n}=\sum _{k=0}^{n-1}\sum _{i=0}^{n-1}{a}_i{b}_{i+k\text{mod}n}\le \sum _{k=0}^{n-1}\sum _{i=0}^{n-1}{a}_i{b}_i=n\sum _i{a}_i{b}_i.$

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

$\frac{1}{b-a}\underset{a}{\overset{b}{\int }}f(x)g(x)dx\ge (\frac{1}{b-a}\underset{a}{\overset{b}{\int }}f(x)dx)(\frac{1}{b-a}\underset{a}{\overset{b}{\int }}g(x)dx)$

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

See also

• Hardy–Littlewood inequality
• Rearrangement inequality