Chebyshev's sum inequality

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

[math]\displaystyle{ a_1 \geq a_2 \geq \cdots \geq a_n \quad }[/math] and [math]\displaystyle{ \quad b_1 \geq b_2 \geq \cdots \geq b_n, }[/math]

then

[math]\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)\!. }[/math]

Similarly, if

[math]\displaystyle{ a_1 \leq a_2 \leq \cdots \leq a_n \quad }[/math] and [math]\displaystyle{ \quad b_1 \geq b_2 \geq \cdots \geq b_n, }[/math]

then

[math]\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)\!. }[/math]

Proof

Consider the sum

[math]\displaystyle{ S = \sum_{j=1}^n \sum_{k=1}^n (a_j - a_k) (b_j - b_k). }[/math]

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:

[math]\displaystyle{ 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]\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)\!. }[/math]

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

[math]\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_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-valued, integrable functions over [a, b], both non-increasing or both non-decreasing, then

[math]\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) }[/math]

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

See also

• Hardy–Littlewood inequality
• Rearrangement inequality