In mathematics, Chebyshev's sum inequality, named after Pafnuty Chebyshev, states that if
a1\geqa2\geq … \geqan
b1\geqb2\geq … \geqbn,
{1\overn}
n | |
\sum | |
k=1 |
akbk\geq\left({1\over
n | |
n}\sum | |
k=1 |
ak\right)\left({1\over
n | |
n}\sum | |
k=1 |
bk\right).
Similarly, if
a1\leqa2\leq … \leqan
b1\geqb2\geq … \geqbn,
{1\overn}
n | |
\sum | |
k=1 |
akbk\leq\left({1\over
n | |
n}\sum | |
k=1 |
ak\right)\left({1\over
n | |
n}\sum | |
k=1 |
bk\right).
Consider the sum
S=
n | |
\sum | |
j=1 |
n | |
\sum | |
k=1 |
(aj-ak)(bj-bk).
The two sequences are non-increasing, therefore and have the same sign for any . Hence .
Opening the brackets, we deduce:
0\leq2n
n | |
\sum | |
j=1 |
ajbj-2
n | |
\sum | |
j=1 |
aj
n | |
\sum | |
j=1 |
bj,
hence
1 | |
n |
n | |
\sum | |
j=1 |
ajbj\geq\left(
1 | |
n |
n | |
\sum | |
j=1 |
a | ||||
|
n | |
\sum | |
j=1 |
bj\right).
An alternative proof is simply obtained with the rearrangement inequality, writing that
n-1 | |
\sum | |
i=0 |
ai
n-1 | |
\sum | |
j=0 |
bj=
n-1 | |
\sum | |
i=0 |
n-1 | |
\sum | |
j=0 |
aibj
n-1 | |
=\sum | |
i=0 |
n-1 | |
\sum | |
k=0 |
aibi+k~mod~n=
n-1 | |
\sum | |
k=0 |
n-1 | |
\sum | |
i=0 |
aibi+k~mod~n\leq
n-1 | |
\sum | |
k=0 |
n-1 | |
\sum | |
i=0 |
aibi=n\sumiaibi.
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
1 | |
b-a |
b | |
\int | |
a |
f(x)g(x)dx\geq\left(
1 | |
b-a |
b | |
\int | |
a |
f(x)dx\right)\left(
1 | |
b-a |
b | |
\int | |
a |
g(x)dx\right)
with the inequality reversed if one is non-increasing and the other is non-decreasing.