Chebyshev's sum inequality explained

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

a1\geqa2\geq\geqan

and

b1\geqb2\geq\geqbn,

then

{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

and

b1\geqb2\geq\geqbn,

then

{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).

[1]

Proof

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
j\right)\left(1
n
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.

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

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.

See also

Notes and References

  1. Book: Hardy, G. H.. 0944909. Littlewood. J. E.. Pólya. G.. Inequalities. Cambridge Mathematical Library. Cambridge University Press. Cambridge. 1988. 0-521-35880-9.