In mathematics , Chebyshev's sum inequality , named after Pafnuty Chebyshev , states that if
a
1
≥
a
2
≥
⋯
≥
a
n
{\displaystyle a_{1}\geq a_{2}\geq \cdots \geq a_{n}}
and
b
1
≥
b
2
≥
⋯
≥
b
n
,
{\displaystyle b_{1}\geq b_{2}\geq \cdots \geq b_{n},}
then
1
n
∑
k
=
1
n
a
k
b
k
≥
(
1
n
∑
k
=
1
n
a
k
)
(
1
n
∑
k
=
1
n
b
k
)
.
{\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).}
Similarly, if
a
1
≤
a
2
≤
⋯
≤
a
n
{\displaystyle a_{1}\leq a_{2}\leq \cdots \leq a_{n}}
and
b
1
≥
b
2
≥
⋯
≥
b
n
,
{\displaystyle b_{1}\geq b_{2}\geq \cdots \geq b_{n},}
then
1
n
∑
k
=
1
n
a
k
b
k
≤
(
1
n
∑
k
=
1
n
a
k
)
(
1
n
∑
k
=
1
n
b
k
)
.
{\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).}
[1] Proof [ ] Consider the sum
S
=
∑
j
=
1
n
∑
k
=
1
n
(
a
j
−
a
k
)
(
b
j
−
b
k
)
.
{\displaystyle 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 b j b k j , k S ≥ 0
Opening the brackets, we deduce:
0
≤
2
n
∑
j
=
1
n
a
j
b
j
−
2
∑
j
=
1
n
a
j
∑
j
=
1
n
b
j
,
{\displaystyle 0\leq 2n\sum _{j=1}^{n}a_{j}b_{j}-2\sum _{j=1}^{n}a_{j}\,\sum _{j=1}^{n}b_{j},}
hence
1
n
∑
j
=
1
n
a
j
b
j
≥
(
1
n
∑
j
=
1
n
a
j
)
(
1
n
∑
j
=
1
n
b
j
)
.
{\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).}
An alternative proof is simply obtained with the rearrangement inequality , writing that
∑
i
=
0
n
−
1
a
i
∑
j
=
0
n
−
1
b
j
=
∑
i
=
0
n
−
1
∑
j
=
0
n
−
1
a
i
b
j
=
∑
i
=
0
n
−
1
∑
k
=
0
n
−
1
a
i
b
i
+
k
mod
n
=
∑
k
=
0
n
−
1
∑
i
=
0
n
−
1
a
i
b
i
+
k
mod
n
≤
∑
k
=
0
n
−
1
∑
i
=
0
n
−
1
a
i
b
i
=
n
∑
i
a
i
b
i
.
{\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_{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 [0,1], both non-increasing or both non-decreasing, then
∫
0
1
f
(
x
)
g
(
x
)
d
x
≥
∫
0
1
f
(
x
)
d
x
∫
0
1
g
(
x
)
d
x
,
{\displaystyle \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.
See also [ ] Notes [ ]
^ Hardy, G. H.; Littlewood, J. E.; Pólya, G. (1988). Inequalities . Cambridge Mathematical Library. Cambridge: Cambridge University Press. ISBN 0-521-35880-9 MR 0944909 .
