Chebyshev's sum inequality
Consider the sum
The two sequences are non-increasing, therefore aj − ak and bj − bk have the same sign for any j, k. Hence S ≥ 0.
Opening the brackets, we deduce:
An alternative proof is simply obtained with the rearrangement inequality.
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
with the inequality reversed if one is non-increasing and the other is non-decreasing.