Informally, this means that in these types of sums, the largest sum is achieved by pairing large values with large values, and the smallest sum is achieved by pairing small values with large values. This can be formalised in the case that the are distinct, meaning that
The upper bound in (1) is attained only for permutations that keep the order of that is,
or equivalently Such a can permute the indices of -values that are equal; in the case every permutation keeps the order of If then the only such is the identiy.
Correspondingly, the lower bound in (1) is attained only for permutations that reverse the order of meaning that
If then for all is the only permutation to do this.
The rearrangement inequality can be regarded as intuitive in the following way. Imagine there is a heap of $10 bills, a heap of $20 bills and one more heap of $100 bills. You are allowed to take 7 bills from a heap of your choice and then the heap disappears. In the second round you are allowed to take 5 bills from another heap and the heap disappears. In the last round you may take 3 bills from the last heap. In what order do you want to choose the heaps to maximize your profit? Obviously, the best you can do is to gain dollars. This is exactly what the upper bound of the rearrangement inequality (1) says for the sequences and In this sense, it can be considered as an example of a greedy algorithm.
Assume that and Consider a rectangle of width and height subdivided into columns of widths and the same number of rows of heights so there are small rectangles. You are supposed to take of these, one from each column and one from each row. The rearrangement inequality (1) says that you optimize the total area of your selection by taking the rectangles on the diagonal or the antidiagonal.
The lower bound and the corresponding discussion of equality follow by applying the results for the upper bound to
Therefore, it suffices to prove the upper bound in (1) and discuss when equality holds.
Since there are only finitely many permutations of there exists at least one for which the middle term in (1)
is maximal. In case there are several permutations with this property, let σ denote one with the highest number of integers from satisfying
We will now prove by contradiction, that has to keep the order of (then we are done with the upper bound in (1), because the identity has that property). Assume that there exists a such that for all and Hence and there has to exist a with to fill the gap. Therefore,
which arises from by exchanging the values and has at least one additional point which keeps the order compared to namely at satisfying and also attains the maximum in (1) due to (4). This contradicts the choice of
If then we have strict inequalities in (2), (3), and (4), hence the maximum can only be attained by permutations keeping the order of and every other permutation cannot be optimal.
As above, if suffices to treat the upper bound in (1). For a proof by mathematical induction, we start with Observe that
which is equivalent to
hence the upper bound in (1) is true for
If then we get strict inequality in (5) and (6) if and only if Hence only the identity, which is the only permutation here keeping the order of gives the maximum.
As an induction hypothesis assume that the upper bound in the rearrangement inequality (1) is true for with and that in the case there is equality only when the permutation of keeps the order of
Consider now and Take a from the finite number of permutations of such that the rearrangement in the middle of (1) gives the maximal result. There are two cases:
If then and, using the induction hypothesis, the upper bound in (1) is true with equality and keeps the order of in the case
If then there is a with Define the permutation
which arises from by exchanging the values of and There are now two subcases:
If or then this exchange of values of has no effect on the middle term in (1) because gives the same sum, and we can proceed by applying the first case to Note that in the case the permutation keeps the order of if and only if does.
If and then which is equivalent to and shows that is not optimal, hence this case cannot happen due to the choice of
A straightforward generalization takes into account more sequences. Assume we have finite ordered sequences of nonnegative real numbers
and a permutation of and another permutation of Then
Note that, unlike the standard rearrangement inequality (1), this statement requires the numbers to be nonnegative. A similar statement is true for any number of sequences with all numbers nonnegative.