Rearrangement inequality explained

\sigma of the numbers 1 , 2 , \ldots n we haveInformally, this means that in these types of sums, the largest sum is achieved by pairing large x values with large y values, and the smallest sum is achieved by pairing small values with large values. This can be formalised in the case that the x _{1,} \ldots, x _{n} are distinct, meaning thatx _{1} < … < x _{n,} then:

1. The upper bound in is attained only for permutations

\sigma that keep the order of y _{1,} \ldots, y _{n,} that is, $$y\_\; \backslash le\; \backslash cdots\; \backslash le\; y\_,$$ or equivalently (y _{1,\ldots,y} _{n)} = (y _{\sigma(1)} ,\ldots,y _{\sigma(n)} ). Such a \sigma can permute the indices of y -values that are equal; in the case y _{1} = … = y _{n} every permutation keeps the order of y _{1,} \ldots, y _{n.} If y _{1} < … < y _{n,} then the only such \sigma is the identiy.

1. Correspondingly, the lower bound in is attained only for permutations

\sigma that reverse the order of y _{1,} \ldots, y _{n,} meaning that $$y\_\; \backslash ge\; \backslash cdots\; \backslash ge\; y\_.$$ If y _{1} < … < y _{n,} then \sigma(i) = n - i + 1 for all i = 1 , \ldots, n, is the only permutation to do this.Note that the rearrangement inequality makes no assumptions on the signs of the real numbers, unlike inequalities such as the arithmetic-geometric mean inequality.

Applications

Many important inequalities can be proved by the rearrangement inequality, such as the arithmetic mean – geometric mean inequality, the Cauchy–Schwarz inequality, and Chebyshev's sum inequality.

As a simple example, consider real numbers

x _{1} \le … \le x _{n} : By applying with y _{i} : = x _{i} for all i = 1 ,\ldots,n, it follows that$$x\_1\; x\_n\; +\; \backslash cdots\; +\; x\_n\; x\_1\backslash le\; x\_1\; x\_\; +\; \backslash cdots\; +\; x\_n\; x\_\backslash le\; x\_1^2\; +\; \backslash cdots\; +\; x\_n^2$$for every permutation \sigma of 1 , \ldots, n.

Intuition

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

7 ⋅ 100 + 5 ⋅ 20 + 3 ⋅ 10 dollars. This is exactly what the upper bound of the rearrangement inequality says for the sequences 3 < 5 < 7 and 10 < 20 < 100 . In this sense, it can be considered as an example of a greedy algorithm.

Geometric interpretation

Assume that

0 < x _{1} < … < x _{n} and 0 < y _{1} < … < y _{n.} Consider a rectangle of width x _{1} + … + x _{n} and height y _{1} + … + y _{n,} subdivided into n columns of widths x _{1,\ldots,x} _{n} and the same number of rows of heights y _{1,\ldots,y} _{n,} so there are style n ^{2} small rectangles. You are supposed to take n of these, one from each column and one from each row. The rearrangement inequality says that you optimize the total area of your selection by taking the rectangles on the diagonal or the antidiagonal.

Proofs

Proof by contradiction

The lower bound and the corresponding discussion of equality follow by applying the results for the upper bound to$$-\; y\_n\; \backslash le\; \backslash cdots\; \backslash le\; -\; y\_1.$$Therefore, it suffices to prove the upper bound in and discuss when equality holds. Since there are only finitely many permutations of

1 ,\ldots,n, there exists at least one \sigma for which the middle term in $$x\_1\; y\_\; +\; \backslash cdots\; +\; x\_n\; y\_$$is maximal. In case there are several permutations with this property, let σ denote one with the highest number of integers i from \{ 1 ,\ldots,n\} satisfying y _{i} = y _{\sigma(i)} .

We will now prove by contradiction, that

\sigma has to keep the order of y _{1,\ldots,y} _{n} (then we are done with the upper bound in, because the identity has that property). Assume that there exists a j\in\{ 1 ,\ldots,n -1 \} such that y _{i} = y _{\sigma(i)} for all i\in\{ 1 ,\ldots,j -1 \} and y _{j} ≠ y _{\sigma(j)} . Hence y _{j} < y _{\sigma(j)} and there has to exist a k\in\{j +1 ,\ldots,n\} with y _{j} = y _{\sigma(k)} to fill the gap. Therefore,which implies thatExpanding this product and rearranging giveswhich is equivalent to . Hence the permutationwhich arises from \sigma by exchanging the values \sigma(j) and \sigma(k), has at least one additional point which keeps the order compared to \sigma, namely at j satisfying y _{j=y} _{\tau(j)} , and also attains the maximum in due to . This contradicts the choice of \sigma.

If

x _{1} < … < x _{n,} then we have strict inequalities in,, and, hence the maximum can only be attained by permutations keeping the order of y _{1} \le … \le y _{n,} and every other permutation \sigma cannot be optimal.

Proof by induction

As above, it suffices to treat the upper bound in . For a proof by mathematical induction, we start with

n =2 . Observe that $$x\_1\; \backslash le\; x\_2\; \backslash quad\; \backslash text\; \backslash quad\; y\_1\; \backslash le\; y\_2$$implies thatwhich is equivalent tohence the upper bound in is true for n = 2 . If x _{1} < x _{2,} then we get strict inequality in and if and only if y _{1} < y _{2.} Hence only the identity, which is the only permutation here keeping the order of y _{1} < y _{2,} gives the maximum.

As an induction hypothesis assume that the upper bound in the rearrangement inequality is true for

n -1 with n\ge 3 and that in the case x _{1} < … <x _{n-1} there is equality only when the permutation \sigma of 1 ,\ldots,n -1 keeps the order of y _{1,\ldots,y} _{n-1} .

Consider now

x _{1} \le … \le x _{n} and y _{1} \le … \le y _{n.} Take a \sigma from the finite number of permutations of 1 ,\ldots,n such that the rearrangement in the middle of gives the maximal result. There are two cases:

• If

\sigma(n) = n, then y _{n=y} _{\sigma(n)} and, using the induction hypothesis, the upper bound in is true with equality and \sigma keeps the order of y _{1,\ldots,y} _{n-1} ,y _{n} in the case x _{1} < … < x _{n.}

• If

k : = \sigma(n) < n, then there is a j\in\{ 1 , ... ,n -1 \} with \sigma(j) = n. Define the permutation which arises from \sigma by exchanging the values of j and n. There are now two subcases:

1. If

x _{k} = x _{n} or y _{k} = y _{n,} then this exchange of values of \sigma has no effect on the middle term in because \tau gives the same sum, and we can proceed by applying the first case to \tau. Note that in the case x _{1} < … < x _{n,} the permutation \tau keeps the order of y _{1,\ldots,y} _{n} if and only if \sigma does.

1. If

x _{k} < x _{n} and y _{k} < y _{n,} then 0 < (x _{n} - x _{k)} (y _{n} - y _{k),} which is equivalent to x _{k} y _{n} + x _{n} y _{k} < x _{k} y _{k} + x _{n} y _{n} and shows that \sigma is not optimal, hence this case cannot happen due to the choice of \sigma.

Generalizations

Three or more sequences

A straightforward generalization takes into account more sequences. Assume we have finite ordered sequences of nonnegative real numbers$$0\; \backslash le\; x\_1\backslash le\backslash cdots\backslash le\; x\_n\backslash quad\backslash text\backslash quad\; 0\backslash le\; y\_1\backslash le\backslash cdots\backslash le\; y\_n\backslash quad\backslash text\backslash quad\; 0\backslash le\; z\_1\backslash le\backslash cdots\backslash le\; z\_n$$and a permutation

y _{\sigma(1)} ,\ldots,y _{\sigma(n)} of y _{1,...,y} _{n} and another permutation z _{\tau(1)} , ... ,z _{\tau(n)} of z _{1,...,z} _{n.} Then$$x\_1\; y\_\; z\_\; +\; \backslash cdots\; +\; x\_n\; y\_\; z\_\; \backslash le\; x\_1\; y\_1\; z\_1\; +\; \backslash cdots\; +\; x\_n\; y\_n\; z\_n.$$

Note that, unlike the standard rearrangement inequality, this statement requires the numbers to be nonnegative. A similar statement is true for any number of sequences with all numbers nonnegative.

Functions instead of factors

Another generalization of the rearrangement inequality states that for all real numbers

x _{1} \le … \le x _{n} and every choice of continuously differentiable functions f _{i:} [x _{1,x} _{n]} \to \R for i = 1 , 2 , \ldots, n such that their derivatives f' _{1,\ldots,f'} _{n} satisfy$$f\text{'}\_1(x)\; \backslash le\; f\text{'}\_2(x)\; \backslash le\; \backslash cdots\; \backslash le\; f\text{'}\_n(x)\; \backslash quad\; \backslash text\; x\; \backslash in\; [x\_1,x\_n],$$the inequality$$\backslash sum\_^n\; f\_(x\_i)\; \backslash le\; \backslash sum\_^n\; f\_(x\_i)\; \backslash le\; \backslash sum\_^n\; f\_i(x\_i)$$holds for every permutation f _{\sigma(1)} , \ldots, f _{\sigma(n)} of f _{1,} \ldots, f _{n.} Taking real numbers y _{1} \le … \le y _{n} and the linear functions f _{i(x)} : = x y _{i} for real x and i = 1 ,\ldots,n, the standard rearrangement inequality is recovered.

See also

• Hardy–Littlewood inequality
• Chebyshev's sum inequality

Notes and References

1. , Section 10.2, Theorem 368