Rearrangement inequality

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In mathematics, the rearrangement inequality[1] states that

for every choice of real numbers

and every permutation

of x1, . . ., xn. If the numbers are different, meaning that

then the lower bound is attained only for the permutation which reverses the order, i.e. σ(i) = n − i + 1 for all i = 1, ..., n, and the upper bound is attained only for the identity, i.e. σ(i) = i for all i = 1, ..., n.

Note that the rearrangement inequality makes no assumptions on the signs of the real numbers.


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.


The lower bound follows by applying the upper bound to

Therefore, it suffices to prove the upper bound. Since there are only finitely many permutations, there exists at least one for which

is maximal. In case there are several permutations with this property, let σ denote one with the highest number of fixed points.

We will now prove by contradiction, that σ has to be the identity (then we are done). Assume that σ is not the identity. Then there exists a j in {1, ..., n − 1} such that σ(j) ≠ j and σ(i) = i for all i in {1, ..., j − 1}. Hence σ(j) > j and there exists a k in {j + 1, ..., n} with σ(k) = j. Now


Expanding this product and rearranging gives

hence the permutation

which arises from σ by exchanging the values σ(j) and σ(k), has at least one additional fixed point compared to σ, namely at j, and also attains the maximum. This contradicts the choice of σ.


then we have strict inequalities at (1), (2), and (3), hence the maximum can only be attained by the identity, any other permutation σ cannot be optimal.

Proof by induction[edit]

Observe first that


hence the result is true if n = 2. Assume it is true at rank n-1, and let


Choose a permutation σ for which the arrangement gives rise a maximal result.

If σ(n) were different from n, say σ(n) = k, there would exist j < n such that σ(j) = n. But

by what has just been proved. Consequently, it would follow that the permutation τ coinciding with σ, except at j and n, where τ(j) = k and τ(n) = n, gives rise a better result. This contradicts the choice of σ. Hence σ(n) = n, and from the induction hypothesis, σ(i) = i for every i < n.

The same proof holds if one replace strict inequalities by non strict ones.

See also[edit]


  1. ^ Hardy, G.H.; Littlewood, J.E.; Pólya, G. (1952), Inequalities, Cambridge Mathematical Library (2. ed.), Cambridge: Cambridge University Press, ISBN 0-521-05206-8, MR 0046395, Zbl 0047.05302 , Section 10.2, Theorem 368