contradicts the choice of σ. If 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 Observe first that implies 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. Generalizations A straightforward generalization takes into account more sequences. Assume we have ordered sequences of positive real numbers and a permutation of and another permutation of . Then it holds Note that unlike the common rearrangement inequality this statement requires the numbers to be nonnegative. A similar statement is true for any number of sequences with all numbers nonnegative. Another generalization of the rearrangement inequality states that for all real numbers and any choice of functions such that the derivatives satisfy: the inequality holds for every permutation of See also Hardy–Littlewood inequality Chebyshev's sum inequality References Inequalities Articles containing proofs