# Rearrangement inequality

In mathematics, the rearrangement inequality states that

$x_{n}y_{1}+\cdots +x_{1}y_{n}\leq x_{\sigma (1)}y_{1}+\cdots +x_{\sigma (n)}y_{n}\leq x_{1}y_{1}+\cdots +x_{n}y_{n}$ for every choice of real numbers

$x_{1}\leq \cdots \leq x_{n}\quad {\text{and}}\quad y_{1}\leq \cdots \leq y_{n}$ and every permutation

$x_{\sigma (1)},\dots ,x_{\sigma (n)}$ of x1, . . ., xn. If the numbers are different, meaning that

$x_{1}<\cdots 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.

## 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.

## Proof

The lower bound follows by applying the upper bound to

$-x_{n}\leq \cdots \leq -x_{1}.$ Therefore, it suffices to prove the upper bound. Since there are only finitely many permutations, there exists at least one for which

$x_{\sigma (1)}y_{1}+\cdots +x_{\sigma (n)}y_{n}$ 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

$j Therefore,

$0\leq (x_{\sigma (j)}-x_{j})(y_{k}-y_{j}).\quad (2)$ Expanding this product and rearranging gives

$x_{\sigma (j)}y_{j}+x_{j}y_{k}\leq x_{j}y_{j}+x_{\sigma (j)}y_{k}\,,\quad (3)$ hence the permutation

$\tau (i):={\begin{cases}i&{\text{for }}i\in \{1,\ldots ,j\},\\\sigma (j)&{\text{for }}i=k,\\\sigma (i)&{\text{for }}i\in \{j+1,\ldots ,n\}\setminus \{k\},\end{cases}}$ 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.

If

$x_{1}<\cdots 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

$x_{1}>x_{2}\quad {\text{and}}\quad y_{1}>y_{2}$ implies

$(x_{1}-x_{2})(y_{1}-y_{2})>0\quad {\text{or}}\quad x_{1}y_{1}+x_{2}y_{2}>x_{2}y_{1}+x_{1}y_{2},$ hence the result is true if n = 2. Assume it is true at rank n-1, and let

$x_{1}>\cdots >x_{n},\quad {\text{and}}\quad y_{1}>\cdots >y_{n}$ .

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

$x_{k}>x_{n}\quad {\text{and}}\quad y_{j}>y_{n},\quad {\text{hence}}\quad x_{n}y_{n}+x_{k}y_{j}>x_{k}y_{n}+x_{n}y_{j}$ 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.

## Generalization

A Generalization of the Rearrangement inequality states that for all real numbers $x_{1}\leq \cdots \leq x_{n}$ and any choice of functions $f_{i}:[x_{1},x_{n}]\rightarrow \mathbb {R} ,i=1,2,...,n$ such that

$f'_{1}(x)\leq f'_{2}(x)\leq ...\leq f'_{n}(x)\quad \forall x\in [x_{1},x_{n}]$ the inequality

$\sum _{i=1}^{n}f_{i}(x_{n-i+1})\leq \sum _{i=1}^{n}f_{i}(x_{\sigma (i)})\leq \sum _{i=1}^{n}f_{i}(x_{i})$ holds for every permutation $x_{\sigma (1)},\dots ,x_{\sigma (n)}$ of $x_{1},\dots ,x_{n}$ .