Rearrangement Inequality - Proof

Proof

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 k in {j + 1, ..., n} with σ(k) = j. Now

j<k\Rightarrow y_j\le y_k \qquad\text{and}\qquad j=\sigma(k)<\sigma(j)\Rightarrow x_j\le x_{\sigma(j)}.\quad(1)

Therefore,

Expanding this product and rearranging gives

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

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.

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