Rearrangement inequality

摘抄自：  https://en.wikipedia.org/wiki/Rearrangement_inequality#Proof

In mathematics, the rearrangement inequality^[1] states that

{displaystyle 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

{displaystyle x_{1}leq cdots leq x_{n}quad { ext{and}}quad y_{1}leq cdots leq y_{n}}

and every permutation

{displaystyle x_{sigma (1)},dots ,x_{sigma (n)}}

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

{displaystyle x_{1}<cdots <x_{n}quad { ext{and}}quad y_{1}<cdots <y_{n},}

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.

Proof[edit]

The lower bound follows by applying the upper bound to

{displaystyle -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

{displaystyle 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

{displaystyle j<kRightarrow y_{j}leq y_{k}qquad { ext{and}}qquad j<sigma (j)Rightarrow x_{j}leq x_{sigma (j)}.quad (1)}

Therefore,

{displaystyle 0leq (x_{sigma (j)}-x_{j})(y_{k}-y_{j}).quad (2)}

Expanding this product and rearranging gives

{displaystyle x_{sigma (j)}y_{j}+x_{j}y_{k}leq x_{j}y_{j}+x_{sigma (j)}y_{k}\,,quad (3)}

hence the permutation

{displaystyle au (i):={egin{cases}i&{ ext{for }}iin {1,ldots ,j},\sigma (j)&{ ext{for }}i=k,\sigma (i)&{ ext{for }}iin {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

{displaystyle x_{1}<cdots <x_{n}quad { ext{and}}quad y_{1}<cdots <y_{n},}

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.

Generalization[edit]

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

{displaystyle f'_{1}(x)leq f'_{2}(x)leq ...leq f'_{n}(x)quad forall xin [x_{1},x_{n}]}

the inequality

{displaystyle 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 {displaystyle x_{sigma (1)},dots ,x_{sigma (n)}} of {displaystyle x_{1},dots ,x_{n}}^[2].