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rearrangement inequality (Theorem)

Let $x_1,x_2,\ldots,x_n$ and $y_1,y_2,\ldots,y_n$ two sequences of positive real numbers. Then the sum $$x_1y_1+x_2y_2+\cdots+x_ny_n$$ is maximized when the two sequences are ordered in the same way (i.e. $x_1\le x_2\le \cdots \le x_n$ and $y_1\le y_2\le\cdots\le y_n$ ) and is minimized when the two sequences are ordered in the opposite way (i.e. $x_1\le x_2\le \cdots \le x_n$ and $y_1\ge y_2\ge\cdots\ge y_n$ ).


This can be seen intuitively as: If $x_1,x_2,\ldots,x_n$ are the prices of $n$ kinds of items, and $y_1,y_2,\ldots,y_n$ the number of units sold of each, then the highest profit is when you sell more items with high prices and fewer items with low prices (same ordering), and the lowest profit happens when you sell more items with lower prices and less items with high prices (opposite orders).




"rearrangement inequality" is owned by drini. [ owner history (1) ]
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See Also: Chebyshev's inequality

Keywords:  Inequality, Sequences

Attachments:
proof of rearrangement inequality (Proof) by pbruin
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Cross-references: orders, ordering, units, number, opposite, sum, real numbers, positive, sequences
There are 4 references to this entry.

This is version 4 of rearrangement inequality, born on 2001-10-17, modified 2003-04-24.
Object id is 276, canonical name is RearrangementInequality.
Accessed 7309 times total.

Classification:
AMS MSC26D15 (Real functions :: Inequalities :: Inequalities for sums, series and integrals)

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