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Chebyshev's inequality (Theorem)

If $ x_1,x_2,\ldots,x_n$ and $ y_1,y_2,\ldots,y_n$ are two sequences (at least one of them consisting of positive numbers):

  • if $ x_1<x_2<\cdots<x_n$ and $ y_1<y_2<\cdots<y_n$ then
    $\displaystyle \left(\frac{x_1+x_2+\cdots+x_n}{n}\right)\left(\frac{y_1+y_2+\cdots+y_n}{n}\right) \le\frac{x_1y_1+x_2y_2+\cdots+x_ny_n}{n}.$
  • if $ x_1<x_2<\cdots<x_n$ and $ y_1>y_2>\cdots>y_n$ then
    $\displaystyle \left(\frac{x_1+x_2+\cdots+x_n}{n}\right)\left(\frac{y_1+y_2+\cdots+y_n}{n}\right) \ge\frac{x_1y_1+x_2y_2+\cdots+x_ny_n}{n}.$



"Chebyshev's inequality" is owned by drini. [ full author list (2) | owner history (1) ]
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See Also: rearrangement inequality, proof of rearrangement inequality, Kolmogorov's inequality, Chebyshev's inequality

Keywords:  Inequality

Attachments:
proof of Chebyshev's inequality (Proof) by pbruin
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Cross-references: numbers, positive, sequences
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This is version 2 of Chebyshev's inequality, born on 2001-10-17, modified 2006-12-11.
Object id is 277, canonical name is ChebyshevsInequality.
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Classification:
AMS MSC26D15 (Real functions :: Inequalities :: Inequalities for sums, series and integrals)

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Extraneous condition by pbruin on 2002-11-11 09:48:27
I think you can remove the condition that at least one of the sequences x_1, x_2, ..., x_n and y_1, y_2, ..., y_n consists of positive numbers (I didn't use it in my proof, but it might be wrong). Also, IIRC x_1, x_2, ..., x_n isn't really a sequence (sequences consist of infinitely many numbers x_1, x_2, ...).
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