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I wonder if there is a name or reference for the following fact. It is not the proof I am looking for.

Let $s_1, s_2, ...,s_n$ be non-negative real numbers ordered in a non-increasing way. Let $b_1,b_2,...,b_n$ be non-negative real numbers ordered in a non-decreasing way (so opposite of the $s_i$).

Then the average value $$\frac{s_1+s_2+\cdots +s_n}{n}$$ is at least as large as the weighted average $$\frac{b_1 s_1+\cdots+b_n s_n}{b_1+\cdots+b_n}.$$

Thanks!

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    $\begingroup$ Chebyshev's sum inequality $\endgroup$
    – Mateusz Kwaśnicki
    Commented Mar 8, 2018 at 19:57
  • $\begingroup$ Please use TeX on this site. $\endgroup$
    – GH from MO
    Commented Mar 8, 2018 at 20:12
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    $\begingroup$ I thought it looked like the "Rearrangement Inequality" (en.wikipedia.org/wiki/Rearrangement_inequality); it's not quite, but the inequality here can be recovered from it by averaging over all $n!$ rearrangements. (The Wikipedia page notes that Chebyshev's sum inequality is a consequence of Rearrangement.) $\endgroup$
    – Noam D. Elkies
    Commented Mar 9, 2018 at 4:24

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This is known as Chebyshev's Sum Inequality. (I've only ever seen it used in the context of competition math, but Wikipedia gives a reference to Hardy-Littlewood-Polya.)

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    $\begingroup$ I've seen it used at least once by K. Matomäki (in "On signs of Fourier coefficients of cusp forms", Math. Proc. Cambridge Philos. Soc. 152 (2012), no. 2, 207–222, if I remember correctly). $\endgroup$
    – Denis Chaperon de Lauzières
    Commented Mar 9, 2018 at 9:07
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It is called often called "Chebyshev's other inequality". And this makes more sense than the name "Chebyshev's sum inequality" proposed by Wikipedia, since even Wikepedia knows that there is a version for integrals.

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    $\begingroup$ Integrals aren’t sums? $\endgroup$
    – Geoffrey Irving
    Commented Mar 9, 2018 at 0:06
  • $\begingroup$ So one of my physics teachers told us. But I could understand it only after a transposition. $\endgroup$
    – Lutz Mattner
    Commented Mar 9, 2018 at 0:14
  • $\begingroup$ @LutzMattner : ... so integrals are permutations of sums? :-) $\endgroup$
    – Eric Towers
    Commented Mar 9, 2018 at 2:07

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