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Let $x_{1},\ldots,x_{n}$ be nonnegative numbers such that $m \leq x_{i} \leq M$. Let $$ S=\sum_{i=1}^{n}{x_{i}} $$ and $$ Q=\sum_{i=1}^{n}{x_{i}^{2}}. $$

Then $$ Q \leq S(M+m)-nMm. $$

This has been recently discovered by at least two different people (including myself) but I am sure that it has come up many times before. Does it have a standard name or reference?

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    $\begingroup$ I guess this is equivalent to the inequality $x^2 \leq x(m+M) - Mm,$ since it must be true term by term ($n=1$ case seems to imply the claim in general). So that inequality might have a name if such a thing exists. The fact that $M \geq x$ then immediately implies the claim, so it might not, because it could be seen as being too easy, since it doesn't seem like there's any trick. $\endgroup$
    – J. E. Pascoe
    Commented Sep 17, 2014 at 8:13
  • $\begingroup$ @J.E.Pascoe That's a nice proof - thanks! Still looking for the provenance, though... $\endgroup$
    – Felix Goldberg
    Commented Sep 17, 2014 at 8:23
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    $\begingroup$ If $A$ is an $n \times n$ matrix, this inequality describes a relationship between ${\rm tr}\ A$ and ${\rm tr}\ A^2$. $\endgroup$
    – David Handelman
    Commented Sep 17, 2014 at 13:58
  • $\begingroup$ @DavidHandelman Can you please elaborate a bit? 10x! $\endgroup$
    – Felix Goldberg
    Commented Sep 17, 2014 at 15:00
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    $\begingroup$ If $A$ has eigenvalues (counting multiple ones, including what are sometimes called algebraic ones, really bad terminology) $(x_i)$, then ${\rm tr}\ A = \sum x_i$ and ${\rm tr}\ A^2 = \sum x_i^2$. $\endgroup$
    – David Handelman
    Commented Sep 17, 2014 at 21:04

2 Answers 2

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As @J.E.Pascoe notes the question reduces to the case $n=1$. The fact that $m\leq x$ and $x\leq M$ imply that $x^2+Mm \leq mx + xM$ is a special case of the rearrangement inequality.

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  • $\begingroup$ Irinically, my original overwrought proof used the Gruss inequality - which is a complement to Chebyshev. So it all comes together. :) $\endgroup$
    – Felix Goldberg
    Commented Sep 17, 2014 at 8:48
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I first came across this inequality (n=1 case) from the short note http://www.ams.org/journals/proc/1965-016-05/S0002-9939-1965-0180561-X/S0002-9939-1965-0180561-X.pdf

which says that if $m\le A\le M$, then $(M-A)(A-m)$ is again positive semidefinite. This inequality has intimate connection with the Kantrovich inequality.

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