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The following question has been posted on mathematics stackexchange: inequalities problem, perhaps arising from a question on expectations.

Let $x_{1},x_{2},\cdots,x_{n}$ are real numbers, and such $$\begin{cases} x_{1}+x_{2}+\cdots+x_{n}=0\\ x^2_{1}+x^2_{2}+\cdots+x^2_{n}=n \end{cases}$$ Let $\alpha_{m}=\displaystyle\dfrac{1}{n}\sum_{i=1}^{n}x^m_{i}$

See Mitrinovic D.S Analytic inequalities (Springer 1970) Page 347.

M.LAKSHMANAMURTI proved that $$\alpha_{m}\le\dfrac{(n-1)^{m-1}+(-1)^m}{n(n-1)^{(m/2)-1}}.$$

I am interested in the details of the proof or a published reference.

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    $\begingroup$ Wow, there's "posti mathsatck" again. Why do you insist on writing that? $\endgroup$
    – Gerry Myerson
    Commented Sep 15, 2018 at 12:33
  • $\begingroup$ I have a feeling that people here are better at math, so I think it's going to work out here.，Thanks $\endgroup$
    – math110
    Commented Sep 15, 2018 at 12:48
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    $\begingroup$ Let me try again. Why do you insist on writing the incomprehensible phrase, "posti mathsatck"? "posti" is not a word. "mathsatck" is not a word. Together, they do not form a meaningful grouping of letters. So why do you keep writing them? $\endgroup$
    – Gerry Myerson
    Commented Sep 15, 2018 at 12:51

2 Answers 2

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This is addressed in the following paper:

Rivin, Igor, Counting cycles and finite dimensional $L^{p}$ norms, Adv. Appl. Math. 29, No. 4, 647-662 (2002). ZBL1013.05042.

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  • $\begingroup$ +1. Such a nice paper along with a slick use of the Newton's identity! $\endgroup$
    – Hans
    Commented May 24, 2020 at 4:38
  • $\begingroup$ @Hans Thank you for the kind words! $\endgroup$
    – Igor Rivin
    Commented May 24, 2020 at 4:42
  • $\begingroup$ You are welcome. In the proof of Theorem 6, why does $z>\sqrt{1-N_{n,p}^2}$ imply there are at most two distinct values of $x_i$? I understand that implies $z>|x_2|$. But $x_1\ge z\ge |x_2|$ holds in any case. I do not see the deduction process. Could you please explain? $\endgroup$
    – Hans
    Commented May 25, 2020 at 0:16
  • $\begingroup$ @Hans I will try to get into my time machine and remember my thoughts :) $\endgroup$
    – Igor Rivin
    Commented May 26, 2020 at 12:54
  • $\begingroup$ So, what secret did Giuseppe Luigi Lagrange divulge to you in your time machine? ;-) $\endgroup$
    – Hans
    Commented May 29, 2020 at 19:09
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I have obtained a copy of the paper by Lakshmanamurti (1950); however, it appears that I cannot post a link to it here. Using Lagrange multipliers in a standard fashion, the problem is quickly reduced to the case when the cardinality (say $c$) of the set $\{x_1,\dots,x_n\}$ is $2$ or $3$. To deal with these two cases, pretty elaborate calculus tools are used. I think the case $c=2$ can be dealt with more efficiently, compared with the way it was done in that paper.

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