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Statement:If $a_1,a_2,a_3\cdots a_n$ be $n$ unequal and positive quantities and if $m>r>0$ , then $$\frac{a_1^{m}+a_2^{m}\cdots +a_n^{m}}{n}> \frac{a_1^{r}+a_2^{r}\cdots +a_n^{r}}{n}. \frac{a_1^{m-r}+a_2^{m-r}\cdots +a_n^{m-r}}{n}$$

My book proves that $$a^8+b^8+c^8>a^3b^3c^3(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})$$ using the above Statement. However, I can't find the usual proofs of the Statement on the internet or in my book. Any solution will be appreciated.If you have the proof please attached the links.
Thanks in advance.

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closed as off-topic by José Carlos Santos, Namaste, Shailesh, Xander Henderson, ArsenBerk Oct 27 '18 at 17:12

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It's just the Chebyshove's inequality, which we can prove by Rearrangement.

Indeed,$$\left(a_1^{r},a_2^{r},\cdots,a_n^{r}\right)$$ and $$\left(a_1^{m-r},a_2^{m-r},\cdots,a_n^{m-r}\right)$$ they are the same ordered.

Thus, for all permutation $\sigma\in S_n$ we obtain: $$\sum_{k=1}^na^m=\sum_{k=1}^na_k^ra_k^{m-r}\geq\sum_{k=1}^na_k^ra_{\sigma(k)}^{m-r},$$ which gives $$n\sum_{k=1}^na^m\geq\sum_{k=1}^na_k^r\sum_{k=1}^na_k^{m-r}$$ and we are done!

About the Chebyshov's inequality see here: https://en.wikipedia.org/wiki/Chebyshev%27s_sum_inequality

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  • $\begingroup$ i am not familiar with Chebyshove's inequality.Can you explain it or provide me some links. And thanks for your information @Michael Rozenberg $\endgroup$ – emonhossain Oct 27 '18 at 9:25
  • $\begingroup$ @emonhossain I added something. See now. $\endgroup$ – Michael Rozenberg Oct 27 '18 at 9:25
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    $\begingroup$ @MichaelRozenberg I think they want a specific link to the result. Can you provide one? I'd also be interested, as I'm having trouble finding an inequality by the name of Chebyshev's Inequality which I can make sense of in this context. $\endgroup$ – Sam Streeter Oct 27 '18 at 9:29
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    $\begingroup$ @Sam Streeter I added something for you. $\endgroup$ – Michael Rozenberg Oct 27 '18 at 9:34
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    $\begingroup$ Great, thanks @MichaelRozenberg! $\endgroup$ – Sam Streeter Oct 27 '18 at 9:38

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