# prove the statement [closed]

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.

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

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## 1 Answer

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

• i am not familiar with Chebyshove's inequality.Can you explain it or provide me some links. And thanks for your information @Michael Rozenberg – emonhossain Oct 27 '18 at 9:25
• @emonhossain I added something. See now. – Michael Rozenberg Oct 27 '18 at 9:25
• @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. – Sam Streeter Oct 27 '18 at 9:29
• @Sam Streeter I added something for you. – Michael Rozenberg Oct 27 '18 at 9:34
• Great, thanks @MichaelRozenberg! – Sam Streeter Oct 27 '18 at 9:38