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, ArsenBerk Oct 27 '18 at 17:12

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – José Carlos Santos, Namaste, Shailesh, Xander Henderson, ArsenBerk
If this question can be reworded to fit the rules in the help center, please edit the question.


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

  • $\begingroup$ 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
  • 1
    $\begingroup$ @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
  • 1
    $\begingroup$ @Sam Streeter I added something for you. $\endgroup$ – Michael Rozenberg Oct 27 '18 at 9:34
  • 1
    $\begingroup$ Great, thanks @MichaelRozenberg! $\endgroup$ – Sam Streeter Oct 27 '18 at 9:38

Not the answer you're looking for? Browse other questions tagged or ask your own question.