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.
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