I think the proper problem was involving $x_{1},x_{2},\ldots,x_{s}$. Am I wrong, or it's straightforward using Chebyshev?

dyd wrote: I think the proper problem was involving $x_{1},x_{2},\ldots,x_{s}$. Am I wrong, or it's straightforward using Chebyshev? Yes, it's Tcebyshev and AM-GM inequality. Actually it's true for arbitrary positive $m \geq n$. And moreover if $m \geq n \geq 1$ then \[\frac{x_{1}^{m}+x_{2}^{m}+\dots+x_{s}^{m}}{x_{1}^{n}+x_{2}^{n}+\dots+x_{s}^{n}}\geq \left( \frac{x_{1}+x_{2}+\dots+x_{s}}{s}\right)^{m-n}.\] It is very well-known corollary from power mean inequality.

Simply with weighed AM-GM $\sum_{cyc}x_{1}^{m}= \sum_{cyc}\frac{(m+n(s-1))x_{1}^{m}+(m-n)x_{2}^{m}+\dots+(m-n)x_{s}^{m}}{ms}\ge \sum_{cyc}\left( x_{1}^{m(m+n(s-1))}x_{2}^{m(m-n)}x_{3}^{m(m-n)}\dots x_{s}^{m(m-n)}\right)^{\frac{1}{ms}}=$$\sum_{cyc}\left( x_{1}^{\frac{m+n(s-1)}{s}}x_{2}^{\frac{m-n}{s}}\dots x_{s}^{\frac{m-n}{s}}\right)=\sum_{cyc}x_{1}^{n}$ using $x_{1}^{\frac{m-n}{s}}x_{2}^{\frac{m-n}{s}}\dots x_{s}^{\frac{m-n}{s}}=1$

or more simply \[LHS\geq \frac{1}{(s-1)!}\sum_{sym}x_{1}^{\frac{m+n(s-1)}{s}}{\left(x_{2}\cdot x_{3}\cdots x_{s}\right)}^{\frac{m-n}{s}}=RHS\cdot{\left(x_{1}\cdots x_{s}\right)}^{\frac{m-n}{s}}\] for bunching