<s>[INST] Prove that $\frac{{x_1}^2+{x_2}^2+\cdots+{x_n}^2}{n}x_1x_2\cdots x_n\le\left(\frac{x_1+x_2+\cdots+x_n}{n}\right)^{n+2}$ 
Show that for any non negative real numbers $x_1,x_2,\cdots x_n,$
$$\frac{{x_1}^2+{x_2}^2+\cdots+{x_n}^2}{n}x_1x_2\cdots x_n\le\left(\frac{x_1+x_2+\cdots+x_n}{n}\right)^{n+2}$$

My work:
Let$$S(n)=\frac{{x_1}^2+{x_2}^2+\cdots+{x_n}^2}{n}x_1x_2\cdots x_n\le\left(\frac{x_1+x_2+\cdots+x_n}{n}\right)^{n+2}$$
By theorem of triviality, if any of $x_i$'s are $0$ the inequality is certainly true. So assume all numbers are $\gt0$
$S(1)$ says ${x_1}^3\le {x_1}^3$ which is certainly true.
$S(2)$ says $({x_1}^2+{x_2}^2)x_1x_2\le\frac18(x_1+x_2)^4$ which reduces to $0\le(x_1-x_2)^4$ which is certainly true.
Assume $S(k)$ is true. Now we just needs to prove that $S(k+1)$ is true. But I'm having a hard time in doing that.
Any help is greatly appreciated. Or is there any better method than induction$?$
 [/INST] Another simple solution by induction, motivated by OP's questions and by the currently most voted answer (which is also by induction but a bit complicated).
\begin{equation}
\frac{{x_1}^2+{x_2}^2+\cdots+{x_n}^2}{n}x_1x_2\cdots x_n\le\left(\frac{x_1+x_2+\cdots+x_n}{n}\right)^{n+2}
\end{equation}
First, the inequality is clearly true for $n=2$, as it can be reduced to $(x_1-x_2)^4\ge 0$. Denote $p = x_1\dots x_n, q = (x_1^2+\dots + x_n^2)/n, s = (x_1+\dots+x_n)/n$, and $x=x_{n+1}$. Assume that the inequality is true for $n$ numbers, that is, $s^{n+2} \ge pq$. We will show that it is also true for $n+1$ numbers, that is, $f(x) \ge 0$, where
\begin{equation}
  f(x) = \left(\frac{x+ns}{n+1}\right)^{n+3} - \frac{px(x^2 + nq)}{n+1}.
 \end{equation}
WLOG, assume that $x_{n+1}$ is the maximum component. Then, it suffices to show that $f(x) \ge 0$ whenever $x\ge s$. Taking derivatives:
\begin{align}
 f'(x) &= \frac{n+3}{n+1}\left(\frac{x+ns}{n+1}\right)^{n+2} - \frac{p(3x^2 + nq)}{n+1},\\
 f''(x) &= \frac{(n+3)(n+2)}{(n+1)^2}\left(\frac{x+ns}{n+1}\right)^{n+1} - \frac{6px}{n+1}.
\end{align}
We can observe that $f(s)\ge 0$ and $f'(s) \ge 0$. Indeed, as $s^{n+2} \ge pq$ and $s^n\ge p$ (AM-GM), we have
\begin{align}
 f(s) &= s^{n+3} - \frac{ps(s^2 + nq)}{n+1} = \frac{ns(s^{n+2}-pq) + s^3(s^n-p)}{n+1} \ge 0,\\
 f'(s) &= \frac{(n+3)s^{n+2}}{n+1} - \frac{p(3s^2 + nq)}{n+1} = \frac{n(s^{n+2}-pq)}{n+1} + \frac{3s^2(s^n-p)}{n+1} \ge 0.
\end{align}
Next, notice that $\left(\frac{x+ns}{n+1}\right)^{n+1} \ge xs^n \ge xp$ (AM-GM), we have
\begin{equation}
 f''(x) \ge \frac{(n+3)(n+2)px}{(n+1)^2} - \frac{6px}{n+1} = \frac{n(n-1)px}{(n+1)^2} \ge 0.
\end{equation}
Therefore, $f'$ is increasing (on $[s,+\infty)$) and thus $f'(x)  \ge f'(s) \ge 0$, hence $f$ is increasing and thus $f(x)\ge f(s) \ge 0$. QED.
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