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$?$

theorem :
Let $x_1,x_2,x_3,y_1,y_2,y_3\in(0,\infty)$ then if we have :
$$x_1+x_2+x_3\geq y_1+y_2+y_3$$
And for $i\neq j,1\leq i\leq 3,1\leq j\leq 3$:
$$|x_i-x_j|\leq |y_i-y_j|$$
Then we have :
$$x_1x_2x_3\geq y_1y_2y_3$$
Case $n=3$ :
Let $y_1=\left(3-\frac{6\left(ab+bx+xa\right)}{\left(a+b+x\right)^{2}}\right)^{\frac{1}{3}}\sqrt{xa},y_2=\left(3-\frac{6\left(ab+bx+xa\right)}{\left(a+b+x\right)^{2}}\right)^{\frac{1}{3}}\sqrt{xb},y_3=\left(3-\frac{6\left(ab+bx+xa\right)}{\left(a+b+x\right)^{2}}\right)^{\frac{1}{3}}\sqrt{ab}$
Let : $x_1=x_2=x_3=\left(a+b+x\right)/3$
We have the easy inequalities :
$$0\geq\left(3-\frac{6\left(\sqrt{xa}+\sqrt{xb}+\sqrt{ab}\right)}{3\left(a+b+x\right)}\right)^{\frac{1}{2}}\left(\sqrt{xa}+\sqrt{xb}+\sqrt{ab}\right)-\left(a+b+x\right)\geq \left(3-\frac{6\left(ab+bx+xa\right)}{\left(a+b+x\right)^{2}}\right)^{\frac{1}{3}}\left(\sqrt{xa}+\sqrt{xb}+\sqrt{ab}\right)-\left(a+b+x\right)$$
Remains to apply the theorem to get the case $n=3$ with a refinement (bonus)
The general case is similar .Also think to Maclaurin's inequalities .
Some details :
A classical idea is using the Bernoulli's inequality as we have with $a,b,c>0$ :
$$-2\left(2-\frac{6\left(ab+bc+ca\right)}{\left(a+b+c\right)^{2}}\right)+3\left(2-\frac{2\left(\sqrt{ab}+\sqrt{bc}+\sqrt{ac}\right)}{\left(a+b+c\right)}\right)\leq \left(3-\frac{6\left(\sqrt{ca}+\sqrt{cb}+\sqrt{ab}\right)}{3\left(a+b+c\right)}\right)^{3}-\left(3-\frac{6\left(ab+bc+ca\right)}{\left(a+b+c\right)^{2}}\right)^{2}$$
$$
