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$?$
 A: Let $k\ge2$ be fixed and suppose that $S(k)$ holds true.
$$S(k):({x_1}^2+{x_2}^2+\cdots+{x_k}^2)x_1x_2\cdots x_k\le(k)\left(\frac{x_1+x_2+\cdots+x_k}{k}\right)^{k+2}$$ It remains to prove:
$$S(k+1):({x_1}^2+{x_2}^2+\cdots+{x_k}^2+x^2)x_1x_2\cdots x_k\cdot x\le(k+1)\left(\frac{x_1+x_2+\cdots+x_k+x}{k+1}\right)^{k+3}$$ Using the fact that $x=x_{k+1}$
Now, Put $A=\frac{x_1+x_2+\cdots+x_k}{k}$ and $P=x_1x_2\cdots x_k$
$\implies$ $$S(k):({x_1}^2+{x_2}^2+\cdots+{x_k}^2)P\le kA^{k+2}$$
And it remains to prove that $$S(k+1):({x_1}^2+{x_2}^2+\cdots+{x_k}^2+x^2)Px\le (k+1)\left(\frac{kA+x}{k+1}\right)^{k+3}$$ The left hand side of $S(k+1)$ is
$$({x_1}^2+{x_2}^2+\cdots+{x_k}^2)Px+Px^3\le kA^{k+2}x+Px^3$$ The above inequality came by using the fact of $S(k)$
So to prove $S(k+1)$, it suffices to prove that
$$kA^{k+2}x+Px^3\le (k+1)\left(\frac{kA+x}{k+1}\right)^{k+3}$$
By $AM\ge GM$, $P\le A^k$, so it suffices to prove that
$$kA^{k+2}x+A^kx^3\le (k+1)\left(\frac{kA+x}{k+1}\right)^{k+3}$$
Now restrict to the situation where the sum $x_1+x_2+\cdots+x_k+x$ is held constant,  and prove
the result with this added constraint. The general result then follows immediately; observe that for any constant $c$, the statement $S(n)$ holds for $x_1,\cdots,x_n$ if and only if it holds for $cx_1,\cdots,cx_n$ (the factor $c^{n+2}$ appears on each side). So, consider only those $(x_1,\cdots,x_k,x)\in\mathbb{R}^{k+1}$ for which $x_1+x_2+\cdots+x_k+x=k+1$,that is,
$$kA+x=k+1$$
So, to prove $S(k + 1)$, it suffices to show $$kA^{k+2}x+A^kx^3\le k+1$$
The left hand of the above inequality is a function of $A$ (and $x = k + 1 − kA$, also a function of $A$), and so this expression is maximized using calculus:
$$\frac{d}{dA}=[kA^{k+2}x+A^kx^3]=k(k+2)A^{k+1}x+kA^{k+2}\frac{dx}{dA}+kA^{k-1}x^3+A^k\cdot3x^2\frac{dx}{dA}$$
$$=k(k+2)A^{k+1}x+kA^{k-1}x^3-k(kA^{k+2}+A^k\cdot3x^2)$$
Putting $A = tx$, this expression becomes (after a bit of algebra)
$$(1-t)(kt^2-2t+1)kt^{k-1}x^{k+2}$$
Since $k\ge2$, the above has roots at only $t = 0$ and $t = 1$, and so the derivative is positive for $0 < t < 1$ and negative for $t > 1$. Thus, $kA^{k+2}x+A^kx^3$ achieves a maximum when $t = 1$, that is, when $A = x = 1$. Hence, $$kA^{k+2}x+A^kx^3\le k+1$$ and so $S(k + 1)$ follows, completing the inductive step.
Thus, by mathematical induction, for all $n ≥ 1$, the statement $S(n)$ is true.
A: WLOG, assume that $x_1+\dots+x_n = n$. We need to show that $f(x_1,\dots,x_n)\le n$, where
\begin{equation}
  f(x_1,\dots,x_n) = x_1\dots x_n(x_1^2+\dots+x_n^2).
 \end{equation}
Since the constraint set is closed and bounded, $f$ attains its maximum. In the following, we let $(x_1,\dots,x_n)$ denote such a maximum solution.  Then, $x_i\neq 0 \forall i$, because otherwise $f = 0$, which is clearly not the maximum value.
Assume, to later arrive at a contradiction, that there exist two components of $(x_1,\dots,x_n)$ that are not equal to each other, say, $x_1\neq x_2$. Put $P = x_3\dots x_n, Q = x_3^2+\dots+x_n^2$, and $t=\frac{x_1+x_2}{2}$. Then we have
\begin{align}
  f(x_1,\dots,x_n) - f(t,t,x_3,\dots,x_n) &= Px_1x_2(x_1^2+x_2^2+Q) - Pt^2(2t^2+Q)\\
  &= -P\left[2t^4 - x_1x_2(x_1^2+x_2^2) + Q(t^2 - x_1x_2)\right]\\
  &=-P\left[ \frac{1}{8}(x_1-x_2)^4 + \frac{Q}{4}(x_1-x_2)^2\right] < 0,
 \end{align}
which is absurd as $(x_1,\dots,x_n)$ is a maximum solution. We conclude that $f$ achieves its maximum when $x_1=x_2=\dots = x_n$, QED.

Remarks.

*

*We have shown that $f(x_1,\dots,x_n) \le f(t,t,x_3,\dots,x_n)$, where $t=\frac{x_1+x_2}{2}$, which should immediately yield the conclusion using the classical mixing variable theorem, but the above proof by contradiction does not require this theorem.


*The same technique can be used to prove the following generalization:
\begin{equation}
  (x_1\dots x_n)^p(x_1^q+\dots+x_n^q) \le n\left(\frac{x_1+\dots+x_n}{n}\right)^{np+q},
 \end{equation}
where $x_1,\dots,x_n\ge 0$ and $2p\ge \max\{q(q-1),0\}$.
A: Simple AM-GM + Newton's solution is as follows:
$$(x_1+x_2+\dots+x_n)^2 = \sum x_i^2 + \sum_{i\neq j}x_ix_j: = A + (n-1)B$$
with $A = \sum x_i^2\geq B = \dfrac{1}{n-1}\sum\limits_{i\neq j}x_ix_j,$ which is just pairwise AM-GM or rearrangement if you like. Then,
$$\left(\dfrac{x_1+x_2+\dots +x_n}{n}\right)^{2n}=\left(\dfrac{A+(n-1)B}{n^2}\right)^n\geq \dfrac{1}{n^n}AB^{n-1}.$$
Using this, it suffices to prove then that:
$$\left(\dfrac{B}{n}\right)^{n-1}\geq x_1x_2\dots x_n\left(\dfrac{x_1+x_2+\dots +x_n}{n}\right)^{n-2}.$$
In the language of elementary symmetric polynomials as you mentioned, this is written as:
$$S_2^{n-1}\geq S_nS_1^{n-2}\iff \left(\dfrac{S_2}{S_1}\right)^{n-1}\geq\dfrac{S_n}{S_1} $$
But this follows from repeated applications of Newton's inequality:
$$\dfrac{S_2}{S_1}\geq\dfrac{S_3}{S_2}\geq\dfrac{S_4}{S_3}\
\geq\dots \geq \dfrac{S_n}{S_{n-1}}\implies \left(\dfrac{S_2}{S_1}\right)^{n-1}\geq\prod_{k=1}^{n-1}\dfrac{S_{k+1}}{S_k} = \dfrac{S_n}{S_1}.$$
A: 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 $a,b,c>0$ :
$$-2\left(2-\frac{6\left(ab+bc+ca\right)}{\left(a+b+c\right)^{2}}\right)+3\left(2-\frac{2\left(\sqrt{ab}+\sqrt{bc}+\sqrt{ac}\right)}{\left(a+b+c\right)}\right)\leq \left(3-\frac{6\left(\sqrt{ca}+\sqrt{cb}+\sqrt{ab}\right)}{3\left(a+b+c\right)}\right)^{3}-\left(3-\frac{6\left(ab+bc+ca\right)}{\left(a+b+c\right)^{2}}\right)^{2}$$
$$
A: Another simple solution by induction, motivated by OP's questions and by the currently most voted answer (which is also by induction but a bit complicated).
\begin{equation}
\frac{{x_1}^2+{x_2}^2+\cdots+{x_n}^2}{n}x_1x_2\cdots x_n\le\left(\frac{x_1+x_2+\cdots+x_n}{n}\right)^{n+2}
\end{equation}
First, the inequality is clearly true for $n=2$, as it can be reduced to $(x_1-x_2)^4\ge 0$. Denote $p = x_1\dots x_n, q = (x_1^2+\dots + x_n^2)/n, s = (x_1+\dots+x_n)/n$, and $x=x_{n+1}$. Assume that the inequality is true for $n$ numbers, that is, $s^{n+2} \ge pq$. We will show that it is also true for $n+1$ numbers, that is, $f(x) \ge 0$, where
\begin{equation}
  f(x) = \left(\frac{x+ns}{n+1}\right)^{n+3} - \frac{px(x^2 + nq)}{n+1}.
 \end{equation}
WLOG, assume that $x_{n+1}$ is the maximum component. Then, it suffices to show that $f(x) \ge 0$ whenever $x\ge s$. Taking derivatives:
\begin{align}
 f'(x) &= \frac{n+3}{n+1}\left(\frac{x+ns}{n+1}\right)^{n+2} - \frac{p(3x^2 + nq)}{n+1},\\
 f''(x) &= \frac{(n+3)(n+2)}{(n+1)^2}\left(\frac{x+ns}{n+1}\right)^{n+1} - \frac{6px}{n+1}.
\end{align}
We can observe that $f(s)\ge 0$ and $f'(s) \ge 0$. Indeed, as $s^{n+2} \ge pq$ and $s^n\ge p$ (AM-GM), we have
\begin{align}
 f(s) &= s^{n+3} - \frac{ps(s^2 + nq)}{n+1} = \frac{ns(s^{n+2}-pq) + s^3(s^n-p)}{n+1} \ge 0,\\
 f'(s) &= \frac{(n+3)s^{n+2}}{n+1} - \frac{p(3s^2 + nq)}{n+1} = \frac{n(s^{n+2}-pq)}{n+1} + \frac{3s^2(s^n-p)}{n+1} \ge 0.
\end{align}
Next, notice that $\left(\frac{x+ns}{n+1}\right)^{n+1} \ge xs^n \ge xp$ (AM-GM), we have
\begin{equation}
 f''(x) \ge \frac{(n+3)(n+2)px}{(n+1)^2} - \frac{6px}{n+1} = \frac{n(n-1)px}{(n+1)^2} \ge 0.
\end{equation}
Therefore, $f'$ is increasing (on $[s,+\infty)$) and thus $f'(x)  \ge f'(s) \ge 0$, hence $f$ is increasing and thus $f(x)\ge f(s) \ge 0$. QED.
A: My try :
Lemma :
Let $x>0$ then define :
$$f(x)=\ln(x)$$
Then a trivial consequence is :
$$f'''(x)>0$$
Now the problem is with the constraint of the OP:
$$\ln\left(\frac{{x_1}^2+{x_2}^2+\cdots+{x_n}^2}{n}\right) +\ln(x_1)+\ln(x_2)+\cdots +\ln(x_n)-\ln\left(\frac{x_1+x_2+\cdots+x_n}{n}\right)^{n+2}\leq 0$$
Now using the lemma above and the Equal Variable Method with the constraint :
$$x_1+x_2\cdots+x_n=constant,x_1^2+x_2^2\cdots+x_n^2=constant$$
You can maximize the function and get another inequality really simpler
Edit after  RiverLi's comment :
We need to show for $n\geq 3$ an integer  $x,a>0$:
$$f(x)=\frac{\left(\left(n-1\right)\left(\frac{x}{a}\right)^{2}+1\right)}{n}\left(\frac{x}{a}\right)^{\left(n-1\right)}-\left(\frac{\left(n-1\right)x\cdot\frac{1}{a}+1}{n}\right)^{\left(n+2\right)}\leq 0\tag{E}$$
we have taking logarithm on both side in $E$ and differentiating  where $a=1$:
$$h(x)=\frac{\left(n-1\right)\left(a-x\right)\left(a^{2}-2ax+\left(n-1\right)x^{2}\right)}{x\left(a+\left(n-1\right)x\right)\left(a^{2}+x^{2}\left(n-1\right)\right)}$$
For $0<x\leq a $ and $n\geq 2$ we have the following trivial inequalities :
$$a-x\geq 0$$
And :
$$a^{2}-2ax+\left(n-1\right)x^{2}\geq 0$$
So the derivative is positive and the function is increasing .Now remarking for $x=a$ we get zero as value the inequality is established .
