Positive real numbers $x_1,x_2,\ldots,x_n$ satisfy $\sum_{i=1}^{n}{x_i}=n$ , prove: $\displaystyle\sum_{sym}{\prod_{i=1}^{n}{x_i^i}}\leqslant n!$ 
If positive real numbers $x_1,x_2,\ldots,x_n$ satisfy $\displaystyle\sum_{i=1}^{n}{x_i}=n$ , prove or falsify:


$$\sum_{\text{sym}}{\prod_{i=1}^{n}{x_i^i}}\leqslant n!$$

Here I'll explain the notation 'sym',
$$
\sum_\text{sym}{f(x_1,x_2,\ldots,x_n)=\sum_{\sigma\in S_n}{f(x_{\sigma(1)},x_{\sigma(2)},\ldots,x_{\sigma(n)})}}
$$
where $S_n$ is the permutation group of degree $n$ .
For example,
$$
\sum_{\text{sym}}{x^3y^2z}=x^3y^2z+x^3yz^2+x^2y^3z+x^2yz^3+xy^3z^2+xy^2z^3
$$
The above example corresponds to the LHS of the case $n=3$ of this problem.
I haven't push forward this question much, all I can prove is the case $n=2$ , which is immediately true by AM-GM. For $n=3$ , I tried to homogenise the inequality and use the 'SOS' method(Sum Of Squares), but it's apparent that this attempt can't be generalised.
I've managed to solved the $n=3$ case by the $pqr$ method. First, note the identity
$$
\sum_{\text{sym}}{a^3b^2c}=abc(a+b+c)(ab+bc+ca)-3(abc)^2
$$
Let $p:=a+b+c,~q:=ab+bc+ca,~r:=abc$ , the inequality is equivalent to
$$
pqr-3r^2\leqslant 6
$$
By Schur's Inequality of degree 3, we have
$$
a(a-b)(a-c)+b(b-a)(b-c)+c(c-a)(c-b)=p^3-4pq+9r\geqslant0
$$
From this we may obtain $q\leqslant(p^3+9r)/(4p)$ , and as constrained we have $p=3$ ,these reduce the inequality to
$$
3r^2-3r~\left(\frac{27+9r}{12}\right)+6=3(r-1)(r-8)\geqslant0
$$
By AM-GM, we obtain $r\leqslant(p/3)^3=1$ ,which proves the inequality.
Here's a proof on the site 'zhihu', I've reposted it as community wiki. This proof is invalid because its lemma is wrong.
Thanks to Sangchul Lee, this inequality is not true for $n\geqslant6$ , which means the cases $n=4$ and $n=5$ might be very difficult to prove.
 A: A simulation suggests that the inequality is true for $n \leq 5$.
However, the inequality is not true for $n \geq 6$. For example, let $n = 6$, introduce the parameter $\varepsilon \in (0, 1)$, and define $(x_1, \ldots, x_6)$ by
$$ x_1 = 1 - 2\varepsilon, \qquad x_2 = x_3 = 1 + \varepsilon, \qquad x_4 = x_5 = x_6 = 1. $$
Clearly $\sum_{i=1}^{6} x_i = 6$. However,
\begin{align*}
\sum_{\sigma \in S_6} \prod_{i=1}^{6} x_{\sigma(i)}^i
&= \sum_{\sigma \in S_6} \prod_{i=1}^{6} x_i^{\sigma(i)} \\
&= \sum_{\sigma \in S_6} (1 - 2\varepsilon)^{\sigma(1)}(1 + \varepsilon)^{\sigma(2)+\sigma(3)} \\
&= 720 + 2520 \varepsilon^3 + \mathcal{O}(\varepsilon^4) \\
&> 6!
\end{align*}
for any sufficiently small $\varepsilon$. In fact, even moderately small values such as $\varepsilon = 0.2$ work. Below is a comparison of the values between the above sum as a function of $\varepsilon$ and $6!$:

For $n \geq 7$, an even simpler choice $x_1 = 1-\varepsilon$, $x_2 = 1+\varepsilon$, $x_3 = \ldots = x_n = 1$ works.
A: A proof for $n=4$ and non-negative variables.
We need to prove that $$abcd\sum_{sym}a^3b^2c\leq24,$$ where $a+b+c+d=4$.
Indeed, let $a+b+c+d=4u$, $ab+ac+ad+bc+bd+cd=6v^2$, $abc+abd+acd+bcd=4w^3$ and $abcd=t^4$.
Thus, we need to prove that $$t^4(96uv^2w^3-48w^6-48u^2t^4+24v^2t^4)\leq24u^{10}$$ or
$$u^{10}-4uv^2w^3t^4+2w^6t^4+2u^2t^8-v^2t^8\geq0.$$
Since $u^2\geq v^2$ and $t^8\geq0$, by AM-GM we obtain:
$$u^{10}-4uv^2w^3t^4+2w^6t^4+2u^2t^8-v^2t^8\geq u^{10}-4uv^2w^3t^4+2w^6t^4+u^2t^8\geq$$
$$\geq4\sqrt[4]{u^{10}\cdot(w^6t^4)^2\cdot u^2t^8}-4uv^2w^3t^4=4uw^3t^4(u^2-v^2)\geq0.$$
A: This a proof on the site 'zhihu' , and I'll repost it here in English. (All I've done is translation, please inform me if it's illegal.)
Lemma. For all integers $p,q\geqslant1$ , let
$$
f(x):=x^p(m-x)^q+x^q(m-x)^p~(0\leqslant x\leqslant m)
$$
Then $f(x)$ is monotonically increasing on the interval $\left[0,m/2\right]$ and monotonically decreasing on the interval $\left[m/2,m\right]$ .
Proof. By taking derivatives, we obtain
$$
f^\prime(x)=(p+q)(m-2x)x^{p-1}(m-x)^{q-1}
$$
The sign of $f^\prime(x)$ depends solely on the sign of $(m-2x)$ , which proves the statement. $\square$
Theorem. Given positive real numbers $x_1,x_2,\ldots,x_n$ such that $\sum_{i=1}^{n}{x_i}=n$ , the following inequality holds true.
$$
\sum_{\sigma\in S_n}\prod_{i=1}^{n}{x_i^{\sigma(i)}}\leqslant n!
$$
Proof. If $x_i=1$ for all $i$ , the equality occurs. Now for cases where $x_i$ don't all equal to $1$ , we may divide them into three categories depending on whether they are $<1,=1,>1$ .
Select $x_u<1,~x_v>1$ , let $d:=\min\{1-x_u,x_v-1\}$ . Define a transformation as substituting $x_u$ with $x_u+d$ and $x_v$ with $x_v-d$ .
Note that this transformation maintains the constraint $\sum_{i=1}^{n}{x_i}=n$ , and it leads to at least one of $x_u$ and $x_v$ turning into $1$ . By the constraint $\sum_{i=1}^{n}{x_i}=n$ , we can tell that whenever there exists some $x_i<1$ , there must exist some $x_j>1$ . It can be concluded from above that after finitely many times of transformation, we may obtain the case where $x_i=1$ for all $i$ , in which the equality occurs.
Therefore, it suffices to show that $\sum_{\sigma\in S_n}\prod_{i=1}^{n}{x_i^{\sigma(i)}}$ is monotonically increasing after each transformation.
Consider a specific pair of $\sigma_1$ and $\sigma_2$ such that
$$
\sigma_2(i)=\begin{cases}\sigma_1(v)&i=u\\\sigma_1(u)&i=v\\\sigma_1(i)&\text{otherwise}\end{cases}
$$
Denote $X:=\prod_{i=1}^{n}{x_i^{\sigma_1(i)}}+\prod_{i=1}^{n}{x_i^{\sigma_2(i)}}$ and $Y:=\prod_{i=1}^{n}{y_i^{\sigma_1(i)}}+\prod_{i=1}^{n}{y_i^{\sigma_2(i)}}$ , where
$$
y_i=\begin{cases}x_i+d&i=u\\x_i-d&i=v\\x_i&\text{otherwise}\end{cases}
$$
Firstly, we'll show that $X\leqslant Y$ . By eliminating their common factor, it's equivalent to show that
$$
x_u^{\sigma_1(u)}x_v^{\sigma_1(v)}+x_u^{\sigma_2(u)}x_v^{\sigma_2(v)}
\leqslant
y_u^{\sigma_1(u)}y_v^{\sigma_1(v)}+y_u^{\sigma_2(u)}y_v^{\sigma_2(v)}
$$
Let $m:=x_u+x_v,~p=\sigma_1(u),~q=\sigma(v)$ , the above equality is equivalent to $f(x_u)\leqslant f(y_u)$ , where $f$ is defined in the lemma.
Note that $x_u<1<x_v$ and $y_u\leqslant 1\leqslant y_v$ , thus $x_u,y_u\in\left[0,m/2\right]$ . By the lemma, we obtain $f(x_u)\leqslant f(y_u)$ , which proves the inequality $X\leqslant Y$ .
Note that all of the permutations in $S_n$ can be divided into  $n!/2$ pairs such that the two permutations $\sigma_1,~\sigma_2$ in each pair satisfy
$$
\sigma_2(i)=\begin{cases}\sigma_1(v)&i=u\\\sigma_1(u)&i=v\\\sigma_1(i)&\text{otherwise}\end{cases}
$$
where $u$ and $v$ are arbitrary intergers between $1$ and $n$ .
Now that finitely many times of transformation ends up with the  occurence of equality, in the process of which the $\text{LHS}$ is monotonically increasing, the theorem is proven. $\square$
