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For $x=(x_1,\dots,x_n)\in(0,\infty)$, let
\begin{equation*}
f(x):=2\sum_{\si\in S_n}\prod_{k=1}^n x_k^{a_{\si(k)}},
\end{equation*}
where $S_n$ is the set of all permutations of the set $\{1,\dots,n\}$. By the symmetry and the Schur--Ostrowski criterion, it suffices to show that
\begin{equation*}
x_1\le x_2\overset{\text{(?)}}\implies f_1(x)-f_2(x)\ge0, \tag{1}
\end{equation*}
where $f_j(x)$ is the partial derivative of $f(x)$ in $x_j$.
Consider the bijection $S_n\ni\si\leftrightarrow\tsi\in S_n$ defined by the formula
\begin{equation*}
\tsi(k):=
\left\{
\begin{aligned}
\si(2)&\text{ if }k=1,\\
\si(1)&\text{ if }k=2,\\
\si(k)&\text { otherwise}.
\end{aligned}
\right.
\end{equation*}
Then
\begin{align*}
f(x)&:=\sum_{\si\in S_n}x_1^{a_{\si(1)}}x_2^{a_{\si(2)}}\prod_{k=3}^n x_k^{a_{\si(k)}}
+ \sum_{\tsi\in S_n}x_1^{a_{\tsi(1)}}x_2^{a_{\tsi(2)}}\prod_{k=3}^n x_k^{a_{\tsi(k)}}
=\sum_{\si\in S_n}Q P, \tag{2}
\end{align*}
where
\begin{equation*}
Q:=Q_\si(x):=x_1^{b_1}x_2^{b_2}+x_1^{b_2}x_2^{b_1},\quad P:=P_\si(x):=\prod_{k=3}^n x_k^{b_k},
\end{equation*}
\begin{equation*}
b_k:=b_{\si;k}:=a_{\si(k)}.
\end{equation*}
Denoting by $Q_j$ the partial derivative of $Q$ in $x_j$, we have
\begin{equation*}
Q_1-Q_2=b_1 (x_1 x_2)^{b_1-1}(x_2^{b_2-b_1+1}-x_1^{b_2-b_1+1})
+b_2 (x_1 x_2)^{b_2-1}(x_2^{b_1-b_2+1}-x_1^{b_1-b_2+1})\ge0.
\end{equation*}
Now noting that $P$ does not depend on $(x_1,x_2)$, we see that (1) follows from (2), as desired.
