$\newcommand{\al}{\alpha} \newcommand{\be}{\beta} \newcommand{\de}{\delta} \newcommand{\De}{\Delta} \newcommand{\ep}{\varepsilon} \newcommand{\ga}{\gamma} \newcommand{\Ga}{\Gamma} \newcommand{\la}{\lambda} \newcommand{\si}{\sigma} \newcommand{\Si}{\Sigma} \newcommand{\thh}{\theta} \newcommand{\om}{\omega} \newcommand{\R}{\mathbb{R}} \newcommand{\Z}{\mathbb{Z}} \newcommand{\F}{\mathcal{F}} \newcommand{\E}{\operatorname{\mathsf E}} \newcommand{\Var}{\operatorname{\mathsf Var}} \renewcommand{\P}{\operatorname{\mathsf P}} \newcommand{\ii}[1]{\operatorname{\mathsf I}\{#1\}} \newcommand{\tf}{\widetilde{f}} \newcommand{\tsi}{\tilde\si}$
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*}\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.