A Muirhead Like Inequality 
I am looking for a proof of the inequality as follow:

Let $n$ be an integer number $n \ge 2$ and $x_1, \cdots, x_n$ and $y_1,\cdots, y_n$ are nonegative real numbers such that $(x_1,\cdots, x_n)$ majorizes $(y_1,\cdots, y_n)$; Let $0 \leq a_1, a_2,\cdots,a_n \leq 1$ then 
$$\sum_{\text{sym}} {x_1}^{a_1}{x_2}^{a_2}\cdots {x_n}^{a_n}\leq \sum_{\text{sym}} {y_1}^{a_1}{y_2}^{a_2}\cdots {y_n}^{a_n}$$
Note: The inequality above is not Muirhead's Inequality.
Example: Let $0 \leq a_i \leq 1$ then 


*

*$4^{a_1}1^{a_2}+ 4^{a_2}1^{a_1} \le 3^{a_1}2^{a_2}+ 3^{a_2}2^{a_1}$ 

*$5^{a_1}5^{a_2}2^{a_3}+5^{a_1}5^{a_3}2^{a_2}+5^{a_2}5^{a_1}2^{a_3}+5^{a_2}5^{a_3}2^{a_1}+5^{a_3}5^{a_1}2^{a_2}+5^{a_3}5^{a_2}2^{a_1}
\leq 4.5^{a_1}4^{a_2}3.5^{a_3}+4.5^{a_1}4^{a_3}3.5^{a_2}+4.5^{a_2}4^{a_1}3.5^{a_3}+4.5^{a_2}4^{a_3}3.5^{a_1}+4.5^{a_3}4^{a_1}3.5^{a_2}+4.5^{a_3}4^{a_2}3.5^{a_1}$
See also:


*

*Muirhead's Inequality

*Group multiplication of permutations
 A: $\newcommand{\al}{\alpha}
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\newcommand{\om}{\omega}
\newcommand{\R}{\mathbb{R}}
\newcommand{\Z}{\mathbb{Z}}
\newcommand{\F}{\mathcal{F}}
\newcommand{\E}{\operatorname{\mathsf E}} 
\newcommand{\Var}{\operatorname{\mathsf Var}} 
\renewcommand{\P}{\operatorname{\mathsf P}}
\newcommand{\ii}[1]{\operatorname{\mathsf I}\{#1\}} 
\newcommand{\tf}{\widetilde{f}} 
\newcommand{\tsi}{\tilde\si}$ 
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*}
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. 
