
> 
> 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](https://en.wikipedia.org/wiki/Majorization) $(y\_1,\cdots, y\_n)$; 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](https://www.cut-the-knot.org/arithmetic/algebra/MuirheadInequality.shtml).


**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](https://artofproblemsolving.com/wiki/index.php?title=Muirhead%27s_Inequality)
* [Group multiplication of permutations](https://www.cut-the-knot.org/do_you_know/PermGroup.shtml)



A Muirhead Like Inequality
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\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](https://en.wikipedia.org/wiki/Schur-convex_function), 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. 


