I improve [my previous question](https://mathoverflow.net/questions/304849/combination-power-elementary-symmetric-polynomial-inequality). Because this conjecture is exactly natural development of [A Muirhead Like Inequality](https://mathoverflow.net/questions/303013/a-muirhead-like-inequality) and [Muirhead's Inequality](https://www.cut-the-knot.org/arithmetic/algebra/MuirheadInequality.shtml) so I think the conjecture is true. But I can not prove it. So I am looking for a proof of a conjecture as follows.



> 
> **Inequality 1:** Let $n>2$ and $1 \le m \le n$ be integers. Let $x\_1, \dots, x\_n$ and $y\_1,\dots, y\_n$ be nonnegative real numbers such that $(x\_1,\dots, x\_n)$ [majorizes](https://en.wikipedia.org/wiki/Majorization) $(y\_1,\dots, y\_n)$. Then for all reals $0 \leq a\_1, a\_2,\cdots,a\_n \leq 1$,
> 
> 
> $$\sum\limits\_{sym}\left( \sum\limits\_{sym} x\_{i\_1}^{a\_{p\_1}} \cdots x\_{i\_m}^{a\_{p\_m}} \right) \leq \sum\limits\_{sym}\left( \sum\limits\_{sym} y\_{i\_1}^{a\_{p\_1}} \cdots y\_{i\_m}^{a\_{p\_m}} \right) $$
> 
> 
> 


The summations are of course meant to be respectively over all $m$-tuples $(i\_1,\dots,i\_m),(p\_1,\dots,p\_m)$ with pairwise distinct entries.


* When $m=n$ this inequality is [A Muirhead Like Inequality](https://mathoverflow.net/questions/303013/a-muirhead-like-inequality)



> 
> **Inequality 2:** Let $n>2$ and $1 \le m \le n$ be integers. Let $x\_1, \dots, x\_n$ and $y\_1,\dots, y\_n$ be nonnegative real numbers such that $(x\_1,\dots, x\_n)$ [majorizes](https://en.wikipedia.org/wiki/Majorization) $(y\_1,\dots, y\_n)$. Then for all reals $ a\_1, a\_2,\dots,a\_n \geq 0$,
> 
> 
> $$\sum\limits\_{sym}\left( \sum\limits\_{sym} a\_{i\_1}^{x\_{p\_1}} \cdots a\_{i\_m}^{x\_{p\_m}} \right) \geq \sum\limits\_{sym}\left( \sum\limits\_{sym} a\_{i\_1}^{y\_{p\_1}} \cdots a\_{i\_m}^{y\_{p\_m}} \right)$$
> 
> 
> 


* When $m=n$, this inequality is just [Muirhead](https://www.cut-the-knot.org/arithmetic/algebra/MuirheadInequality.shtml).



A symmetric polynomial inequality
Both inequalities are true and can be deduced from their $m=n$ special cases.


**Inequality 1** You can prove that for any choice of $\vec{p}=(p\_1,p\_2,\dots,p\_m)$ we have
$$\sum\_{\text{sym}}x\_{i\_1}^{a\_{p\_1}}\cdots x\_{i\_m}^{a\_{p\_m}}\le \sum\_{\text{sym}}y\_{i\_1}^{a\_{p\_1}}\cdots y\_{i\_m}^{a\_{p\_m}}.$$
This follows from the Muirhead like inequality in your link by taking the exponents to be $$\vec{a}=(a\_{p\_1},a\_{p\_2},\dots,a\_{p\_m},0,\dots,0).$$ Once we have this inequality just sum over all choices of $\vec{p}$ and you obtain your inequality.


**Inequality 2** Similarly you can prove that for any choice of $\vec{i}=(i\_1,\dots,i\_m)$ we have
$$\sum\_{\text{sym}} a\_{i\_1}^{x\_{p\_1}} \cdots a\_{i\_m}^{x\_{p\_m}} \geq \sum\_{\text{sym}} a\_{i\_1}^{y\_{p\_1}} \cdots a\_{i\_m}^{y\_{p\_m}}.$$
This follows from Muirhead's inequality by taking the bases to be $(a\_{i\_1},\dots,a\_{i\_m},1,\dots,1)$. Then sum over all choices of $\vec{i}$ and obtain your inequality.


