A symmetric polynomial inequality

Question

I improve my previous question. Because this conjecture is exactly natural development of A Muirhead Like Inequality and Muirhead's Inequality 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 $(y_1,\dots, y_n)$. 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

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 $(y_1,\dots, y_n)$. 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.

Answer 1

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.

Answer 2

Denote $$F(x_1,\dots,x_n)=\sum\limits_{sym}\left( \sum\limits_{sym} x_{i_1}^{a_{p_1}} \cdots x_{i_m}^{a_{p_m}} \right).$$ Note that the array of $y$'s is obtained from the array of $x$'s by a sequence of replacements of the form $(x_i,x_j)\mapsto (x_i+\delta,x_j-\delta)$ such that $x_i\leqslant x_i+\delta\leqslant x_j-\delta\leqslant x_j$ (and permutations of the variables). Thus for proving that $F(y_1,\dots,y_n)\geqslant F(x_1,\dots,x_n)$ it suffices to prove that $\frac{\partial F}{\partial x_i} \geqslant \frac{\partial F}{\partial x_j}$ whenever $x_i\leqslant x_j$: it is the reformulation of $F$ being increased when aforementioned $\delta$ varies from 0 to $(x_j-x_i)/2$. If we consider $F$ as a function of $x_i$ and $x_j$, it is a positive linear combination of the functions $H_{\alpha}(x_i,x_j)=x_i^\alpha+x_j^\alpha$ and $H_{\alpha,\beta}(x_i,x_j)=x_i^\alpha x_j^\beta+x_j^\alpha x_i^\beta$ for certain $\alpha,\beta \in [0,1]$ (actually we may look at $H_\alpha$ as $H_{\alpha,0}$). The inequality $\frac{\partial F}{\partial x_i} \geqslant \frac{\partial F}{\partial x_j}$ for $H_{\alpha,\beta}$ is simple: $\frac{\partial (x_i^\alpha x_j^\beta)}{\partial x_i} \geqslant \frac{\partial (x_j^\alpha x_i^\beta)}{\partial x_j}$ and viceversa.

For the second question, the function is $G_{a,b}(x_i,x_j):=a^{x_i}b^{x_j}+a^{x_j}b^{x_i}$ for some $a,b>0$ and we need to check that $\frac{\partial G_{a,b}}{\partial x_i} \leqslant \frac{\partial G_{a,b}}{\partial x_j}$ whenever $x_i<x_j$. This rewrites as $(\log b-\log a)(a^{x_i}b^{x_j}-a^{x_j}b^{x_i})\geqslant 0$ that is true.