Combination power elementary symmetric polynomial inequality Combine my first previous question and  second previous question with the Muirhead inequality. I have posed conjectures of two inequalities 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 $0 \leq a_1, a_2,\cdots,a_n \leq 1$, 
$$\sum\limits_{1 \le p_1 <\cdots < p_m \le n}\left(  \sum\limits_{1 \le i_1 <\cdots < i_m \le n} x_{i_1}^{a_{p_1}} \cdots x_{i_m}^{a_{p_m}} \right) \leq \sum\limits_{1 \le p_1 <\cdots < p_m \le n}\left(  \sum\limits_{1 \le i_1 <\cdots < i_m \le n} y_{i_1}^{a_{p_1}} \cdots y_{i_m}^{a_{p_m}} \right) $$

Example for inequality 1 with $n=3, m=2$:
$${x_1}^{a_1}.{x_2}^{a_2}+{x_1}^{a_1}.{x_3}^{a_2}+{x_2}^{a_1}.{x_3}^{a_2}+{x_1}^{a_2}.{x_2}^{a_3}+{x_1}^{a_2}.{x_3}^{a_3}+{x_2}^{a_2}.{x_3}^{a_3}+{x_1}^{a_1}.{x_2}^{a_3}+{x_1}^{a_1}.{x_3}^{a_3}+{x_2}^{a_1}.{x_3}^{a_3} \leq {y_1}^{a_1}.{y_2}^{a_2}+{y_1}^{a_1}.{y_3}^{a_2}+{y_2}^{a_1}.{y_3}^{a_2}+{y_1}^{a_2}.{y_2}^{a_3}+{y_1}^{a_2}.{y_3}^{a_3}+{y_2}^{a_2}.{y_3}^{a_3}+{y_1}^{a_1}.{y_2}^{a_3}+{y_1}^{a_1}.{y_3}^{a_3}+{y_2}^{a_1}.{y_3}^{a_3}$$

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$, 
$$\sum\limits_{1 \le p_1 <\cdots < p_m \le n}\left(  \sum\limits_{1 \le i_1 <\cdots < i_m \le n} a_{i_1}^{x_{p_1}} \cdots a_{i_m}^{x_{p_m}} \right) \geq \sum\limits_{1 \le p_1 <\cdots < p_m \le n}\left(  \sum\limits_{1 \le i_1 <\cdots < i_m \le n} a_{i_1}^{y_{p_1}} \cdots a_{i_m}^{y_{p_m}} \right)$$

Example for inequality 2 with $n=3, m=2$:
$${a_1}^{x_1}.{a_2}^{x_2}+{a_1}^{x_1}.{a_3}^{x_2}+{a_2}^{x_1}.{a_3}^{x_2}+{a_1}^{x_2}.{a_2}^{x_3}+{a_1}^{x_2}.{a_3}^{x_3}+{a_2}^{x_2}.{a_3}^{x_3}+{a_1}^{x_1}.{a_2}^{x_3}+{a_1}^{x_1}.{a_3}^{x_3}+{a_2}^{x_1}.{a_3}^{x_3} \geq 
{a_1}^{y_1}.{a_2}^{y_2}+{a_1}^{y_1}.{a_3}^{y_2}+{a_2}^{y_1}.{a_3}^{y_2}+{a_1}^{y_2}.{a_2}^{y_3}+{a_1}^{y_2}.{a_3}^{y_3}+{a_2}^{y_2}.{a_3}^{y_3}+{a_1}^{y_1}.{a_2}^{y_3}+{a_1}^{y_1}.{a_3}^{y_3}+{a_2}^{y_1}.{a_3}^{y_3}$$


My question: I am looking for a proof of two inequalities above.


 A: These inequalities are true when $m=1$, and they both follow from Karamata's inequality, since $x^a$ is concave when $0\le a\le 1$, and $a^x$ is convex. However they are both false for every $m\geq 2$.
Counterexample to Inequality 1 Set $a_1=1$ and $a_i=0$ for $i>1$. Then your inequality now reads $$\sum_{i=1}^{n-m+1}\binom{n-1}{m-1}\binom{n-i}{m-1}x_i\le \sum_{i=1}^{n-m+1}\binom{n-1}{m-1}\binom{n-i}{m-1}y_i.$$
Notice that since $m\geq 2$ we have $n-m+1\le n-1$ so that $x_n$ and $y_n$ don't appear in the inequality. So let $x_1=x_2=\cdots=x_{n-1}=M$ be a very large number and take $x_n=0$. Then take $y_1=y_2=\cdots=y_{n-1}=M-1$ and $y_n=n-1$. This is easily seen to be a counterexample since every term on the left hand side is greater than the corresponding term on the right.
Counterexample to Inequality 2 We will take $a_1=a_2=\cdots a_{n-1}=1$ and $a_n=2$ as well as $x_1=nM$ and $x_i=0$ for $i\geq 2$, and all $y_i=M$. The left hand side is seen to equal $\binom{n}{m}^2=O(1)$ whereas the right hand side is $O(2^M)$, so the inequality breaks as soon as $M$ is large enough.
