# 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.

• In inequality 1, it appears that the left hand side should only have monomials with three variables ($x_p, x_q$ and $x_u$), but you still have $\dots$'s. On the right, it appears that we have all $a_i$'s occurring. How should I interpret this? Jul 12, 2018 at 14:23
• Supposedly you mean that all products have a fixed number $m<n$ of factors? Jul 12, 2018 at 14:27
• In perfectly correct notation, you would need nested indices, with the terms being something like $\sum\limits_{1 \le p_1 <\cdots < p_m \le n}\left( \sum\limits_{1 \le i_1 <\cdots < i_m \le n} x_{i_1}^{a_{p_1}} \cdots x_{i_m}^{a_{p_m}} \right)$ Jul 12, 2018 at 14:48
• Done. :) $\quad$ Jul 12, 2018 at 15:08
• I don't think the 2nd one is Muirhead. Note that Muirhead is about symmetrical sums, i.e. adding all the permutations of the $a_i$'s! E.g. if there is a term ${a_1}^{x_1}{a_2}^{x_2}$ on the LHS, there should also be ${a_2}^{x_1}{a_1}^{x_2}$. Jul 12, 2018 at 17:18

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.

• I think $n=m$ is true because mathoverflow.net/questions/303013/a-muirhead-like-inequality Jul 17, 2018 at 2:55
• When $n=m$ your inequality does not agree with the one in that link. Jul 17, 2018 at 2:59
• Can You see my comment in the active answer? Jul 17, 2018 at 7:39

This is not an answer, this is a message to @Gjergji Zaimi. Thank You very much. (And thank to dear Wolfgang very much). Your answer is true with the version above. But if You see my comment to You and Wolfgang. I said that when $m=n$ the inequality 1 is A Muirhead Like Inequality. The inequality 2 is A Muirhead. But You and Wolfgang said no.

So I see detail my question again (with Wolfgang help me formulate). I think the old version is not formulate true with my ideas, so may I re-formulate my ideas again 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 $0 \leq a_1, a_2,\cdots,a_n \leq 1$,

$$\sum\limits_{sym}\left( \sum\limits_{sym} x_{i_1}^{a_{p_1}} \cdots x_{i_m}^{a_{p_m}} \right) \leq \sum\limits_{sym}\left( \sum\limits_{sym} y_{i_1}^{a_{p_1}} \cdots y_{i_m}^{a_{p_m}} \right)$$

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 $a_1, a_2,\dots,a_n \geq 0$,

$$\sum\limits_{sym}\left( \sum\limits_{sym} a_{i_1}^{x_{p_1}} \cdots a_{i_m}^{x_{p_m}} \right) \geq \sum\limits_{sym}\left( \sum\limits_{sym} a_{i_1}^{y_{p_1}} \cdots a_{i_m}^{y_{p_m}} \right)$$

Example for Inequality 1 with $n=3$, $m=2$.

$${x_1}^{a_1}.{x_2}^{a_2}+{x_1}^{a_1}.{x_3}^{a_2}+{x_2}^{a_1}.{x_3}^{a_2}+{x_1}^{a_2}.{x_2}^{a_3}+{x_1}^{a_2}.{x_3}^{a_3}+{x_2}^{a_2}.{x_3}^{a_3}+{x_1}^{a_1}.{x_2}^{a_3}+{x_1}^{a_1}.{x_3}^{a_3}+{x_2}^{a_1}.{x_3}^{a_3}+{x_1}^{a_2}.{x_2}^{a_1}+{x_1}^{a_2}.{x_3}^{a_1}+{x_2}^{a_2}.{x_3}^{a_1}+{x_1}^{a_3}.{x_2}^{a_2}+{x_1}^{a_3}.{x_3}^{a_2}+{x_2}^{a_3}.{x_3}^{a_2}+{x_1}^{a_3}.{x_2}^{a_1}+{x_1}^{a_3}.{x_3}^{a_1}+{x_2}^{a_3}.{x_3}^{a_1} \leq {y_1}^{a_1}.{y_2}^{a_2}+{y_1}^{a_1}.{y_3}^{a_2}+{y_2}^{a_1}.{y_3}^{a_2}+{y_1}^{a_2}.{y_2}^{a_3}+{y_1}^{a_2}.{y_3}^{a_3}+{y_2}^{a_2}.{y_3}^{a_3}+{y_1}^{a_1}.{y_2}^{a_3}+{y_1}^{a_1}.{y_3}^{a_3}+{y_2}^{a_1}.{y_3}^{a_3}+{y_1}^{a_2}.{y_2}^{a_1}+{y_1}^{a_2}.{y_3}^{a_1}+{y_2}^{a_2}.{y_3}^{a_1}+{y_1}^{a_3}.{y_2}^{a_2}+{y_1}^{a_3}.{y_3}^{a_2}+{y_2}^{a_3}.{y_3}^{a_2}+{y_1}^{a_3}.{y_2}^{a_1}+{y_1}^{a_3}.{y_3}^{a_1}+{y_2}^{a_3}.{y_3}^{a_1}$$

Example Inequality 2 with $n=3$, $m=2$.

$${a_1}^{x_1}.{a_2}^{x_2}+{a_1}^{x_1}.{a_3}^{x_2}+{a_2}^{x_1}.{a_3}^{x_2}+{a_1}^{x_2}.{a_2}^{x_3}+{a_1}^{x_2}.{a_3}^{x_3}+{a_2}^{x_2}.{a_3}^{x_3}+{a_1}^{x_1}.{a_2}^{x_3}+{a_1}^{x_1}.{a_3}^{x_3}+{a_2}^{x_1}.{a_3}^{x_3}+{a_1}^{x_2}.{a_2}^{x_1}+{a_1}^{x_2}.{a_3}^{x_1}+{a_2}^{x_2}.{a_3}^{x_1}+{a_1}^{x_3}.{a_2}^{x_2}+{a_1}^{x_3}.{a_3}^{x_2}+{a_2}^{x_3}.{a_3}^{x_2}+{a_1}^{x_3}.{a_2}^{x_1}+{a_1}^{x_3}.{a_3}^{x_1}+{a_2}^{x_3}.{a_3}^{x_1} \geq {a_1}^{y_1}.{a_2}^{y_2}+{a_1}^{y_1}.{a_3}^{y_2}+{a_2}^{y_1}.{a_3}^{y_2}+{a_1}^{y_2}.{a_2}^{y_3}+{a_1}^{y_2}.{a_3}^{y_3}+{a_2}^{y_2}.{a_3}^{y_3}+{a_1}^{y_1}.{a_2}^{y_3}+{a_1}^{y_1}.{a_3}^{y_3}+{a_2}^{y_1}.{a_3}^{y_3}+{a_1}^{y_2}.{a_2}^{y_1}+{a_1}^{y_2}.{a_3}^{y_1}+{a_2}^{y_2}.{a_3}^{y_1}+{a_1}^{y_3}.{a_2}^{y_2}+{a_1}^{y_3}.{a_3}^{y_2}+{a_2}^{y_3}.{a_3}^{y_2}+{a_1}^{y_3}.{a_2}^{y_1}+{a_1}^{y_3}.{a_3}^{y_1}+{a_2}^{y_3}.{a_3}^{y_1}$$

• I believe having this as an answer is confusing, even if you begin by explaining that this is not an answer. You rather should add this to the question Jul 17, 2018 at 10:44
• In inequality 1, inner sym for a; outer sym for x, y. Similarly to inequality 2. Jul 17, 2018 at 11:03
• @ĐàoThanhOai Making the sums symmetric yields two completely different conjectures (which have many chances to hold true, supposedly quite easy to prove by majorization). i'd simply suggest that you make up a new question with those two conjectures! Jul 17, 2018 at 13:05
• Dear sir @Wolfgang I post this to here mathoverflow.net/questions/306204/… Jul 17, 2018 at 14:12