# A family of Counterexamples on
Inequality among Symmetric Functions

Jia Xu<EMAIL_ADDRESS>Yong Yao<EMAIL_ADDRESS>Department of Mathematics,
Southwest Minzu University, Chengdu, Sichuan 610041, China  Chengdu Computer
Application Institute, Chinese Academy of Sciences, Chengdu, Sichuan 610213,
China

###### Abstract

Inequalities among symmetric functions are fundamental questions in
mathematics and have various applications in science and engineering. In this
paper, we tackle a conjecture about inequalities among the complete
homogeneous symmetric function $H_{n,\lambda}$, that is, the inequality
$H_{n,\lambda}\leq H_{n,\mu}$ implies majorization order $\lambda\preceq\mu$.
This conjecture was proposed by Cuttler, Greene and Skandera in 2011. The
conjecture is a close analogy with other known results on Muirhead-type
inequalities. In 2021, Heaton and Shankar disproved the conjecture by showing
a counterexample for degree $d=8$ and number of variables $n=3$. They then
asked whether the conjecture is true when the number of variables, $n$, is
large enough? In this paper, we answer the question by proving that the
conjecture does not hold when $d\geq 8$ and $n\geq 2$. A crucial step of the
proof relies on variables reduction. Inspired by this, we propose a new
conjecture for $H_{n,\lambda}\leq H_{n,\mu}$.

###### keywords:

complete homogeneous symmetric function, majorization , symmetric inequalities

###### MSC:

05E05 , 14P99, 90C22

##  1 Introduction

Symmetric functions play indispensable ingredients in combinatorics [5, 21],
and have various applications in diverse fields [19, 20, 22, 26]. An important
collection of tools in the study of symmetric functions is various
inequalities. Thus much research has been carried out in the hope of
discovering and proving inequalities among symmetric functions, to list a few
[1, 3, 7, 8, 11, 12, 13, 15, 16, 17, 18, 23, 24, 25]. Some of them are well
known and wide used, such as arithmetic mean and geometric means, Schur,
Maclaurin and Muirhead-type.

It turns out that all these are special cases of inequalities among the
following fundamental symmetric functions:

  * 1.

Monomial symmetric functions $m_{n,\lambda}$: arithmetic means and geometric
means [15], Hardy, littlewood, Polya [8],…

  * 2.

Elementary symmetric functions $e_{n,\lambda}$: Maclaurin [13], Newton [17],…

  * 3.

Power-sum symmetric functions $p_{n,\lambda}$: R. Gantmacher [6], Ursell
[25],…

  * 4.

Schur functions $s_{n,\lambda}$: Schur [8]

  * 5.

Complete homogeneous symmetric functions $h_{n,\lambda}$: Grommer [7], Hunter
[11],…

Naturally there have been extensive studies on inequalities among the above
fundamental symmetric functions [2, 8, 15], resulting in much progress,
providing very efficient way to check the inequalities, which make various
applications process more efficient.

First, we list some notions and notations before concisely illustrating these
works. Given a symmetric polynomial $f(x)$, the term-normalized symmetric
polynomial is

$F(x):=\frac{f(x)}{f(1,\cdots,1)}.$

The inequation $F_{n,\lambda}\leq F_{n,\mu}$ means that $F_{n,\lambda}(x)\leq
F_{n,\mu}(x)$ for every $x$ in $\mathbb{R}_{+}^{n}\setminus 0$, where
$\mathbb{R}_{+}$ is the set of nonnegative real numbers and $n\geq 2$. Thus
the term-normalized symmetric functions of $m_{n,\lambda}$, $e_{n,\lambda}$,
$p_{n,\lambda}$, $s_{n,\lambda}$ and $h_{n,\lambda}$ can be written by
$M_{n,\lambda}$, $E_{n,\lambda}$, $P_{n,\lambda}$, $S_{n,\lambda}$ and
$H_{n,\lambda}$. The following theorem is a summary of known results on these
term-normalized symmetric functions. The proofs of the these results can be
found in [4, 8, 16, 17, 27].

Known results: [9] Let $\mu,\lambda\in\mathbb{N}^{m}$ such that
$|\mu|=|\lambda|$. Then

$\displaystyle M_{n,\mu}\geq M_{n,\lambda}$ $\displaystyle\
\Longleftrightarrow\ \ \mu\succeq\lambda,$ $\displaystyle E_{n,\mu}\geq
E_{n,\lambda}$ $\displaystyle\ \Longleftrightarrow\ \ \mu\preceq\lambda,$
$\displaystyle P_{n,\mu}\geq P_{n,\lambda}$ $\displaystyle\
\Longleftrightarrow\ \ \mu\succeq\lambda,$ $\displaystyle S_{n,\mu}\geq
S_{n,\lambda}$ $\displaystyle\ \Longleftrightarrow\ \ \mu\succeq\lambda,$
$\displaystyle H_{n,\mu}\geq H_{n,\lambda}$ $\displaystyle\ \Longleftarrow\ \
\mu\succeq\lambda.$

where ”$\succeq$ ” is majorization order (see [14] or Definition 6 of this
paper).

Note that unlike the other, the family of complete homogeneous is still open.
The techniques successfully used for other families do not work well in
general. Hence recently the effort is focused on this, making incremental
progresses, producing conjectures that says that the technique could be used
in large cases. In 2011, Allison Cuttler, Curtis Greene and Mark Skandera [4]
conjectured that $H_{n,\mu}\geq
H_{n,\lambda}\Longrightarrow\mu\succeq\lambda.$ Moreover, they indicated the
conjecture is true when the degree $d=|\lambda|=|\mu|=1,2,\ldots,7$, and lead
the question to $d\geq 8$. In 2021, Alexander Heaton and Isabelle Shankar
found some counterexamples which overturn the conjecture for $d=8,9,10$ (see
[9]). Specially, they bring the $H_{3,(4,4)}-H_{3,(5,2,1)}$ ($d=8,n=3$), and
certified its nonnegativity by utilizing the sum of squares (SOS) method. The
positive semi-definite matrix they found is left on web page (see [10]) due to
the enormous and complex output. Instead, they raised the following question
in [9] and put the hope to much more variables.

Question:“Is the following claim true asymptotically: $H_{n,\mu}\geq
H_{n,\lambda}$ implies $\mu\succeq\lambda$?”

In this paper, we conclude this line of research by showing the technique does
not work for even large cases. We show that for every sufficiently large $n$,
there is a counter example. The precise form of our main result is stated in
Theorem 7. So the answer of the above question is as follows.

Answer: “No.”

Hence there is no hope in tackling the complete homogeneous case using the
previous approach. There is a need for an alternative approach. In this paper,
we suggest such a potential alternative approach, as a conjecture.

Conjecture: Let $\mu,\lambda\in\mathbb{N}^{m},|\mu|=|\lambda|$, then

$H_{n,\mu}\geq H_{n,\lambda}\Longleftrightarrow\ \underset{u+v=n}{\forall}\ \
\underset{t\in\mathbb{R}_{+}}{\forall}\ \
H_{n,\mu}({\mathbf{t}}_{u},{\mathbf{1}}_{v})\geq
H_{n,\lambda}({\mathbf{t}}_{u},{\mathbf{1}}_{v}).$

The above conjecture gives another ideas for studying Muirhead-type
inequalities of complete homogenous polynomials.

##  2 Preliminaries

In order to precisely state the main theorem, we recall some definitions and
notations.

######  Definition 1 (Partition [21]).

Let $d\geq 1$. The $d$-partition $Par(d)$ is defined by

$Par(d)=\left\\{\left(\lambda_{1},\ldots,\lambda_{d}\right)\in\mathbb{N}^{d}:\lambda_{1}\geq\cdots\geq\lambda_{d}\geq
0\ \,\text{and }\lambda_{1}+\cdots+\lambda_{d}=d\right\\}.$

######  Example 2.

Note

$Par(3)=\left\\{\left(3,0,0\right),\left(2,1,0\right),\left(1,1,1\right)\right\\}.$

######  Remark 3.

  1. 1.

We will delete $0$ included in the elements of a partition if there is no
confusion. For example, $(2,1,0)$ can be written briefly as $(2,1)$.

  2. 2.

If there are $m$ consecutive $\lambda_{i}$ that are same, then we can
abbreviate them as $\mathbf{\lambda_{i}}_{m}$. For example, $(1,1,1)$ can be
written as $(\mathbf{1}_{3})$.

######  Definition 4 (Complete homogenous symmetric function [21]).

For a partition $\lambda\in Par(d)$, a complete homogeneous symmetric function
$h_{n,\lambda}$ is written as

$h_{n,\lambda}=\prod_{i=1}^{d}h_{n,\lambda_{i}},$

where

$h_{n,\lambda_{i}}=\sum_{1\leq j_{1}\leq\cdots\leq j_{\lambda_{i}}\leq
n}x_{j_{1}}\cdots x_{j_{\lambda_{i}}},\ \ \ (\text{with}\ \ h_{n,0}=1).$

######  Remark 5.

The term-normalized form of the complete homogeneous symmetric function is

$H_{n,\lambda}=\frac{1}{\binom{n+\lambda_{1}-1}{\lambda_{1}}\cdots\binom{n+\lambda_{d}-1}{\lambda_{d}}}\
h_{n,\lambda}.$

######  Definition 6 (Majorization [14]).

Let $\mu,\lambda\in Par(d)$. We say that $\mu$ majorizes $\lambda,\ \,$ and
write $\mu\succeq\lambda$, if

$\underset{1\leq j\leq d-1}{\forall}\
\sum_{i=1}^{j}\mu_{i}\geq\sum_{i=1}^{j}\lambda_{i}.$

##  3 Main theorem

######  Theorem 7 (Main Result).

For every $n\geq 2$ and $d\geq 8$, there exist $\mu,\lambda\in Par(d)$ such
that $H_{n,\mu}\geq H_{n,\lambda}$ but $\mu$ does not majorizes $\lambda$,
that is,

$\underset{d\geq 8}{\underset{n\geq 2}{\forall}}\ \ \ \underset{\mu,\lambda\in
Par\left(d\right)}{\exists}H_{n,\mu}\geq H_{n,\lambda}\ \text{\ but\
}\mu\not\succeq\lambda.$

Before we plunge into technical details, we will first provide the top-level
structure of the proof, in the hope of helping the reader to grasp the
essence.

Top-level structure:

Let $n\geq 2$ and $d\geq 8$ be arbitrary but fixed. It is sufficient to prove
that there exist $\lambda,\mu\in P_{d}\ $such that $H_{n,\mu}\geq
H_{n,\lambda}\ $ but $\mu\not\succeq\lambda$. In general there are two
different strategies for proving existence: (1) constructive, guess a
potential witness and check it. (2) non-constructive, assume non-existence and
derive contradiction. In this paper, we follow the constructive approach,
since it is more interesting.

  1. 1.

Guess a witness for $\mu,\lambda$.

Since $Par(d)$ expands rapidly while $d$ is growing. For example,
$|Par(17)|=297$ while $|Par(18)|=385$. It takes a little luck to guess the
following witness.

$\begin{array}[c]{llll}\text{Case
}d=2m&:&\mu=(\underset{m}{\underbrace{2,\ldots,2})}=(\mathbf{2}_{m})&\lambda=(3,\underset{2m-3}{\underbrace{1,\ldots,1})}=(3,\mathbf{1}_{2m-3})\\\
\text{Case
}d=2m+1&:&\mu=(\underset{m}{\underbrace{2,\ldots,2},1)}=(\mathbf{2}_{m},1)&\lambda=(3,\underset{2m-2}{\underbrace{1,\ldots,1})}=(3,\mathbf{1}_{2m-2})\end{array}$

  2. 2.

Check that it is indeed a witness.

    1. (a)

$\mu\not\succeq\lambda$.

Trivial.

    2. (b)

$H_{n,\mu}\geq H_{n,\lambda}$

This is non-trivial, requiring much technical details. Again before we plunge
into the detail, here we provide a high-level strategy. We first tackle the
smallest still “open” degree $d=8,$ that is,

$\mu=\left(\mathbf{2}_{4}\right)\ \text{and
}\lambda=\left(3,\mathbf{1}_{5}\right)$

We prove it by transforming the problem into an optimization problem on the
simplex. The details are given in Lemma 11 and its proof is given below.
Briefly, the proof is divided into two parts, interior and boundary of the
simplex. In the interior, we reduce the number of variables into $2$ by
Lagrange’s equation ( see Lemma 9 ). At boundary, we deal with it by proving
an inequality ( see Lemma 10 ).

After this, we extend the result with degree $8$ to arbitrary degree $d$ by
using relaxation method repeatedly. The details are given in Lemma 12 and its
proof is given below.

This concludes the top-level structure of the proof.

######  Remark 8.

  1. 1.

It will be interesting to find different counter-examples.

  2. 2.

In fact, one wonders about the set of all counter-examples. Does it have any
discernable structure?

######  Lemma 9.

Let $\mu=\left(\mathbf{2}_{4}\right)\ $and
$\lambda=\left(3,\mathbf{1}_{5}\right)$. Then we have

$\underset{u,v\geq 0}{\underset{u+v=n}{\forall}}\
\underset{t\in\mathbb{R}_{+}}{\forall}\
H_{n,\mu}\mathbb{(}\mathbf{t}_{u},\mathbf{1}_{v})\ \geq\
H_{n,\lambda}\mathbb{(}\mathbf{t}_{u},\mathbf{1}_{v}).$

###### Proof.

Let $J_{n}=H_{n,\mu}-H_{n,\lambda}$. Note

$\displaystyle\underset{u,v\geq 0}{\underset{u+v=n}{\forall}}\
\underset{t\in\mathbb{R}_{+}}{\forall}J_{n}(\mathbf{t}_{u},\mathbf{1}_{v})\geq
0$ $\displaystyle\Longleftrightarrow$ $\displaystyle\underset{u,v\geq
1}{\underset{u+v=n}{\forall}}\
\underset{t\in\mathbb{R}_{+}}{\forall}J_{n}(\mathbf{t}_{u},\mathbf{1}_{v})\geq
0\ \ \ \ \text{(since if }u=0\ \text{or\ \ }v=0\ \text{then
}J_{n}(\mathbf{t}_{u},\mathbf{1}_{v})=0\text{)}$
$\displaystyle\Longleftrightarrow$ $\displaystyle\underset{k,l\geq
0}{\underset{k+l=n-2}{\forall}}\
\underset{t\in\mathbb{R}_{+}}{\forall}J_{n}(\mathbf{t}_{k+1},\mathbf{1}_{l+1})\geq
0\ \ \text{(obtained by }u=k+1\ \text{and }v=l+1\text{)}$

Direct computation show that

$J_{n}(\mathbf{t}_{k+1},\mathbf{1}_{l+1})=\frac{(k+1)(l+1)}{\binom{n+2}{3}\binom{n}{1}^{5}\binom{n+1}{2}^{4}}\left(t-1\right)^{2}W\left(k,l,t\right)$

for some polynomial $W$. Thus it suffices to show that

$\underset{k,l\geq 0}{\underset{k+l=n-2}{\forall}}W\left(k,l,t\right)\geq 0$

Direct calculation shows that all the coefficients of $W$ are non-negative
(see Appendix). Hence the claim holds. ∎

######  Lemma 10.

Let $\mu=\left(\mathbf{2}_{4}\right)\ $and
$\lambda=\left(3,\mathbf{1}_{5}\right)$. We have the inequality

$\underset{x\in\mathbb{R}_{+}^{n}}{\forall}\ \ \
H_{n+1,\mu}\left(x,0\right)-H_{n+1,\lambda}\left(x,0\right)\ \ \geq\ \
\frac{n^{6}}{(n+3)(n+1)^{5}}\left(H_{n,\mu}\left(x\right)-H_{n,\lambda}\left(x\right)\right).$

###### Proof.

Note $h_{n+1,\mu}\left(x,0\right)=h_{n,\mu}\left(x\right)$ and
$h_{n+1,\lambda}\left(x,0\right)=h_{n,\lambda}\left(x\right)$. Then we have

$\begin{array}[c]{lllllll}\frac{H_{n+1,\mu}\left(x,0\right)}{H_{n,\mu}\left(x\right)}&=&\frac{\frac{h_{n+1,\mu}\left(x,0\right)}{\binom{n+2}{2}^{4}}}{\frac{h_{n,\mu}\left(x\right)}{\binom{n+1}{2}^{4}}}&=&\left(\frac{\binom{n+1}{2}}{\binom{n+2}{2}}\right)^{4}&=&\left(\frac{n}{n+2}\right)^{4},\\\
&&&&&&\\\
\frac{H_{n+1,\lambda}\left(x,0\right)}{H_{n,\lambda}\left(x\right)}&=&\frac{\frac{h_{n+1,\lambda}\left(x,0\right)}{\binom{n+3}{3}^{1}\binom{n+1}{1}^{5}}}{\frac{h_{n,\lambda}\left(x\right)}{\binom{n+2}{3}^{1}\binom{n}{1}^{5}}}&=&\left(\frac{\binom{n+2}{3}}{\binom{n+3}{3}}\right)^{1}\left(\frac{\binom{n}{1}}{\binom{n+1}{1}}\right)^{5}&=&\left(\frac{n}{n+3}\right)^{1}\left(\frac{n}{n+1}\right)^{5}.\end{array}$

One can verify that

$\left(\frac{n}{n+3}\right)^{1}\left(\frac{n}{n+1}\right)^{5}\ \ <\ \
\left(\frac{n}{n+2}\right)^{4}.$

Thus

$\displaystyle H_{n+1,\mu}\left(x,0\right)\ \ $ $\displaystyle>\ \
\left(\frac{n}{n+3}\right)^{1}\left(\frac{n}{n+1}\right)^{5}H_{n,\mu}\left(x\right)$
$\displaystyle H_{n+1,\lambda}\left(x,0\right)\ \ $ $\displaystyle=\ \
\left(\frac{n}{n+3}\right)^{1}\left(\frac{n}{n+1}\right)^{5}H_{n,\lambda}\left(x\right)$

Thus

$H_{n+1,\mu}\left(x,0\right)-H_{n+1,\lambda}\left(x,0\right)\ \ \geq\ \
\frac{n^{6}}{(n+3)(n+1)^{5}}\left(H_{n,\mu}\left(x\right)-H_{n,\lambda}\left(x\right)\right).$

∎

######  Lemma 11.

Let $\mu=\left(\mathbf{2}_{4}\right)$ and
$\lambda=\left(3,\mathbf{1}_{5}\right)$. We have

$\ H_{n,\mu}\geq H_{n,\lambda}\ \ (n\geq 2).$

###### Proof.

Let $J_{n}=H_{n,\mu}-H_{n,\lambda}$. We will prove $J_{n}\geq 0$ by induction
on $n$.

Base case: The following calculation verifies that the claim is true when
$n=2$.

Direct computation show that

$J_{2}=H_{2,\mu}-H_{2,\lambda}=\frac{h_{2,\mu}}{\binom{2+1}{2}^{4}}-\frac{h_{2,\lambda}}{\binom{2+2}{3}^{1}\binom{2}{1}^{5}}=(x_{1}-x_{2})^{2}P(x_{1},x_{2}),$

where

$P\left(x_{1},x_{2}\right)=\frac{1}{10368}\left(47(x_{1}^{6}+x_{2}^{6})+120(x_{1}^{5}x_{2}+x_{1}x_{2}^{5})+177(x_{1}^{4}x_{2}^{2}+x_{1}^{2}x_{2}^{4})+176x_{1}^{3}x_{2}^{3}\right).$

Thus, $J_{2}\geq 0$ holds.

Induction step: Given that $J_{n-1}(x)\geq 0$ holds for $n\geq 3$, we will
show that $J_{n}(x)\geq 0$ holds too.

Since $J_{n}(x)$ is homogeneous, it suffices to show that

$\min_{x\in\Delta_{n}}J_{n}(x)\geq 0,$

where

$\Delta_{n}=\\{x\in\mathbb{R}_{+}^{n}:\ x_{1}+\cdots+x_{n}=1\\}.$

Note that $\Delta_{n}$ is compact, hence there exists $p\in\Delta_{n}$ such
that $J_{n}(p)=\min\limits_{x\in\Delta_{n}}J_{n}(x)$. It remains to prove
$J_{n}(p)\geq 0$, and will be done in the following two cases.

  1. 1.

$p\in\Delta_{n}^{\circ}$ (the interior of $\Delta_{n}$). We claim that
$p=(\mathbf{t}_{u},\mathbf{r}_{v})$ for some $t,r$ and $u+v=n$.

Since $p$ is an extreme point, it follows from Lagrange multiplier theorem
that there is a real $\lambda$ such that $p$ satisfies the following
equations.

$\frac{\partial J_{n}}{\partial x_{i}}(p)=\lambda\frac{\partial
h_{n,1}}{\partial x_{i}}(p),\ i=1,2,\ldots,n.$

Since

$\displaystyle\frac{\partial\ h_{n,1}}{\partial x_{i}}$ $\displaystyle=1,$
$\displaystyle\frac{\partial\ h_{n,2}}{\partial x_{i}}$
$\displaystyle=x_{i}+h_{n,1},$ $\displaystyle\frac{\partial\ h_{n,3}}{\partial
x_{i}}$ $\displaystyle=x_{i}^{2}+h_{n,1}x_{i}+h_{n,2}$
$\displaystyle\frac{\partial J_{n}}{\partial x_{i}}$
$\displaystyle=\frac{\partial\left(H_{n,\mu}-H_{n,\lambda}\right)}{\partial
x_{i}},$
$\displaystyle=\frac{\partial\left(\frac{h_{n,2}^{4}}{\binom{n+1}{2}^{4}}-\frac{h_{n,1}^{5}h_{n,3}}{\binom{n+2}{3}^{1}\binom{n}{1}^{5}}\right)}{\partial
x_{i}},$

this in turn implies that each of the $p_{i}$ is a root of the quadratic
equation

$ax_{i}^{2}+bx_{i}+c=0,$

where

$\displaystyle a$ $\displaystyle=-\binom{n+2}{3}^{-1}\binom{n}{1}^{-5}$
$\displaystyle b$
$\displaystyle=4\binom{n+1}{2}^{-4}h_{n,2}^{3}(p)-\binom{n+2}{3}^{-1}\binom{n}{1}^{-5}$
$\displaystyle c$
$\displaystyle=4\binom{n+1}{2}^{-4}h_{n,2}^{3}(p)-\binom{n+2}{3}^{-1}\binom{n}{1}^{-5}(h_{n,2}(p)+5h_{n,3}(p))-\lambda$

Thus $p_{1},\cdots,p_{n}$ take at most two different numbers. Without loss of
generality, suppose $\\{p_{1},\cdots,p_{n}\\}=\\{t,r\\}$. $J_{n}$ is
symmetric, so $p$ can be written as follows.

$p=(\underset{u}{\underbrace{t,\cdots,t}},\underset{v}{\underbrace{r,\cdots,r}})=(\mathbf{t}_{u},\mathbf{r}_{v}),\
\ u,v\in\mathbb{N},\ u+v=n.$

Noticed that $J_{n}(\mathbf{t}_{u},\mathbf{1}_{v})\geq 0\Longleftrightarrow
J_{n}(\mathbf{t}_{u},\mathbf{r}_{v})\geq 0$ due to homogeneity of $J_{n}$.
Hence by Lemma 9, we have

$\displaystyle J_{n}(\mathbf{t}_{u},\mathbf{1}_{v})\geq 0$
$\displaystyle\Longrightarrow J_{n}(\mathbf{t}_{u},\mathbf{r}_{v})\geq
0\Longrightarrow J_{n}(p)\geq 0.$

  2. 2.

$p\in\partial\Delta_{n}$ (the boundary of $\Delta_{n}$). Let
$p=(p_{1},\cdots,p_{n-1},0)$ by symmetry. Thus, $J_{n}(p)\geq 0$ is trivial.
Since from Lemma 10 and the induction hypothesis, we have

$J_{n}(x_{1},\cdots,x_{n-1},0)\geq\frac{(n-1)^{6}}{(n+2)n^{5}}\
J_{n-1}(x_{1},\cdots,x_{n-1})\geq 0.$

According to the principle of induction, the proof is done. ∎

######  Lemma 12.

We have

$H_{n,(\mathbf{2}_{m})}\geq H_{n,(3,{\mathbf{1}}_{2m-3})},\
H_{n,(\mathbf{2}_{m},1)}\geq H_{n,(3,{\mathbf{1}}_{2m-2})}\ \ (m\geq 4).$

where $\mathbf{2}_{m}=\underset{m}{\underbrace{2,\cdots,2}},\ \
{\mathbf{1}}_{v}=\underset{v}{\underbrace{1,\cdots,1}}$.

###### Proof.

From Lemma 11, we have

$\frac{H_{n,(\mathbf{2}_{4})}}{H_{n,(3,\mathbf{1}_{5})}}\geq 1,$ (1)

where $n\geq 2$ and $m=4$.

Generally, let
$F_{n,m}=\frac{H_{n,(\mathbf{2}_{m})}}{H_{n,(3,\mathbf{1}_{2m-3})}}$. We claim
that

$F_{n,m}\geq F_{n,m-1}$ (2)

It is because

$\displaystyle\frac{F_{n,m}}{F_{n,m-1}}$
$\displaystyle=\frac{\frac{\binom{n+2}{3}\binom{n}{1}^{2m-3}}{\binom{n+1}{2}^{m}}\frac{(h_{n,2})^{m}}{h_{n,3}\
(h_{n,1})^{2m-3}}}{\frac{\binom{n+2}{3}\binom{n}{1}^{2m-5}}{\binom{n+1}{2}^{m-1}}\frac{(h_{n,2})^{m-1}}{h_{n,3}\
(h_{n,1})^{2m-5}}}$ $\displaystyle=\frac{2n}{n+1}\frac{\sum\limits_{1\leq
i\leq n}x_{i}^{2}+\sum\limits_{1\leq i<j\leq n}x_{i}x_{j}}{(\sum\limits_{1\leq
i\leq n}x_{i})^{2}}$
$\displaystyle=\frac{n(\sum_{i=1}^{n}x_{i}^{2})+n(\sum_{i=1}^{n}x_{i})^{2}}{(n+1)(\sum_{i=1}^{n}x_{i})^{2}}$
$\displaystyle\geq\frac{(\sum_{i=1}^{n}x_{i})^{2}+n(\sum_{i=1}^{n}x_{i})^{2}}{(n+1)(\sum_{i=1}^{n}x_{i})^{2}}\
\ \ \ \text{from}\ \
n\left(\sum_{i=1}^{n}x_{i}^{2}\right)\geq\left(\sum_{i=1}^{n}x_{i}\right)^{2}$
$\displaystyle=1.$

By using inequality (2) repeatedly and combining formula (1), we have

$F_{n,m}\geq F_{n,{m-1}}\geq\cdots\geq
F_{n,4}=\frac{H_{n,(\mathbf{2}_{4})}}{H_{n,(3,\mathbf{1}_{5})}}\geq 1.$

Hence

$H_{n,(\mathbf{2}_{m})}\geq H_{n,(3,{\mathbf{1}}_{2m-3})}.$

Further, note that $H_{n,\lambda}$ is multiplicative, then

$\frac{H_{n,({\mathbf{2}}_{m},1)}}{H_{n,(3,{\mathbf{1}}_{2m-2})}}=\frac{H_{n,({\mathbf{2}}_{m})}}{H_{n,(3,{\mathbf{1}}_{2m-3})}}=F_{n,m}\geq
1,$

Hence

$H_{n,(\mathbf{2}_{m},1)}\geq H_{n,(3,{\mathbf{1}}_{2m-2})}.$

∎

Now let us complete the proof of Theorem 7.

###### Proof.

[Proof of Theorem 7]

  1. 1.

$d\geq 8$ and even: Let $d=2m$ where $m\geq 4$. Take
$\mu=(\mathbf{2}_{m}),\lambda=(3,\mathbf{1}_{2m-3})$. From Lemma 12, we have
$H_{n,\mu}=H_{n,(\mathbf{2}_{m})}\geq
H_{n,(3,\mathbf{1}_{2m-3})}=H_{n,\lambda}$, but

$\mu=(\mathbf{2}_{m})=(\underset{m}{\underbrace{2,\ldots,2}})\nsucceq(3,\underset{2m-3}{\underbrace{1,\ldots,1}})=(3,\mathbf{1}_{2m-3})=\lambda.$

  2. 2.

$d\geq 9$ and odd: Let $d=2m+1$ for $m\geq 4$. Take
$\mu=(\mathbf{2}_{m},1),\lambda=(3,\mathbf{1}_{2m-2})$. From Lemma 12, we have
$H_{n,\mu}\geq H_{n,\lambda}$, but $\mu\nsucceq\lambda$.

We have completed the proof. ∎

##  4 A conjecture

In this section, we propose a conjecture for an alternative characterization.
The conjecture (see below) is inspired by the following observation.

######  Proposition 13.

Let $\mu,\lambda\in Par(d)$. We have

$\begin{array}[c]{lllllllllll}M_{n,\mu}&\geq&M_{n,\lambda}&&\Longleftrightarrow&&\underset{u+v=n}{\forall}&\underset{t\in\mathbb{R}_{+}}{\forall}&M_{n,\mu}({\mathbf{t}}_{u},{\mathbf{1}}_{v})&\geq&M_{n,\lambda}({\mathbf{t}}_{u},{\mathbf{1}}_{v})\\\
E_{n,\mu}&\geq&E_{n,\lambda}&&\Longleftrightarrow&&\underset{u+v=n}{\forall}&\underset{t\in\mathbb{R}_{+}}{\forall}&E_{n,\mu}({\mathbf{t}}_{u},{\mathbf{1}}_{v})&\geq&E_{n,\lambda}({\mathbf{t}}_{u},{\mathbf{1}}_{v})\\\
P_{n,\mu}&\geq&P_{n,\lambda}&&\Longleftrightarrow&&\underset{u+v=n}{\forall}&\underset{t\in\mathbb{R}_{+}}{\forall}&P_{n,\mu}({\mathbf{t}}_{u},{\mathbf{1}}_{v})&\geq&P_{n,\lambda}({\mathbf{t}}_{u},{\mathbf{1}}_{v})\\\
S_{n,\mu}&\geq&S_{n,\lambda}&&\Longleftrightarrow&&\underset{u+v=n}{\forall}&\underset{t\in\mathbb{R}_{+}}{\forall}&S_{n,\mu}({\mathbf{t}}_{u},{\mathbf{1}}_{v})&\geq&S_{n,\lambda}({\mathbf{t}}_{u},{\mathbf{1}}_{v})\end{array}$

###### Proof.

$\Longrightarrow$:

It is obvious.

$\Longleftarrow$:

  * 1.

$M$

The following proof is essentially based on comparing degrees. It is
straightforward to show

$\deg_{t}M_{n,\alpha}({\mathbf{t}}_{u},{\mathbf{1}}_{v})=\sum_{i=1}^{u}\alpha_{i}$

Now observe

$\displaystyle\ \ \underset{u+v=n}{\forall}\
\underset{t\in\mathbb{R}_{+}}{\forall}M_{n,\mu}({\mathbf{t}}_{u},{\mathbf{1}}_{v})\geq
M_{n,\lambda}({\mathbf{t}}_{u},{\mathbf{1}}_{v})$
$\displaystyle\Longrightarrow$ $\displaystyle\ \ \underset{u+v=n}{\forall}\
\deg_{t}M_{n,\mu}({\mathbf{t}}_{u},{\mathbf{1}}_{v})\geq\deg_{t}M_{n,\lambda}({\mathbf{t}}_{u},{\mathbf{1}}_{v})\
\ \ \ \ \text{by comparing them when }t\rightarrow\infty$
$\displaystyle\Longleftrightarrow$ $\displaystyle\ \
\underset{u+v=n}{\forall}\ \sum_{i=1}^{u}\mu_{i}\geq\sum_{i=1}^{u}\lambda_{i}$
$\displaystyle\Longleftrightarrow$ $\displaystyle\ \ \mu\succeq\lambda\ \
\text{from the definition of }\succeq$ $\displaystyle\Longleftrightarrow$
$\displaystyle\ \ M_{n,\mu}\geq M_{n,\mu}$

  * 2.

$S$

The following proof is the same as the proof for $M$. It is straightforward to
show

$\deg_{t}S_{n,\alpha}({\mathbf{t}}_{u},{\mathbf{1}}_{v})=\sum_{i=1}^{u}\alpha_{i}$

Now observe

$\displaystyle\ \ \underset{u+v=n}{\forall}\
\underset{t\in\mathbb{R}_{+}}{\forall}S_{n,\mu}({\mathbf{t}}_{u},{\mathbf{1}}_{v})\geq
S_{n,\lambda}({\mathbf{t}}_{u},{\mathbf{1}}_{v})$
$\displaystyle\Longrightarrow\ \ \underset{u+v=n}{\forall}\
\deg_{t}S_{n,\mu}({\mathbf{t}}_{u},{\mathbf{1}}_{v})\geq\deg_{t}S_{n,\lambda}({\mathbf{t}}_{u},{\mathbf{1}}_{v})\
\ \ \ \ \text{by comparing them when }t\rightarrow\infty$
$\displaystyle\Longleftrightarrow\ \ \underset{u+v=n}{\forall}\
\sum_{i=1}^{u}\mu_{i}\geq\sum_{i=1}^{u}\lambda_{i}$
$\displaystyle\Longleftrightarrow\ \ \mu\succeq\lambda\ \ \text{from the
definition of }\succeq$ $\displaystyle\Longleftrightarrow\ \ S_{n,\mu}\geq
S_{n,\mu}$

  * 3.

$E$

The following proof is almost the same as the proof for $M$. However there is
a subtle difference. It is straightforward to show

$\deg_{t}E_{n,\alpha}({\mathbf{t}}_{u},{\mathbf{1}}_{v})=\sum_{i=1}^{u}\alpha_{i}^{\prime}$

where $\alpha^{\prime}$ denotes the conjugate of the partition $\alpha$, that
is, $\alpha_{j}^{\prime}=\max\\{i|\alpha_{i}>j\\}$.

Now observe

$\displaystyle\ \ \underset{u+v=n}{\forall}\
\underset{t\in\mathbb{R}_{+}}{\forall}E_{n,\mu}({\mathbf{t}}_{u},{\mathbf{1}}_{v})\geq
E_{n,\lambda}({\mathbf{t}}_{u},{\mathbf{1}}_{v})$
$\displaystyle\Longrightarrow\ \ \underset{u+v=n}{\forall}\
\deg_{t}E_{n,\mu}({\mathbf{t}}_{u},{\mathbf{1}}_{v})\geq\deg_{t}E_{n,\lambda}({\mathbf{t}}_{u},{\mathbf{1}}_{v})\
\ \ \ \ \text{by comparing them when }t\rightarrow\infty$
$\displaystyle\Longleftrightarrow\ \ \underset{u+v=n}{\forall}\
\sum_{i=1}^{u}\mu_{i}^{\prime}\geq\sum_{i=1}^{u}\lambda_{i}^{\prime}$
$\displaystyle\Longleftrightarrow\ \ \mu^{\prime}\succeq\lambda^{\prime}\ \
\text{from the definition of }\succeq$ $\displaystyle\Longleftrightarrow\ \
\mu\preceq\lambda\ \ \text{from Theorem 3.2 of
\cite[cite]{[\@@bibref{Number}{CGS2011}{}{}]}}$
$\displaystyle\Longleftrightarrow\ \ E_{n,\mu}\geq E_{n,\mu}$

  * 4.

$P$

It was proved in P.753 of [4], using a proof technique quite different from
the proof for $M$, because the degree comparison does not provide any
information since

$\deg_{t}P_{n,\alpha}({\mathbf{t}}_{u},{\mathbf{1}}_{v})=d.$

∎

The above results naturally leads to the following conjecture.

######  Conjecture 14.

We conjecture that

$H_{n,\mu}\geq H_{n,\lambda}\ \ \Longleftrightarrow\ \
\underset{u+v=n}{\forall}\
\underset{t\in\mathbb{R}_{+}}{\forall}H_{n,\mu}({\mathbf{t}}_{u},{\mathbf{1}}_{v})\geq
H_{n,\lambda}({\mathbf{t}}_{u},{\mathbf{1}}_{v}).$

######  Remark 15.

The proof technique used for $M$, $S$ and $E$ does not work since
$\deg_{t}H_{n,\alpha}({\mathbf{t}}_{u},{\mathbf{1}}_{v})=d$. The proof
technique used for $P$ does not seem to work either.

######  Remark 16.

We checked the conjeture on all possible $\mu,\lambda$ with increasing degrees
and number of variables. We used the following tools.

  1. 1.

LHS: difference substitution method (DS) [30, 28, 29, 31, 32].

  2. 2.

RHS: Sturm sequence.

We have verified this by explicit computation up through $d=12$ and $n=12$,
and have not found any counter-example. We invite the reader to help complete
the proof or disproof of the conjecture.

##  5 Appendix

$\displaystyle W(k,l,t)=$
$\displaystyle\left(k+2\right)\left(k+1\right)^{3}({k}^{4}+2\,{k}^{3}l+{k}^{2}{l}^{2}+12\,{k}^{3}+17\,{k}^{2}l+5\,k{l}^{2}+49\,{k}^{2}+43\,kl$
$\displaystyle+5\,{l}^{2}+82\,k+32\,l+47){t}^{6}+2\,\left(k+2\right)\left(k+1\right)^{3}(3\,{k}^{3}l+6\,{k}^{2}{l}^{2}+3\,k{l}^{3}+2\,{k}^{3}$
$\displaystyle+32\,{k}^{2}l+37\,k{l}^{2}+7\,{l}^{3}+21\,{k}^{2}+106\,kl+52\,{l}^{2}+64\,k+109\,l+60){t}^{5}$
$\displaystyle+\left(l+1\right)\left(k+1\right)^{2}(15\,{k}^{4}l+30\,{k}^{3}{l}^{2}+15\,{k}^{2}{l}^{3}+11\,{k}^{4}+173\,{k}^{3}l+208\,{k}^{2}{l}^{2}$
$\displaystyle+46\,k{l}^{3}+121\,{k}^{3}+677\,{k}^{2}l+426\,k{l}^{2}+35\,{l}^{3}+442\,{k}^{2}+1074\,kl+272\,{l}^{2}$
$\displaystyle+662\,k+599\,l+354){t}^{4}+4\,\left(l+1\right)^{2}\left(k+1\right)^{2}(5\,{k}^{3}l+10\,{k}^{2}{l}^{2}+5\,k{l}^{3}+6\,{k}^{3}$
$\displaystyle+53\,{k}^{2}l+53\,k{l}^{2}+6\,{l}^{3}+51\,{k}^{2}+157\,kl+51\,{l}^{2}+125\,k+125\,l+88){t}^{3}$
$\displaystyle+\left(l+1\right)^{2}\left(k+1\right)(15\,{k}^{3}{l}^{2}+30\,{k}^{2}{l}^{3}+15\,k{l}^{4}+46\,{k}^{3}l+208\,{k}^{2}{l}^{2}+173\,k{l}^{3}$
$\displaystyle+11\,{l}^{4}+35\,{k}^{3}+426\,{k}^{2}l+677\,k{l}^{2}+121\,{l}^{3}+272\,{k}^{2}+1074\,kl+442\,{l}^{2}$
$\displaystyle+599\,k+662\,l+354){t}^{2}+2\,\left(l+2\right)\left(l+1\right)^{3}(3\,{k}^{3}l+6\,{k}^{2}{l}^{2}+3\,k{l}^{3}+7\,{k}^{3}$
$\displaystyle+37\,{k}^{2}l+32\,k{l}^{2}+2\,{l}^{3}+52\,{k}^{2}+106\,kl+21\,{l}^{2}+109\,k+64\,l+60)t$
$\displaystyle+\left(l+2\right)\left(l+1\right)^{3}({k}^{2}{l}^{2}+2\,k{l}^{3}+{l}^{4}+5\,{k}^{2}l+17\,k{l}^{2}+12\,{l}^{3}+5\,{k}^{2}+43\,kl$
$\displaystyle+49\,{l}^{2}+32\,k+82\,l+47)$

Acknowledgements. The authors are grateful to Bi-can Xia for drawing their
attention to some relevant references and to Hoon Hong for helpful
conversations.This work was supported by the Fundamental Research Funds for
the Central Universities, Southwest Minzu University (2020NYB40).

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