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Extending Muirhead’s Inequality

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Abstract

We present a method to prove an inequality concerning a linear combination of symmetric monomial functions. This is based on Muirhead’s inequality combining with a graph theoretical setting. As an application we prove some interesting inequalities motivated from extremal combinatorics.

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  • 27 September 2021

    The original version is updated due to the appearance of special characters in the HTML version.

References

  1. Frankl, P., Tokushige, N.: Extremal problems for finite sets. Student Mathematical Library, 86. American Mathematical Society, Providence (2018)

  2. Garling, D.: Inequalities: a journey into linear analysis. Cambridge University Press, Cambridge (2007)

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Acknowledgements

The authors thank the referees for their very careful reading and helpful suggestions. The second author is supported by JSPS KAKENHI 19K03398. The last author is supported by JSPS KAKENHI 18K03399.

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Correspondence to Norihide Tokushige.

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Appendix

Appendix

Write \(a_i\) for \(\alpha _{p_i}\), and let \(q_i=1-p_i\).

Fact 1

Let \(t\ge 1\) be an integer, and let \(l_j\) be the line \(y=(r-1)x+j\). Then the walk \(W_{p_i}\) hits the line \(l_t\) with probability \(a_i^t\), and the walk \(W'\) hits the line \(l_{rt}\) with probability \(\beta ^t\).

Proof

Suppose that the probability that the walk \(W_{p_i}\) hits the line \(l_t\) is given by \(X^t\) for some \(X\in (0,1)\). After the first step, the walk is at (0, 1) with probability \(p_i\), and at (1, 0) with probability \(q_i\). From (0, 1) the probability for the walk hitting \(l_t\) is \(X^{t-1}\), and from (1, 0) the probability is \(X^{t-1+r}\). Then it follows

$$\begin{aligned} X^t=p_iX^{t-1}+q_iX^{t-1+r}, \end{aligned}$$

that is,

$$\begin{aligned} X=p_i+q_iX^r. \end{aligned}$$

Thus \(X=a_i\), and the walk hits the line \(l_t\) with probability \(a_i^t\).

Next suppose that the probability that the walk \(W'\) hits the line \(l_{rt}\) is given by \(Y^t\) for some \(Y\in (0,1)\). After the first r steps, it is at \((x,r-x)\) with probability

$$\begin{aligned} \sum _{I\in \left( {\begin{array}{c}[r]\\ x\end{array}}\right) }\prod _{i\in I}p_i\prod _{j\in [r]\setminus I}q_j, \end{aligned}$$

where \([r]=\{1,2,\ldots ,r\}\). From \((x,r-x)\) the probability for the walk hitting \(l_{rt}\) is \(Y^{x+t-1}\). This yields

$$\begin{aligned} Y^t=\sum _{x=0}^rY^{x+t-1} \sum _{I\in \left( {\begin{array}{c}[r]\\ x\end{array}}\right) }\prod _{i\in I}p_i\prod _{j\in [r]\setminus I}q_j, \end{aligned}$$

that is,

$$\begin{aligned} Y=\sum _{x=0}^rY^{x} \sum _{I\in \left( {\begin{array}{c}[r]\\ x\end{array}}\right) }\prod _{i\in I}p_i\prod _{j\in [r]\setminus I}q_j =\prod _{i=1}^r(p_i+q_iY). \end{aligned}$$

Thus \(Y=\beta \), and the walk hits the line \(l_{rt}\) with probability \(\beta ^t\). \(\square \)

Define a polynomial f(x) by

$$\begin{aligned} f(X):=-X+\prod _{i=1}^r(p_i+q_iX). \end{aligned}$$

By definition \(f(\beta )=0\).

Fact 2

If \(0<y<1\) and \(f(y)\le 0\), then \(\beta \le y\).

Proof

This follows because \(f(0)>0\), \(f(1)=0\), \(f'(1)=-1+\sum _{i=1}^r q_i> -1+r\cdot \frac{1}{r}=0\) (here we used \(p_i<1-\frac{1}{r}\)), and \(f''(x)>0\) for \(x>0\). \(\square \)

Fact 3

The inequality \(a:=a_1\cdots a_r\le \beta \) follows from (1).

Proof

By Fact 7 it suffices to show \(f(a)\le 0\). Since \(a_i=p_i+q_ia_i^r\) we have

$$\begin{aligned} a=a_1\cdots a_r=\prod _{i=1}^r(p_i+q_ia_i^r), \end{aligned}$$

and

$$\begin{aligned} f(a)=-a+\prod _{i=1}^r(p_i+q_ia) =-\prod _{i=1}^r(p_i+q_ia_i^r)+\prod _{i=1}^r(p_i+q_ia). \end{aligned}$$

So we need to show

$$\begin{aligned} \prod _{i=1}^r(p_i+q_ia)\le \prod _{i=1}^r(p_i+q_ia_i^r). \end{aligned}$$
(14)

Solving \(a_i=p_i+(1-p_i)a_i^r\) for \(p_i\) gives

$$\begin{aligned} p_i=\frac{a_i-a_i^r}{1-a_i^r}. \end{aligned}$$

Then

$$\begin{aligned} p_i+q_ia&= \frac{a_i-a_i^r}{1-a_i^r} +\left( 1-\frac{a_i-a_i^r}{1-a_i^r}\right) a=\frac{1}{1-a_i^r}\left( a_i-a_i^r+(1-a_i)a\right) ,\\ p_i+q_ia_i^r&=a_i =\frac{1}{1-a_i^r}\left( a_i(1-a_i^r)\right) . \end{aligned}$$

Noting that \(0<a_i<1\) we see that (14) is equivalent to

$$\begin{aligned}&\prod _{i=1}^r(a_i-a_i^r+(1-a_i)a) \le \prod _{i=1}^ra_i({1-a_i^r})\\ \Longleftrightarrow \quad&\prod _{i=1}^r\left( 1-a_i^{r-1}+(1-a_i)\frac{a}{a_i}\right) \le \prod _{i=1}^r(1-a_i^r)\\ \Longleftrightarrow \quad&\prod _{i=1}^r\left( 1+a_i+\cdots +a_i^{r-2} +\frac{a}{a_i}\right) \le \prod _{i=1}^r\left( 1+a_i+\cdots +a_i^{r-1} \right) , \end{aligned}$$

and multiplying both sides by \(a=a_1\cdots a_r\) we get (1). \(\square \)

Fig. 1
figure 1

The down map sending \(\tilde{\lambda }=77544200\) to \(\tilde{\mu }=66654110\)

Fig. 2
figure 2

Example for Lemma 4 and Lemma 5

Fig. 3
figure 3

Hasse diagram of \(\Lambda (\beta )\) and \(\Lambda (\beta ')\) as posets for \(r=7\) and \(d=15\)

Fig. 4
figure 4

The reduced graph for t and \(t'\) (\(r=8\) and \(d=29\))

Fig. 5
figure 5

An optimal flow in the reduced graph (\(r=8\) and \(d=29\))

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Kato, M., Kosuda, M. & Tokushige, N. Extending Muirhead’s Inequality. Graphs and Combinatorics 37, 1923–1941 (2021). https://doi.org/10.1007/s00373-021-02356-z

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