# On Muirhead's Inequality

Antoine Mhanna

► To cite this version:

Antoine Mhanna. On Muirhead's Inequality. Arhimede Mathematical Journal, 2022, 9 (1), pp.23-31.
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Distributed under a Creative Commons Attribution 4.0 International License
# ON MUIRHEAD'S INEQUALITY

ANTOINE MHANNA¹*

ABSTRACT. We revisit Muirhead's inequality together with some related known inequalities. We give a short proof to the Power-Mean inequality and Minkowski's determinant inequality.

## 1. INTRODUCTION AND PRELIMINARIES

The first section introduces some notations together with known results and inequalities starting by Muirhead's inequality. For $a = (a_1, \dots, a_n)$, $x = (x_1, \dots, x_n)$ two real $n$-tuples and $\sigma$ a permutation in $S_n$, we denote by $[x_\sigma^a]$ the product $x_{\sigma(1)}^{a_1} x_{\sigma(2)}^{a_2} \cdots x_{\sigma(n)}^{a_n}$ whenever the product is well defined.

The vector $\tilde{a}_j$ is the vector $a$ without the $a_j$ entry and $[x_\sigma^{\tilde{a}_j}]$ is the product without $x_{\sigma(j)}^{a_j}$ that equals to $\frac{x_{\sigma(1)}^{a_1} x_{\sigma(2)}^{a_2} \cdots x_{\sigma(n)}^{a_n}}{x_{\sigma(j)}^{a_j}}$.

In general we only consider nonnegative entry wise vectors except mentioning otherwise.

Throughout the paper let $a = (a_1, \dots, a_n) \in \mathbb{R}^{n+}$ and $b = (b_1, \dots, b_n) \in \mathbb{R}^{n+}$ two vectors with nonnegative entries arranged in decreasing order: $a_1 \ge \dots \ge a_n$ and $b_1 \ge \dots \ge b_n$.

We say $b$ majorizes $a$ or $a \prec b$ if and only if

$$ \sum_{i=1}^{k} a_i \le \sum_{i=1}^{k} b_i \quad \text{for all } k, 1 \le k \le n \text{ and } \sum_{i=1}^{n} a_i = \sum_{i=1}^{n} b_i. $$

**Muirhead's inequality:** If $a \prec b$ then for any nonnegative $n$-tuple $x$

$$ (1.1) \qquad \sum_{\sigma \in S_n} [x_{\sigma}^a] \le \sum_{\sigma \in S_n} [x_{\sigma}^b]. $$

The inequality was first proved for integer exponents in [2], see also [1, 3]. Mainly all proofs of (1.1) use induction and rely on the following general property.

**Proposition 1.** Let $a \prec b$ with precisely $m$ different entries, then there exist at most $m-2$ vectors $c^{(l)}$ $l \ge 1$, such that $a = c^{(0)} \prec c^{(1)} \prec c^{(2)} \prec \dots \prec c^{(j)} \prec b = c^{(j+1)}$ where $j \le m-1$ and for $l \ge 0$, $c^{(l)}$ differs from $c^{(l+1)}$ by only two entries.

Proposition 1 is an important result that can be used to prove the next two classical statements.

<footnote>Date: 2020</footnote>

<footnote>* Corresponding author.</footnote>

<footnote n="2010">2010 Mathematics Subject Classification. 11B83; 26D15; 05E05.</footnote>

<footnote>Key words and phrases. Muirhead's inequality; algebraic inequalities; Minkowski's determinant inequality.</footnote>
**Corollary 1.** [3, Chapter 5] Let $a = (a_1, \dots, a_n)$ and $b = (b_1, \dots, b_n)$ be two nonnegative real vectors arranged in decreasing order,

(1) If $a < b$ then $\prod_{i=k}^{n} a_i \ge \prod_{i=k}^{n} b_i$ for every $k, 1 \le k \le n$.

(2) If $\prod_{i=1}^{k} a_i \le \prod_{i=1}^{k} b_i$ for every $k = 1, \dots, n$ then $\sum_{i=1}^{k} a_i \le \sum_{i=1}^{k} b_i$ for every $k = 1, \dots, n$.

*Proof.* For (1) the case $n = 2$ is elementary, applying Proposition 1 we get $\prod_{i=k}^{n} c_i^{(l)} \ge \prod_{i=k}^{n} c_i^{(l+1)}$ for every $k$ and $l, 1 \le k \le n, 0 \le l \le j$. To prove (2) we go by induction, the result is true for $n = 1$ assuming it holds for $n - 1$ variables and $\sum_{i=1}^{n} a_i > \sum_{i=1}^{n} b_i$ we have

$$ \sum_{i=1}^{n-1} a_i \le \sum_{i=1}^{n-1} b_i \le \sum_{i=1}^{n} b_i < \sum_{i=1}^{n-1} a_i + a_n. $$

So there exist $h \le a_n$ such that $\sum_{i=1}^{n} b_i = \sum_{i=1}^{n-1} a_i + a_n - h$ applying (1) for $k = 1$ to $a = (a_1, a_2, \dots, a_n - h)$ and $b$ yields to a contradiction. $\square$

Next we present a proof of Muirhead's inequality similar to other known proofs see [1, 3]. Apparently the only difference is that here the sequence $c^{(l)}$ (from Proposition 1) is reversely constructed, see Example 1.

*Proof of (1.1).* The case $n = 2$ is proved by inspection: $(a_1, a_2) < (b_1, b_2)$ and $b_2 \le a_2 \le a_1 \le b_1$ with $a_2 - b_2 = b_1 - a_1 = r$ we get:

$$ x_1^{b_2} x_2^{b_2} (x_1^{a_1-b_2} x_2^{a_2-b_2} + x_2^{a_1-b_2} x_1^{a_2-b_2} - x_1^{b_1-b_2} - x_2^{b_1-b_2}) = x_1^{b_2} x_2^{b_2} \cdot T $$

and

$$ T = (x_2^r - x_1^r)(x_1^{a_1-b_2} - x_2^{a_1-b_2}) \le 0. $$

Since the sum is symmetric in $x_i$, the inequality is true if for some $j$, $a_j = b_j$ from the previous cases (lower dimension); this is because we can write (1.1) as

$$ \sum_{i=1}^{n} x_i^{a_j} \left[ \sum_{\sigma, \sigma(j)=i} [x_{\sigma}^{\tilde{b}_j}] - \sum_{\sigma, \sigma(j)=i} [x_{\sigma}^{\tilde{a}_j}] \right] \ge 0. $$

Thus we only need to give a vector $c = (c_1, \dots, c_n)$ such that $a < c < b$ with $a_k = c_k$ for some $k$ and $c_h = b_h$ for some index $h$. We assume $n \ge 3$ and $a_i \ne b_i$ for every $i$.

Take $p$ to be minimal with $a_i - b_i = r_i > 0$ for all $p \le i \le n$ and take $q$ to be minimal with $a_i - b_i = d_i < 0$ for all $q \le i \le p - 1$. If $|r_n| \le |d_q|$ set

$$ c = \begin{cases} c_q = a_q + r_n & \text{if } |r_n| \ge |d_q| \text{ set } c = \begin{cases} c_n = a_n + d_q & \\ c_q = b_q & \\ c_i = a_i \text{ otherwise,} & \end{cases} \end{cases} \square $$
**Example 1.** We give an example to illustrate the construction of sequence $c^{(1)}$, say $a = (18, 16, 6, 5, 3, 3)$ and $b = (20, 15, 8, 5, 2, 1)$. We have $a \prec b$ and omitting equal entries we may apply the previous proof to get successively $c^{(1)} = (18, 16, 8, 5, 3, 1)$, $c^{(2)} = (19, 16, 8, 5, 2, 1)$.

## 2. SOME APPLICATIONS

We first state some applications of (1.1) that recently appeared in Amer. Math. Monthly ([6]) and College Mathematics Journal ([7]) respectively.

**Problem 12066:** Let $n$ and $k$ be two integers greater than 1. Let $A$ be an $n \times n$ positive definite Hermitian matrix. Prove:

$$ (\det(A))^{1/n} \le \left( \frac{(\mathrm{Tr}(A))^k - \mathrm{Tr}(A^k)}{n^k - n} \right)^{1/k}. $$

We first rewrite the inequality as:

$$ \left(\sum_{i=1}^{n} a_i^n\right)^k - \sum_{i=1}^{n} a_i^{kn} \ge (n^k - n) \prod_{i=1}^{n} a_i^k $$

where the eigenvalue of $A$ are $\lambda_i = a_i^n$. Using the multinomial expansion:

$$ \left(\sum_{i=1}^{n} a_i^n\right)^k - \sum_{i=1}^{n} a_i^{kn} = \sum_{\substack{k_1+k_2+\cdots+k_n=k \\ 0 \le k_t < k}} \binom{k}{k_1, \dots, k_n} \prod_{t=1}^{n} a_t^{nk_t} $$

and by symmetry we see that the sequence of powers or the vector $l = (nk_1, \dots, nk_n)$ majorizes the $n$-tuple $r = (k, \dots, k)$ for any partition with $k_1 + \cdots + k_n = k$, in fact if $n \ge k$ we have nothing to prove if $n < k$, assume without loss of generality (upon rearranging the entries of $l$) that $k_1 > \cdots > k_n$ and there is $j$ less than $n$ such that $jk > n(k_1 + \cdots + k_j)$ so $n(k_{j+1} + \cdots + k_n) > (n-j)k$ which means that for some $i$, $j+1 \le i \le n$, $nk_i > k$ a contradiction since we rearranged the vector $l$ in decreasing order and supposed $jk > n(k_1 + \cdots + k_j)$ where then for some $m \in [1, j]$, $nk_m < k$. The inequality holds with equality if and only if all eigenvalues are equal.

Another application of this inequality is the following:

**Problem 1176:** Let $A$ be an $n \times n$ positive semidefinite Hermitian matrix having eigenvalues $(\lambda_1, \dots, \lambda_n)$, $\lambda_1 \ge \cdots \ge \lambda_n$ and $\lambda_1 > 0$. For $p$ and $q$ nonnegative integers prove:

$$ \frac{\mathrm{Tr}(A^p) + \mathrm{Tr}(A^{p+1}) + \dots \mathrm{Tr}(A^{p+q})}{\mathrm{Tr}(A^{p+1}) + \mathrm{Tr}(A^{p+2}) + \dots \mathrm{Tr}(A^{p+q+1})} \le \frac{n}{\mathrm{Tr}(A)}. $$

We want to prove

$$ \frac{\sum_{i=1}^{n} \lambda_i^p}{\sum_{i=1}^{n} \lambda_i^{p+q}} \le \frac{n}{\sum_{i=1}^{n} \lambda_i^q} $$

for any nonnegative reals $p$ and $q$, $(p, q \ne 0)$ which implies the result by using the next property. This is just true by Muirhead's inequality as

$$ (\sum_{i=1}^{n} \lambda_i^p)(\sum_{i=1}^{n} \lambda_i^q) \le n \sum_{i=1}^{n} \lambda_i^{p+q}. $$
Equality occurs only if all eigenvalues are equal.

**Proposition 2 ([8]).** Let $p_1, \dots, p_n$ respectively $q_1, \dots, q_n$ be $n$ real respectively positive real numbers. If $m = \min_i \frac{p_i}{q_i}$ and $M = \max_i \frac{p_i}{q_i}$ then

$$m \le \frac{p_1 + \cdots + p_n}{q_1 + \cdots + q_n} \le M.$$

For sake of completeness we present a short proof of the classical Power-Mean inequality see [5] and the references therein;

**Power-Mean Inequality:** Let $a_1 \ge a_2 \ge \cdots \ge a_n \ge 0$, $\lambda \ge 1$ and for $1 \le i \le n$, $\alpha_i \ge 0$ with $\sum_{i=1}^n \alpha_i = 1$. We have

$$\sum_{i=1}^{n} \alpha_{i} a_{i} \le \left( \sum_{i=1}^{n} \alpha_{i} a_{i}^{\lambda} \right)^{\frac{1}{\lambda}}.$$

*Proof.* Clearly the result holds for $n = 1$ suppose it is true for any $n$ and we want to prove it for $n + 1$ variables. If for example $a_1 = a_2$ the inequality holds by assumption, otherwise $a_1 > a_2$ taking the derivative in $a_1$ of $f(a_1) = (\sum_{i=1}^n \alpha_i a_i^\lambda)^{\frac{1}{\lambda}} - \sum_{i=1}^n \alpha_i a_i$ we obtain

$$f'(a_1) = \alpha_1 \left( \frac{a_1^{\lambda-1}}{\left(\sum_{i=1}^{n} \alpha_i a_i^{\lambda}\right)^{\frac{\lambda-1}{\lambda}}} - 1 \right) \ge 0$$

and the inequality follows.

<figure><img src="image_2.png" /></figure>

Along the same lines we can prove the Weighted Arithmetic Geometric inequality.

**Weighted AM-GM Inequality:** Let $a_1 \ge a_2 \ge \cdots \ge a_n \ge 0$ and for $1 \le i \le n$, $\alpha_i \ge 0$ with $\sum_{i=1}^n \alpha_i = 1$. We have

$$(2.1) \qquad \sum_{i=1}^{n} \alpha_{i} a_{i} \ge \prod_{i=1}^{n} a_{i}^{\alpha_{i}}.$$

Another interesting question is to ask when we can replace symmetric sums by cyclic ones. For this we define the cyclic permutation $c$ over the set of indexes $(1, \dots, n)$ by $c(i) = i + 1$. Of course all indexes are to be assumed (mod $n$) and in $[1; n]$. We denote by $c^n$ the iteration of $n$ cyclic permutation with the convention that $c^0$ is the identity permutation. For example:

$$x_{c^3(1)}^{15} x_{c^3(2)}^{8} x_{c^3(3)}^{7} x_{c^3(4)}^{4} = x_{4}^{15} x_{1}^{8} x_{2}^{7} x_{3}^{4}.$$

In the particular case when $b = (b_1, 0, \dots, 0)$ we have:

**Lemma 1.** Suppose $a = (a_1, \dots, a_n) \prec b = (b_1, 0, \dots, 0)$ and $x = (x_1, \dots, x_n)$ with nonnegative entries then:

$$\sum_{i=1}^{n} x_{i}^{b_{1}} \ge \sum_{i=0}^{n-1} [x_{c^{i}}^{a_{i}}].$$

*Proof.* By (2.1) we have:

$$\frac{a_1 x_{c^i(1)}^{b_1} + a_2 x_{c^i(2)}^{b_1} + \cdots + a_n x_{c^i(n)}^{b_1}}{b_1} \ge x_{c^i(1)}^{a_1} \cdots x_{c^i(n)}^{a_n},$$

summing over $i$ we obtain the inequality.

<figure><img src="image_3.png" /></figure>
**Remark 1.** If $a = \frac{p}{q}$ is a nonnegative rational number and $q$ is odd, then for $x_1 \ge x_2$ it is easy to see that $x_1^a - x_2^a \ge 0$ if $x_1 \ge -x_2$.

**Lemma 2.** Let $a = (a_1, a_2) \prec b = (b_1, 0)$ where $a_1, a_2, b_1$ are positive rationnals with odd denominators. If $x = (x_1, \dots, x_n)$ with $x_i + x_j \ge 0$ for every $i \ne j$, ($x$ has at most one negative entry), then

$$ \sum_{i=1}^{n} x_i^{b_1} \ge \sum_{i=1}^{n} x_i^{a_1} x_{i+1}^{a_2}, $$

where $x_{n+1} \equiv x_1$.

*Proof.* For $n = 2$, $x_1^{b_1} + x_2^{b_1} - x_1^{a_1} x_2^{a_2} - x_2^{a_1} x_1^{a_2} = (x_1^{a_2} - x_2^{a_2})(x_1^{a_1} - x_2^{a_1}) \ge 0$, suppose the result holds for $x = (x_1, \dots, x_{n-1})$ we shall prove it for $n$ variables. Assume without loss of generality $x_1$ to be the maximal variable (positive) and take the derivative of the function $f(x_1) = x_1^{b_1} - x_1^{a_1} x_2^{a_2} - x_n^{a_1} x_1^{a_2}$ to get

$$ f'(x_1) = a_1 x_1^{a_1-1} (x_1^{a_2} - x_2^{a_2}) + a_2 x_1^{a_2-1} (x_1^{a_1} - x_n^{a_1}) \ge 0 \quad (\text{Remark 1}). $$

In particular we only need to consider the case $x_1 = x_2$ or $x_1 = x_n$ but in either case the inequaliy reduces to the case of $n-1$ variables which is true by induction. $\square$

The well known Minkowski's determinant inequality -see [4, page 510]- states that for any $A$ and $B$ in $M_n(C)$ positive semi-definite (Hermitian) we have

$$ (2.2) \qquad \mathrm{Det}(A+B) \ge \left(\mathrm{Det}(A)^{\frac{1}{n}} + \mathrm{Det}(B)^{\frac{1}{n}}\right)^n. $$

Denote by $\lambda_i(M)$, $1 \le i \le n$ the eigenvalues of the Hermitian positive semi-definite matrix $M$ in decreasing order. For any $k$, $1 \le k \le n$, $\sum_{i=1}^k \lambda_i(M) = \|M\|_k$ is by definition the Ky-Fan $k$-norm of $M$ see [4]. We can write for $1 \le k \le n$:

$$ \sum_{i=1}^{k} \lambda_i(A+B) \le \sum_{i=1}^{k} \lambda_i(A) + \sum_{i=1}^{k} \lambda_i(B) = \sum_{i=1}^{k} (\lambda_i(A) + \lambda_i(B)) $$

and $\sum_{i=1}^n \lambda_i(A+B) = \sum_{i=1}^n (\lambda_i(A) + \lambda_i(B)) = \mathrm{Tr}(A+B)$. Setting $a_i = \lambda_i(A+B)$, $b_i = \lambda_i(A) + \lambda_i(B)$ and using Corollary 1 we have

$$ (2.3) \qquad \mathrm{Det}(A+B) \ge \prod_{i=1}^{n} (\lambda_i(A) + \lambda_i(B)) \ge \left(\mathrm{Det}(A)^{\frac{1}{n}} + \mathrm{Det}(B)^{\frac{1}{n}}\right)^n. $$

The last bound of (2.3) can be deduced from [9, ex. 219] and is proved as follows: Suppose $\mathrm{Det}(B) \ne 0$ (if not it is easy to conclude), dividing by $\mathrm{Det}(B)$ and letting $\frac{\lambda_i(A)}{\lambda_i(B)} = c_i$ we need to prove

$$ \prod_{i=1}^{n} (c_i + 1) \ge \left(\prod_{i=1}^{n} c_i^{\frac{1}{n}} + 1\right)^n, $$

equivalently

$$ \sum_{k=0}^{n} e_k(c_1, \dots, c_n) \ge \sum_{k=0}^{n} \binom{n}{k} \prod_{i=1}^{n} c_i^{\frac{k}{n}} $$
where $e_k(c_1, \dots, c_n)$ is the elementary symmetric function of order $k$. Clearly
$$e_k(c_1, \dots, c_n) \ge \binom{n}{k} \prod_{i=1}^n c_i^{\frac{k}{n}}$$
with equality if and only if all $c_i$'s are equal.

REFERENCES

[1] Z. Kadelburg, D. Dukić, M. Lukić and I. Matić, *Inequalities of Karamata, Schur and Muirhead and some applications*, The Teaching of Mathematics, **Vol. VIII**, (1), 2005, pp. 31–45.

[2] R. F. Muirhead, *Some Methods applicable to identities and inequalities of symmetric algebraic functions of m letters*, Proc. Edinburgh Math. Soc. **21**, 1903, pp. 144–157.

[3] A. W. Marshall, I. Olkin, B. C. Arnold, *Inequalities: Theory of Majorization and Its Applications, 2nd Edition*, Springer Series in Statistics, New York, 2011.

[4] R. A. Horn, C. R. Johnson, *Matrix Analysis 2nd Edition*, Cambridge University Press, Cambridge, 2013.

[5] Y. Li, X-M. Gu, J. Xiao, *A note on the proofs of generalized Radon inequality*, Mathematica Moravica, **Vol. 22**, (2), 2018, 59–67.

[6] X-Q. Chang, Problem 12066, Amer. Math. Monthly **125**(8) (2018), pp. 756.

[7] X-Q. Chang, Problem 1176, College Mathematics Journal **51**(3) (2020), pp. 225.

[8] www.cut-the-knot.org/arithmetic/algebra/InequalityWithFractions.shtml.

[9] D. Serre, *Matrices: Theory & Applications Additional exercices*, Available at http://www.umpa.ens-lyon.fr/~serre/DPF/exobis.pdf.

<footnote n="1">¹ KFARDEBIAN, LEBANON</footnote>

<footnote>E-mail address: tmhanat@yahoo.com</footnote>