% arXiv v2, derived from the immutable general-purpose manuscript v4. \documentclass[11pt]{article} \usepackage[T1]{fontenc} \usepackage{lmodern} \usepackage{amsmath,amssymb,amsthm} \usepackage[margin=1in]{geometry} \usepackage{cite} \usepackage[hidelinks]{hyperref} \hypersetup{ pdftitle={The Marcus--Minc Transform Inequality}, pdfauthor={Yair Lavi} } \newtheorem{theorem}{Theorem}[section] \newtheorem{proposition}[theorem]{Proposition} \newtheorem{lemma}[theorem]{Lemma} \setlength{\emergencystretch}{3em} \title{The Marcus--Minc Transform Inequality} \author{Yair Lavi} \date{24 September 2026} \begin{document} \maketitle \begin{abstract} We prove the Marcus--Minc conjecture: if $n\ge2$ is an integer, $A$ is a nonnegative $n\times n$ matrix whose row and column sums equal one, and $J_n$ has every entry $1/n$, then \[ \operatorname{per}A\ge \operatorname{per}\!\left(\frac{nJ_n-A}{n-1}\right). \] We also determine all equality cases. \end{abstract} \noindent\textbf{Keywords.} Permanent; doubly stochastic matrix; matrix inequality; Marcus-Minc. \noindent\textbf{2020 Mathematics Subject Classification.} 15A15, 05A20, 15B51. \section{Introduction and main results} For an integer $n\ge2$, put $[n]=\{1,\ldots,n\}$, and let $S_n$ be the set of all permutations of $[n]$. For an $n\times n$ matrix $D=(d_{ij})$, its permanent is \[ \operatorname{per}D=\sum_{\pi\in S_n}\prod_{i=1}^n d_{i,\pi(i)}. \tag{1.1} \] A real matrix is \textbf{doubly stochastic} if its entries are nonnegative and the entries in each row and in each column sum to one. We denote the set of $n\times n$ doubly stochastic matrices by $\Omega_n$. Let $\mathbf1$ be the all-ones column vector, let \[ J_n=\frac1n\mathbf1\mathbf1^{\mathsf T}, \] and define the Marcus--Minc transform $\tau_n:\Omega_n\to\Omega_n$ by \[ \tau_n(A)=\frac{nJ_n-A}{n-1}. \tag{1.2} \] The map is well defined: every entry of a doubly stochastic matrix is at most one, so $nJ_n-A$ is nonnegative, and each of its row and column sums is $n-1$. A \textbf{permutation matrix} is a zero--one matrix having exactly one $1$ in each row and column. In 1967 Marcus and Minc \cite{ref1} conjectured that $\operatorname{per}A\ge\operatorname{per}\tau_n(A)$ for every $A\in\Omega_n$, and that for $n\ge3$ equality holds only at $J_n$; we call this the Marcus--Minc conjecture. As reported in \cite{ref7,ref9}, they proved the inequality for $n=2$, for symmetric positive semidefinite $A$, and for $A$ sufficiently close to $J_n$. Wang proved the inequality in order three and showed that the original equality claim fails \cite{ref2}: equality also holds for the six matrices $(\mathbf1\mathbf1^{\mathsf T}-P)/2$, with $P$ a permutation matrix. Foregger proved it in order four \cite{ref3}. Hwang proved it for partly decomposable matrices \cite{ref4}, and Malek proved it near $J_n$ and under a spectral-sector condition \cite{ref5,ref6}. Cheon and Wanless recorded the general assertion as Conjecture 3 in their 2005 update of Minc's catalogue \cite{ref6}. The problem also appears in later work \cite{ref7,ref8,ref9}. Our first result proves the Marcus--Minc conjecture and classifies equality in every dimension. \begin{theorem} For every integer $n\ge2$ and every $A\in\Omega_n$, \[ \operatorname{per}A\ge\operatorname{per}\tau_n(A). \tag{1.3} \] The equality cases are exactly the following: \begin{enumerate} \item if $n=2$, every $A\in\Omega_2$; \item if $n=3$, the matrix $J_3$ and the six matrices $(\mathbf1\mathbf1^{\mathsf T}-P)/2$, with $P$ a permutation matrix; \item if $n\ge4$, only $J_n$. \end{enumerate} \end{theorem} We also give new proofs for the known cases $n=3,4$; $n=2$ is immediate. The proof comes from two estimates with different dimensional scales. To state them, define the Frobenius norm of $X=(x_{ij})$ by \[ \|X\|_F=\left(\sum_{i,j}x_{ij}^2\right)^{1/2}, \] and set \[ d_n=\frac{n!}{n^n},\qquad r^2=\|A-J_n\|_F^2,\qquad a_n=\frac1{2n}+\frac1{3n^2}. \tag{1.4} \] For real $t$, also put \[ T_n(t)=\sum_{k=0}^n\frac{(nt)^k(1-t)^{n-k}}{k!}, \qquad \beta_n=\frac{T_n(1/(n-1))-1}{n-1}. \tag{1.5} \] \begin{theorem} For every $n\ge2$ and every $A\in\Omega_n$, \[ \frac{\operatorname{per}A}{d_n}\ge e^{a_nr^2}, \qquad \frac{\operatorname{per}\tau_n(A)}{d_n}\le1+\beta_nr^2. \tag{1.6} \] \end{theorem} For $n\ge5$ one has $\beta_n<1/(2n)0. \] It is \textbf{homogeneous of degree $m$} if every monomial with nonzero coefficient has total degree $m$. For a homogeneous polynomial of degree $m$ in $m$ variables with nonnegative coefficients, define its capacity by \[ \operatorname{Cap}(p)= \inf_{x_1,\ldots,x_m>0} \frac{p(x_1,\ldots,x_m)}{x_1\cdots x_m}. \tag{2.1} \] The notation $[x_1\cdots x_m]p$ means the coefficient of the monomial $x_1\cdots x_m$ in $p$. We use the following coefficient bound of Gurvits \cite[Theorem~2.4 and Corollary~2.5]{ref10}. A proof is included because the zero-capacity boundary case requires care and because the bound is the only stable-polynomial input to our argument. \begin{proposition}[Gurvits] If $p$ is real stable, homogeneous of degree $m$ in $m$ variables, and has nonnegative coefficients, then \[ [x_1\cdots x_m]p\ge\frac{m!}{m^m}\operatorname{Cap}(p). \tag{2.2} \] \end{proposition} \begin{proof} If $p=0$ or $m=1$, the claim is immediate; hence assume $p\ne0$ and $m\ge2$. We first recall two closure facts used below. Specializing one variable of a real-stable polynomial at a real value preserves real stability or gives zero: approach that value from the upper half-plane and apply Hurwitz's theorem. Partial differentiation has the same property. Indeed, if $d=\deg_{x_m}p\ge1$, the leading coefficient $p_d(x_1,\ldots,x_{m-1})$ is nonvanishing when all its arguments lie in the upper half-plane: the nonvanishing polynomials $p(z_1,\ldots,z_{m-1},it)/(it)^d$ converge locally uniformly to $p_d$ as $t\to\infty$, and Hurwitz's theorem applies. Thus $p(z_1,\ldots,z_{m-1},\cdot)$ is nonconstant there. Gauss--Lucas places all roots of its derivative in the closed lower half-plane, so the partial derivative is real stable. If $d=0$, that derivative is zero. For integers $s\ge2$, put $g(s)=((s-1)/s)^{s-1}$, and set $g(1)=1$. Let $h$ be a one-variable polynomial of degree at most $s$, with nonnegative coefficients and all roots real and nonpositive. We first show \[ h'(0)\ge g(s)\inf_{t>0}\frac{h(t)}t. \tag{2.3} \] If $h(0)>0$, write $h(t)=h(0)\prod_{i=1}^s(1+c_it)$ with $c_i\ge0$, adding zero values of $c_i$ when the degree is smaller than $s$. Set $u=\sum_i c_i=h'(0)/h(0)$. The arithmetic--geometric mean inequality gives \[ h(t)\le h(0)\left(1+\frac{ut}{s}\right)^s. \] If $u>0$ and $s\ge2$, evaluate the right-hand side divided by $t$ at $t=s/[u(s-1)]$ to obtain (2.3). If $u=0$, both sides of (2.3) are zero; if $s=1$, (2.3) is the direct identity for a linear polynomial. If $h(0)=0$, nonnegativity of the coefficients gives $\inf_{t>0}h(t)/t=h'(0)$, including the case in which $t^2$ divides $h$. Now fix positive values of $x_1,\ldots,x_{m-1}$ and apply (2.3) to $h(t)=p(x_1,\ldots,x_{m-1},t)$. Real stability, followed by specialization at positive real values, makes every root of this one-variable polynomial real; nonnegative coefficients make every root nonpositive. If \[ q(x_1,\ldots,x_{m-1})= \left.\frac{\partial p}{\partial x_m}\right|_{x_m=0}, \] then the definition of capacity and (2.3) give \[ q(x_1,\ldots,x_{m-1}) \ge g(m)\operatorname{Cap}(p)x_1\cdots x_{m-1}. \] Hence $\operatorname{Cap}(q)\ge g(m)\operatorname{Cap}(p)$. The closure facts above allow the same step to be iterated. At each stage the symbol $s$ in $g(s)$ is the number of variables still present. After the last differentiation the resulting constant is $[x_1\cdots x_m]p$, and \[ \prod_{s=2}^m g(s)=\frac{m!}{m^m}. \] This proves (2.2), including zero capacity. \end{proof} \subsection{A one-column bound} Fix $A=(a_{ij})\in\Omega_n$ whose entries are initially all positive, and fix a column index $j$. Write \[ c_i=a_{ij}, \qquad L_i(x)=\sum_{k\ne j}a_{ik}x_k. \] Consider \[ p_A(x)=\prod_{i=1}^n\left(\sum_{k=1}^n a_{ik}x_k\right). \] Every linear factor is nonzero when all variables have positive imaginary parts, so $p_A$ is real stable. Differentiation and specialization give the nonzero real-stable polynomial \[ q_j(x)=\left.\frac{\partial p_A}{\partial x_j}\right|_{x_j=0} =\sum_{i=1}^n c_i\prod_{s\ne i}L_s(x). \tag{2.4} \] It is homogeneous of degree $n-1$ in the $n-1$ variables indexed by $k\ne j$, and the coefficient of $\prod_{k\ne j}x_k$ is $\operatorname{per}A$. Since $\sum_i c_i=1$, weighted arithmetic--geometric mean applied to (2.4) gives \[ q_j(x)\ge \prod_{i=1}^n\left(\prod_{s\ne i}L_s(x)\right)^{c_i} =\prod_{s=1}^n L_s(x)^{1-c_s}. \tag{2.5} \] For each row $s$, the remaining row sum is $\sum_{k\ne j}a_{sk}=1-c_s$. A second weighted arithmetic--geometric mean inequality gives \[ L_s(x)^{1-c_s} \ge(1-c_s)^{1-c_s}\prod_{k\ne j}x_k^{a_{sk}}. \tag{2.6} \] Multiplying (2.6) over $s$ and using the remaining column sums shows \[ \operatorname{Cap}(q_j)\ge\prod_{i=1}^n(1-c_i)^{1-c_i}. \] Proposition 2.1, with $m=n-1$, therefore yields \[ \operatorname{per}A\ge d_{n-1}\prod_{i=1}^n(1-a_{ij})^{1-a_{ij}}. \tag{2.7} \] The full one-column bound (2.7) is implicit in the proof of Laurent and Schrijver \cite[proof of Corollary~1d, equation~(39)]{ref11}; our use of it is the explicit quadratic estimate (2.11). For an arbitrary $A\in\Omega_n$, apply (2.7) to $(1-\delta)A+\delta J_n$ and let $\delta\downarrow0$. We use the continuous convention $0^0=1$. This proves (2.7) on the boundary without assuming continuity or attainment of capacity. \subsection{From one column to the global lower bound} Define $f:[0,1]\to\mathbb R$ by \[ f(x)=(1-x)\log(1-x), \] with its continuous value $f(1)=0$, and define the squared deviation of column $j$ from the uniform column by \[ F_j=\sum_{i=1}^n\left(a_{ij}-\frac1n\right)^2. \] Because $d_{n-1}(1-1/n)^{n-1}=d_n$, taking logarithms in (2.7) gives \[ \log\frac{\operatorname{per}A}{d_n} \ge\sum_{i=1}^n f(a_{ij})-nf(1/n). \tag{2.8} \] The convergent endpoint-valid expansion \[ f(x)=-x+\sum_{k\ge2}\frac{x^k}{k(k-1)} \tag{2.9} \] makes the gain over the uniform column explicit. Put $s_2=\sum_i a_{ij}^2=1/n+F_j$. Jensen's inequality with weights $a_{ij}$ gives \[ \sum_i a_{ij}^3\ge s_2^2, \qquad \sum_i a_{ij}^3-\frac1{n^2} \ge\frac{2F_j}{n}+F_j^2. \tag{2.10} \] For every $k\ge4$, ordinary convexity gives $\sum_i a_{ij}^k\ge n^{1-k}$. Substituting (2.9) into the difference in (2.8), the linear terms cancel, the quadratic term contributes $F_j/2$, and (2.10) controls the cubic term. Consequently \[ \sum_i f(a_{ij})-nf(1/n) \ge\left(\frac12+\frac1{3n}\right)F_j+\frac{F_j^2}{6} \ge\left(\frac12+\frac1{3n}\right)F_j. \tag{2.11} \] Since $\sum_jF_j=r^2$, some column has $F_j\ge r^2/n$. Equations (2.8)--(2.11) prove the first inequality in (1.6): \[ \frac{\operatorname{per}A}{d_n}\ge \exp\!\left[\left(\frac1{2n}+\frac1{3n^2}\right)r^2\right]. \tag{2.12} \] \section{A subpermanent estimate for zero-margin matrices} For subsets $S,T\subseteq[n]$ of the same size, let $X[S,T]$ be the submatrix of $X$ with row set $S$ and column set $T$. For $0\le k\le n$, define the sum of all $k\times k$ subpermanents by \[ \sigma_k(X)= \sum_{\substack{S,T\subseteq[n]\\ |S|=|T|=k}} \operatorname{per}X[S,T], \qquad \sigma_0(X)=1. \tag{3.1} \] A matrix has \textbf{zero margins} if every row sum and every column sum is zero. Its Euclidean operator norm is \[ \|X\|_{2\to2}=\sup_{v\ne0}\frac{\|Xv\|_2}{\|v\|_2}. \] Write $I_n$ for the identity matrix and set \[ Q=I_n-J_n. \] Put $H=\mathbf1^\perp=\{v\in\mathbb R^n:\mathbf1^{\mathsf T}v=0\}$. The map $Q$ is the identity on $H$ and is zero on the line $\mathbb R\mathbf1$; this is what we mean by the orthogonal projection onto $H$. \begin{lemma} If $X$ is a real $n\times n$ matrix with zero margins and $\|X\|_{2\to2}\le1$, then, for $2\le k\le n$, \[ |\sigma_k(X)|\le \frac{\sigma_k(Q)}{n-1}\|X\|_F^2. \tag{3.2} \] \end{lemma} \begin{proof} We use tensor powers of $\mathbb R^n$ with the product inner product \[ \langle x_1\otimes\cdots\otimes x_k, y_1\otimes\cdots\otimes y_k\rangle =\prod_{\ell=1}^k\langle x_\ell,y_\ell\rangle, \] extended bilinearly. If $S=\{i_1,\ldots,i_k\}$ is a $k$-element subset of $[n]$, put \[ v_S=\frac1{\sqrt{k!}}\sum_{\pi\in S_k} e_{i_{\pi(1)}}\otimes\cdots\otimes e_{i_{\pi(k)}}, \qquad u_k=\sum_{|S|=k}v_S, \tag{3.3} \] where $e_1,\ldots,e_n$ are the standard basis vectors. The vectors $v_S$ are orthonormal, and expanding the inner product gives \[ \langle v_S,X^{\otimes k}v_T\rangle=\operatorname{per}X[S,T]. \tag{3.4} \] Zero margins imply $X=QXQ$. Hence, with $w=Q^{\otimes k}u_k\in H^{\otimes k}$, \[ \sigma_k(X)=\langle w,X^{\otimes k}w\rangle, \qquad \|w\|^2=\sigma_k(Q)\ge0. \tag{3.5} \] The second identity follows because $Q^{\otimes k}$ is an orthogonal projection. It also supplies the nonnegativity of every $\sigma_k(Q)$ used below. We now define the tensor property responsible for the estimate. A linear operator $R$ on $H$ is \textbf{self-adjoint} if $\langle Rx,y\rangle=\langle x,Ry\rangle$ for all $x,y\in H$. For $w\in H^{\otimes k}$ and $1\le\ell\le k$, its $\ell$-th \textbf{one-factor marginal} is the unique self-adjoint operator $\rho_\ell$ on $H$ satisfying \[ \operatorname{tr}(\rho_\ell R)= \left\langle w, \bigl(I_H^{\otimes(\ell-1)}\otimes R\otimes I_H^{\otimes(k-\ell)}\bigr)w\right\rangle \tag{3.6} \] for every self-adjoint operator $R$ on $H$; here $\operatorname{tr}$ denotes the sum of the diagonal entries of an operator in an orthonormal basis. We call $w$ \textbf{one-factor isotropic} when each $\rho_\ell$ is a scalar multiple of $I_H$. To see existence, choose an orthonormal basis $(h_a)$ of $H$ and, after putting the $\ell$-th factor first, write $w=\sum_a h_a\otimes z_a$, where $z_a\in H^{\otimes(k-1)}$. The matrix $(\langle z_a,z_b\rangle)_{a,b}$ is self-adjoint and satisfies (3.6), so it defines $\rho_\ell$. If two self-adjoint operators satisfy (3.6), their difference $D$ gives $\operatorname{tr}(D^2)=0$ upon taking $R=D$, and therefore $D=0$. Simultaneously permuting the $n$ coordinate labels fixes $u_k$, commutes with $Q$, and therefore fixes $w$. Thus each $\rho_\ell$ commutes with every coordinate permutation on $H$. Extend $\rho_\ell$ by zero on $\mathbb R\mathbf1$. A matrix commuting with every permutation has one constant on its diagonal and another off its diagonal. Its restriction to $H$ is therefore scalar. Since $\operatorname{tr}\rho_\ell=\|w\|^2$, we obtain \[ \rho_\ell=\frac{\|w\|^2}{n-1}I_H \qquad(1\le\ell\le k). \tag{3.7} \] This is the asserted isotropy; it is a property of $w$, not an assumption on $X$. Choose linear operators $U,V$ on $H$ satisfying $U^{\mathsf T}U=V^{\mathsf T}V=I_H$ and numbers $0\le s_1,\ldots,s_{n-1}\le1$ such that \[ X|_H=UDV^{\mathsf T}, \qquad D=\operatorname{diag}(s_1,\ldots,s_{n-1}). \] Such a factorization is a singular-value decomposition; the displayed identities say that $U$ and $V$ are orthogonal, and the upper bounds on the $s_i$ follow from $\|X\|_{2\to2}\le1$. Put \[ a=(U^{\mathsf T})^{\otimes k}w, \qquad b=(V^{\mathsf T})^{\otimes k}w, \qquad K=D^{\otimes k}. \] The one-factor marginals of both $a$ and $b$ still equal (3.7). A self-adjoint operator $R$ is \textbf{positive semidefinite} if $\langle z,Rz\rangle\ge0$ for every $z$. For self-adjoint operators $R,S$, write $R\preceq S$ when $S-R$ is positive semidefinite. For every diagonal entry of $K$, the assumptions $k\ge2$ and $0\le s_i\le1$ give \[ \prod_{\ell=1}^k s_{i_\ell} \le\left(\prod_{\ell=1}^k s_{i_\ell}^2\right)^{1/k} \le\frac1k\sum_{\ell=1}^k s_{i_\ell}^2. \] Equivalently, \[ 0\preceq K\preceq\frac1k\sum_{\ell=1}^k I_H^{\otimes(\ell-1)}\otimes D^2\otimes I_H^{\otimes(k-\ell)}. \tag{3.8} \] Equations (3.6)--(3.8) yield \[ \langle a,Ka\rangle,\ \langle b,Kb\rangle \le\frac{\|w\|^2}{n-1}\operatorname{tr}D^2. \] Cauchy--Schwarz for the positive semidefinite operator $K$ now gives \[ \begin{aligned} |\sigma_k(X)| &=|\langle a,Kb\rangle|\\ &\le\sqrt{\langle a,Ka\rangle\langle b,Kb\rangle}\\ &\le\frac{\sigma_k(Q)}{n-1}\|X\|_F^2, \end{aligned} \] because $\operatorname{tr}D^2=\|X\|_F^2$. This proves (3.2). \end{proof} For later use, we verify the contraction hypothesis. If $A\in\Omega_n$ and $x\in\mathbb R^n$, Jensen's inequality in each row and the column sums give \[ \sum_i\left(\sum_j a_{ij}x_j\right)^2 \le\sum_{i,j}a_{ij}x_j^2 =\sum_jx_j^2. \tag{3.9} \] Thus $\|A\|_{2\to2}\le1$. With $B=A-J_n$, the row and column sum identities also give $B=QAQ$, so $B$ has zero margins and $\|B\|_{2\to2}\le1$. Lemma 3.1 applies to $B$. \section{The transformed upper bound} Let $B=A-J_n$. Expanding the permanent according to the entries chosen from $tB$ gives, for every real $t$, \[ \operatorname{per}(J_n+tB) =\sum_{k=0}^n\frac{(n-k)!}{n^{n-k}}t^k\sigma_k(B). \tag{4.1} \] Indeed, after choosing the $k$ rows and $k$ columns used by $B$, the remaining $n-k$ rows can be matched to the remaining columns in $(n-k)!$ ways, and each remaining entry of $J_n$ contributes $1/n$. Since $B$ has zero margins, $\sigma_1(B)=0$. Put $q=1/(n-1)$. Formula (1.2) becomes $\tau_n(A)=J_n-qB$. Applying Lemma 3.1 to every term of (4.1), and bounding the odd terms in absolute value rather than assuming a sign, gives \[ \begin{aligned} \operatorname{per}\tau_n(A) &\le d_n+\frac{r^2}{n-1} \sum_{k=2}^n\frac{(n-k)!}{n^{n-k}}q^k\sigma_k(Q)\\ &=d_n+\frac{r^2}{n-1} \bigl(\operatorname{per}(J_n+qQ)-d_n\bigr). \end{aligned} \tag{4.2} \] Here $\sigma_1(Q)=0$ because $Q$ has zero margins, and $\sigma_k(Q)\ge0$ by (3.5). No positivity of an odd-degree coefficient $\sigma_k(B)$ is being assumed. To evaluate the last permanent, write $J_n+tQ=tI_n+(1-t)J_n$. Choosing the $k$ rows in which the identity term is used gives \[ \operatorname{per}(J_n+tQ) =\sum_{k=0}^n \binom{n}{k}t^k(n-k)! \left(\frac{1-t}{n}\right)^{n-k}. \tag{4.3} \] Dividing (4.3) by $d_n$ gives exactly $T_n(t)$ from (1.5). Therefore (4.2) is the second inequality in (1.6): \[ \frac{\operatorname{per}\tau_n(A)}{d_n}\le1+\beta_nr^2. \tag{4.4} \] \section{Separation in orders at least four} Assume first that $n\ge5$, and retain $q=1/(n-1)$. Formula (1.5) can be written as the truncated-exponential identity \[ T_n(q)=(1-q)^n \sum_{k=0}^n\frac1{k!} \left(\frac{nq}{1-q}\right)^k. \tag{5.1} \] All omitted terms of the exponential series are positive. Hence \[ T_n(q)< \exp\!\left(n\left[\log(1-q)+\frac{q}{1-q}\right]\right). \tag{5.2} \] The expression in brackets satisfies \[ \log(1-q)+\frac{q}{1-q} =\int_0^q\frac{t}{(1-t)^2}\,dt \le\frac{q^2}{2(1-q)^2}. \tag{5.3} \] Set $x=n/[2(n-2)^2]$. Then $00. \end{aligned} \tag{5.5} \] This proves $\beta_n<1/(2n)$ and the positivity of $\epsilon_n$ in (1.8) for every $n\ge5$. If $A\ne J_n$, then $r>0$, so (2.12), (4.4), and (5.4) give the strict chain \[ \frac{\operatorname{per}A}{d_n} \ge e^{r^2/(2n)} >1+\frac{r^2}{2n} >1+\beta_nr^2 \ge\frac{\operatorname{per}\tau_n(A)}{d_n}. \tag{5.6} \] At $A=J_n$, the two permanents both equal $d_n$. For $n=4$, direct evaluation of (1.5) gives \[ T_4(t)=1+2t^2+\frac83t^3+5t^4, \qquad T_4(1/3)=\frac{112}{81}. \] Consequently \[ \beta_4=\frac{31}{243}, \qquad a_4=\frac7{48}, \qquad a_4-\beta_4=\frac{71}{3888}>0, \] which proves (1.7) and makes the comparison in (1.3) strict away from $J_4$. Finally, $e^{a_nr^2}\ge1+a_nr^2$. Subtracting the two bounds in (1.6) therefore gives \[ G_n(A)\ge d_n(a_n-\beta_n)r^2. \tag{5.7} \] For $n=4$, (5.7) is (1.9). For $n\ge5$, replacing $a_n$ by the smaller $1/(2n)$ and using (5.4) gives (1.9) with the explicit $\epsilon_n$ from (1.8). This establishes (1.3) and its equality case for every $n\ge4$. \section{Orders three and two} \subsection{Order three} Let $A\in\Omega_3$, put $B=A-J_3$, and retain $r^2=\|B\|_F^2$. Direct expansion of a $3\times3$ zero-margin matrix gives \[ \sigma_2(B)=\frac12r^2, \qquad \operatorname{per}B=\frac23\sum_{i,j=1}^3b_{ij}^3. \tag{6.1} \] Apply (4.1) at $t=1$ and $t=-1/2$. Since $\tau_3(A)=J_3-B/2$, one obtains \[ G_3(A)=\frac18\left(r^2+6\sum_{i,j=1}^3b_{ij}^3\right). \tag{6.2} \] Consider one row $(x,y,z)$ of $A$. It consists of nonnegative numbers with $x+y+z=1$. Its contribution to the expression in parentheses in (6.2) is \[ \begin{aligned} &\sum_{u\in\{x,y,z\}}(u-1/3)^2 +6\sum_{u\in\{x,y,z\}}(u-1/3)^3\\ &\qquad=2\bigl(1-4(xy+xz+yz)+9xyz\bigr). \end{aligned} \tag{6.3} \] Reorder the three entries so that $x\ge y\ge z$. The following elementary form of Schur's identity, using $x+y+z=1$, is \[ 1-4(xy+xz+yz)+9xyz =(x-y)^2(x+y-z)+z(x-z)(y-z). \tag{6.4} \] Both terms on the right are nonnegative: in particular, $x+y-z=1-2z\ge1/3$. Thus every row contribution in (6.3) is nonnegative, which proves $G_3(A)\ge0$. Equation (6.4) also determines equality. It first forces $x=y$ and then forces either $z=0$ or $z=x$. Thus each equality row is either $(1/3,1/3,1/3)$ or a permutation of $(1/2,1/2,0)$. If exactly one or two rows are half-rows, the half-rows would have to contribute respectively $1/3$ or $2/3$ to every column after the uniform rows are removed. Their contribution to each column is a multiple of $1/2$, so this is impossible. All rows are therefore uniform, giving $J_3$, or all rows are half-rows. In the latter case every column contains exactly one zero, the zero positions form a permutation matrix $P$, and \[ A=\frac{\mathbf1\mathbf1^{\mathsf T}-P}{2}. \] Conversely, $J_3$ and each of these six matrices make every row contribution in (6.3) zero. This proves the order-three statement and its full equality classification. \subsection{Order two} Every matrix in $\Omega_2$ has the form \[ A=\begin{pmatrix}t&1-t\\1-t&t\end{pmatrix}, \qquad 0\le t\le1. \] The transform replaces $t$ by $1-t$, while \[ \operatorname{per}A=t^2+(1-t)^2. \] Hence every $A\in\Omega_2$ is an equality case. \subsection{Completion of the proof} \begin{proof}[Proof of Theorem~1.1] Section 5 proves (1.3) for $n\ge4$, with equality only at $J_n$. Section 6.1 proves (1.3) in order three, with exactly the seven equality cases listed in the theorem. 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