[proofplan] Since the statement now assumes $n \ge 2$, the two distinguished arguments $A$ and $B$ and the remaining arguments $A_3,\dots,A_n$ are well-defined. We regard the determinant as a homogeneous hyperbolic polynomial on the real [vector space](/page/Vector%20Space) of symmetric matrices. Its closed hyperbolicity cone is the cone of positive semidefinite matrices, so the stated matrices lie in the domain where the closed-cone form of the hyperbolic-polynomial [Alexandrov--Fenchel inequality](/theorems/4126) applies. Applying that inequality to the determinant with the fixed arguments $A_3,\dots,A_n$ gives exactly the mixed-discriminant inequality after identifying the determinant polarization with $D$. [/proofplan]

[step:Identify the mixed discriminant with the complete polarization of the determinant] Let $V := \operatorname{Sym}_n(\mathbb{R})$ denote the real vector space of real symmetric $n \times n$ matrices, and define the determinant polynomial \begin{align*} h: V &\to \mathbb{R} \\ M &\mapsto \det M. \end{align*} Let $H: V^n \to \mathbb{R}$ denote the complete polarization of $h$, defined by \begin{align*} H(M_1,\dots,M_n) := \frac{1}{n!}\frac{\partial^n}{\partial t_1\cdots \partial t_n}\bigg|_{t_1=\cdots=t_n=0} h(t_1M_1+\cdots+t_nM_n). \end{align*} By the definition of the mixed discriminant, for every $M_1,\dots,M_n \in V$, \begin{align*} D(M_1,\dots,M_n)=H(M_1,\dots,M_n). \end{align*} Thus it is enough to prove \begin{align*} H(A,B,A_3,\dots,A_n)^2 \ge H(A,A,A_3,\dots,A_n)H(B,B,A_3,\dots,A_n). \end{align*} [/step]

[step:Verify that the determinant is hyperbolic with positive semidefinite cone]Let $I_n \in V$ be the identity matrix. For each $M \in V$, the polynomial $s \mapsto h(M+sI_n)$ has only real roots, because the spectral theorem gives real eigenvalues $\lambda_1(M),\dots,\lambda_n(M)$ and hence \begin{align*} h(M+sI_n)=\prod_{i=1}^n (\lambda_i(M)+s). \end{align*} Also $h(I_n)=1>0$. Therefore $h$ is hyperbolic with respect to $I_n$. The hyperbolicity cone containing $I_n$ is the cone $\mathcal{P} \subset V$ of positive definite real symmetric matrices, and its closure $\overline{\mathcal{P}}$ is the cone of positive semidefinite real symmetric matrices. Since $A,B,A_3,\dots,A_n$ are positive semidefinite by hypothesis, all of them lie in $\overline{\mathcal{P}}$.[/step]

[guided]We need to put the matrix statement into the setting where the Alexandrov--Fenchel theorem for hyperbolic polynomials applies. The ambient vector space is \begin{align*} V := \operatorname{Sym}_n(\mathbb{R}), \end{align*} the real vector space of real symmetric $n \times n$ matrices, and the polynomial is \begin{align*} h: V &\to \mathbb{R} \\ M &\mapsto \det M. \end{align*} We use $I_n \in V$ for the identity matrix. To verify hyperbolicity with respect to $I_n$, we must check two facts: $h(I_n)>0$ and, for every $M \in V$, the one-variable polynomial $s \mapsto h(M+sI_n)$ has only real zeros. The first condition is \begin{align*} h(I_n)=\det I_n=1>0. \end{align*} For the second condition, the spectral theorem for real symmetric matrices gives real eigenvalues $\lambda_1(M),\dots,\lambda_n(M)$ of $M$. Since $M+sI_n$ has eigenvalues $\lambda_1(M)+s,\dots,\lambda_n(M)+s$, we obtain \begin{align*} h(M+sI_n)=\det(M+sI_n)=\prod_{i=1}^n (\lambda_i(M)+s). \end{align*} The zeros are $s=-\lambda_i(M)$, all real. Hence $h$ is hyperbolic with respect to $I_n$. The hyperbolicity cone containing $I_n$ consists exactly of matrices whose eigenvalues are all positive, namely the positive definite cone $\mathcal{P}$. Its closure in $V$ consists exactly of matrices whose eigenvalues are all non-negative, namely the positive semidefinite cone $\overline{\mathcal{P}}$. The theorem assumes $A,B,A_3,\dots,A_n$ are positive semidefinite, so each of these matrices lies in $\overline{\mathcal{P}}$, which is the closed cone required for the hyperbolic-polynomial Alexandrov--Fenchel inequality.[/guided]

[step:Apply the Alexandrov--Fenchel inequality for hyperbolic polarizations]We use the following form of the Alexandrov--Fenchel inequality for hyperbolic polynomial polarizations: if $p: W \to \mathbb{R}$ is a homogeneous polynomial of degree $n$ hyperbolic with respect to an element $e \in W$, if $P: W^n \to \mathbb{R}$ is its complete polarization, and if $\overline{\Lambda_e(p)}$ denotes the closure of the hyperbolicity cone containing $e$, then for all $u,v,w_3,\dots,w_n \in \overline{\Lambda_e(p)}$, \begin{align*} P(u,v,w_3,\dots,w_n)^2 \ge P(u,u,w_3,\dots,w_n)P(v,v,w_3,\dots,w_n). \end{align*} Apply this theorem to $W := V$, $p := h$, $e := I_n$, $P := H$, and \begin{align*} u := A, \qquad v := B, \qquad w_i := A_i \quad \text{for } 3 \le i \le n. \end{align*} The preceding step proves that $h$ is hyperbolic with respect to $I_n$ and that $A,B,A_3,\dots,A_n \in \overline{\mathcal{P}} = \overline{\Lambda_{I_n}(h)}$. Therefore the theorem gives \begin{align*} H(A,B,A_3,\dots,A_n)^2 \ge H(A,A,A_3,\dots,A_n)H(B,B,A_3,\dots,A_n). \end{align*} Using the identity $H=D$ from the first step, this becomes \begin{align*} D(A,B,A_3,\dots,A_n)^2 \ge D(A,A,A_3,\dots,A_n)D(B,B,A_3,\dots,A_n), \end{align*} which is the desired inequality.[/step]

[guided]The decisive input is the Alexandrov--Fenchel inequality for complete polarizations of hyperbolic polynomials. We use it in the closed-cone form: if $p: W \to \mathbb{R}$ is a homogeneous polynomial of degree $n$ hyperbolic with respect to $e \in W$, if $P: W^n \to \mathbb{R}$ is the complete polarization of $p$, and if $\overline{\Lambda_e(p)}$ is the closure of the hyperbolicity cone containing $e$, then every choice of elements $u,v,w_3,\dots,w_n \in \overline{\Lambda_e(p)}$ satisfies \begin{align*} P(u,v,w_3,\dots,w_n)^2 \ge P(u,u,w_3,\dots,w_n)P(v,v,w_3,\dots,w_n). \end{align*} The closed-cone formulation is essential here because the theorem assumes the matrices are positive semidefinite, not necessarily positive definite. We now verify the inputs for this theorem. The vector space is $W := V = \operatorname{Sym}_n(\mathbb{R})$. The polynomial is $p := h$, where \begin{align*} h: V &\to \mathbb{R} \\ M &\mapsto \det M, \end{align*} and its complete polarization is $P := H$. The hyperbolicity direction is $e := I_n$. The previous step proved that $h$ is hyperbolic with respect to $I_n$ and that the closure of the hyperbolicity cone is \begin{align*} \overline{\Lambda_{I_n}(h)} = \overline{\mathcal{P}}, \end{align*} the cone of positive semidefinite real symmetric matrices. Define the arguments for the hyperbolic-polynomial inequality by \begin{align*} u := A, \qquad v := B, \qquad w_i := A_i \quad \text{for } 3 \le i \le n. \end{align*} Because $A,B,A_3,\dots,A_n$ are positive semidefinite by hypothesis, all these arguments lie in $\overline{\mathcal{P}} = \overline{\Lambda_{I_n}(h)}$. Thus every hypothesis of the closed-cone Alexandrov--Fenchel inequality is satisfied, and we get \begin{align*} H(A,B,A_3,\dots,A_n)^2 \ge H(A,A,A_3,\dots,A_n)H(B,B,A_3,\dots,A_n). \end{align*} Finally, the first step identified $H$ with the mixed discriminant $D$ on every $n$-tuple of real symmetric matrices. Substituting $H=D$ gives \begin{align*} D(A,B,A_3,\dots,A_n)^2 \ge D(A,A,A_3,\dots,A_n)D(B,B,A_3,\dots,A_n), \end{align*} which is the theorem's asserted inequality.[/guided]