TITLE: A determinant inequality
QUESTION [17 upvotes]: Notation: Suppose $\mathbf{A}$ and $\mathbf{B}$ are positive definite matrices in $\mathbb{R}^{n\times n}$ such that $\mathbf{A} \succeq \mathbf{B}$  (Loewner order). Let $\mathcal{S}(n,k)$ be the set of all $k$-subsets of $\{1,2,\dots,n\}$. For any $\mathcal{Q} \subset [n] \triangleq \{1,2,...,n\}$, $\mathbf{M}_\mathcal{Q}$ is the matrix obtained by deleting the rows and columns of matrix $\mathbf{M}$ whose indices are in $\mathcal{Q}$. Similarly, $\mathbf{x}_\mathcal{Q}$ is the vector obtained by deleting the rows of vector $\mathbf{x}$ whose indices are in $\mathcal{Q}$.
Conjecture: For any such $\mathbf{A}$, $\mathbf{B}$ and $\mathbf{x} \in \mathbb{R}^n$ and for all $k \in [n-1]$:
$$
\frac{\sum_{\mathcal{Q} \in \mathcal{S}(n,k)} \det(\mathbf{A}_\mathcal{Q} + \mathbf{x}_\mathcal{Q}\mathbf{x}_\mathcal{Q}^\top)}{\sum_{\mathcal{Q} \in \mathcal{S}(n,k)} \det(\mathbf{A}_\mathcal{Q})}
\leq
\frac{\sum_{\mathcal{Q} \in \mathcal{S}(n,k)} \det(\mathbf{B}_\mathcal{Q} + \mathbf{x}_\mathcal{Q}\mathbf{x}_\mathcal{Q}^\top)}{\sum_{\mathcal{Q} \in \mathcal{S}(n,k)} \det(\mathbf{B}_\mathcal{Q})}
\tag{1}
$$
Progress so far:

We have $\det(\mathbf{A}_\mathcal{Q}+\mathbf{x}_\mathcal{Q}\mathbf{x}_\mathcal{Q}^\top) = \det(\mathbf{A}_\mathcal{Q})\,(1+\mathbf{x}_\mathcal{Q}^\top \mathbf{A}_\mathcal{Q}^{-1}\mathbf{x}_\mathcal{Q})$. Therefore, (1) can be rewritten as (2):
$$
\tag{2}
\sum_{\mathcal{Q} \in \mathcal{S}(n,k)}\frac{ \det(\mathbf{A}_\mathcal{Q})\,}{\sum_{\mathcal{Q} \in \mathcal{S}(n,k)} \det(\mathbf{A}_\mathcal{Q})} \,
\mathbf{x}_\mathcal{Q}^\top \mathbf{A}_\mathcal{Q}^{-1}\mathbf{x}_\mathcal{Q}
\leq
\sum_{\mathcal{Q} \in \mathcal{S}(n,k)}\frac{ \det(\mathbf{B}_\mathcal{Q})\,}{\sum_{\mathcal{Q} \in \mathcal{S}(n,k)} \det(\mathbf{B}_\mathcal{Q})} \,
\mathbf{x}_\mathcal{Q}^\top \mathbf{B}_\mathcal{Q}^{-1}\mathbf{x}_\mathcal{Q}
$$
Lemmas: It is easy to show that for any $\mathcal{Q} \subset [n]$:

(2.1) $\quad \mathbf{A} \succeq \mathbf{B} \Rightarrow \mathbf{A}_\mathcal{Q} \succeq \mathbf{B}_\mathcal{Q}$ 
(2.2) $\quad \det(\mathbf{A}_\mathcal{Q}) \geq \det(\mathbf{B}_\mathcal{Q})$
(2.3) $\quad \mathbf{A}^{-1}_\mathcal{Q} \preceq \mathbf{B}_\mathcal{Q}^{-1} \Rightarrow \mathbf{x}_\mathcal{Q}^\top \mathbf{A}_\mathcal{Q}^{-1}\mathbf{x}_\mathcal{Q} \leq \mathbf{x}_\mathcal{Q}^\top \mathbf{B}_\mathcal{Q}^{-1}\mathbf{x}_\mathcal{Q}$
(2.4) $\quad \det(\mathbf{A}_\mathcal{Q})\,\mathbf{x}_\mathcal{Q}^\top \mathbf{A}_\mathcal{Q}^{-1}\mathbf{x}_\mathcal{Q} \geq \det(\mathbf{B}_\mathcal{Q})\,\mathbf{x}_\mathcal{Q}^\top \mathbf{B}_\mathcal{Q}^{-1}\mathbf{x}_\mathcal{Q} $

Through recursion, it suffices to show that (1) or (2) hold for the special case of $\mathbf{A} = \mathbf{B} + \mathbf{p}\mathbf{p}^\top$ for some $\mathbf{p} \in \mathbb{R}^n$.
A very special case is posted here ($\mathbf{B} = \mathbf{I}$, $\mathbf{A} = \mathbf{I} + \mathbf{p}\mathbf{p}^\top$).
I haven't encountered any counterexample after running "many" simulations (I know this is not a strong argument).

Update: Let me explain the motivation as requested.
Motivation:
Suppose $\{\mathbf{a}_i\}_{i=1}^{m}$ are some vectors in $\mathbb{R}^n$ ($m \geq n$) and $\mathbf{A}_0 \succ \mathbf{0}$ is a given matrix in $\mathbb{R}^{n \times n}$. Now consider,
$$
\begin{align}
f_k : 2^{[m]} &\to \mathbb{R}, \\
    \mathcal{S} & \mapsto c_k(\mathbf{A}_0 + \sum_{i \in \mathcal{S}}\mathbf{a}_i\mathbf{a}_i^\top)
\end{align}
$$
where $c_k(\mathbf{A})$ is the coefficient of $x^k$ in the characteristic polynomial of $\mathbf{A}$, i.e., $\det(x\mathbf{I} - \mathbf{A})$. 
Conjecture: $f_k$ is monotone log-submodular (multiplicative submodular) for all $k \in \{0,1,\dots,n\}$.
I have proved the monotonicity (maybe for $|f_k|$).
Special Cases: This holds for $k=n-1$ (trace), $k=0$ (determinant is log-submodular) and $k=n$ (constant, $f_n(\mathcal{S}) = 1$).
Now (1) emerges from the proof of log-submodularity (multiplicative submodularity) of $f_k$ after expressing $c_k$ as the sum of determinants of principal minors.
Applications:

I came across this when working on graph Laplacian matrices. $f_k$ for Laplacian matrices (and $\mathbf{a}_i = \mathbf{e}_s - \mathbf{e}_r$ where $\{\mathbf{e}_s\}_{s=1}^n$ is the standard basis) is related to the weighted number of spanning trees. I recently showed that the weighted number of spanning trees is a monotone log-submodular function of the edge set (see a draft here). Other coefficients can be also be nicely related to the weighted number of spanning trees as shown by Alexander Kelmans (as a generalization of Kirchhoff's matrix tree theorem). For Laplacian matrices, (2) has a beautiful interpretation in terms of the expected value of the effective resistance ("distance") between two vertices after performing some random operations on the graph.
(1) and (2) also arise in $k$-DPPs (determinantal point process).

REPLY [6 votes]: As suspected, the desired inequality actually holds for all hyperbolic polynomials; the inequality in the OP follows as corollary (Corollary 1) to Theorem 2 (which seems to be new). 
We will need the following remarkable theorem to obtain our result.


Theorem 1 (Bauschke, Güler, Lewis, Sendov, 2001) Let $p$ be a homogenous hyperbolic polynomial; let $v$ be a vector in the strict interior of the hyperbolicity cone $\Lambda_{++}$ of $p$. Then, 
    \begin{equation*}
   g(x) := \frac{p(x)}{Dp(x)[v]}
\end{equation*}
    is concave on $\Lambda_{++}$.


This theorem helps prove the more general inequality (also conjectured by Denis Serre above).


Theorem 2. Let $p$ be a homogenous hyperbolic polynomial with hyperbolicity cone $\Lambda_{++}$. Let $a, b, c \in \Lambda_{++}$. Then, $p$ satisfies the (conic log-submodularity) inequality:
      \begin{equation*}
    \tag{0}
    p(a)p(a+b+c) \le p(a+b)p(a+c).
  \end{equation*}


Proof. 
Let $c \in \Lambda_{++}$ be arbitrary. Consider the function $f(a) := \frac{p(a+c)}{p(a)}$. Inequality (0) amounts to showing that $f(a)$ is monotonically decreasing on the cone $\Lambda_{++}$. Equivalently, we consider $\log f$ and show that its derivative is negative in the direction $v$. That is, for an arbitrary direction vector $v\in \Lambda_{++}$, we show that
  \begin{equation}
    \tag{1}
    \frac{Dp(a+c)[v]}{p(a+c)} - \frac{Dp(a)[v]}{p(a)} \le 0\quad\Longleftrightarrow\quad 
    \frac{p(a+c)}{Dp(a+c)[v]} \ge \frac{p(a)}{Dp(a)[v]}.
  \end{equation}
  But from Theorem 1, we know that $\frac{p(x)}{Dp(x)[v]}$ is concave. Moreover, since $p$ is homogenous, from its concavity we obtain its superadditivity
  \begin{equation*}
    \frac{p(a+c)}{Dp(a+c)[v]} \ge \frac{p(a)}{Dp(a)[v]} + \frac{p(c)}{Dp(c)[v]},
  \end{equation*}
  which is stronger than the desired monotonicity inequality (1) (since all terms are nonnegative).


Corollary 1. Let $E_k(A) = e_k \circ \lambda(A)$ denote the $k$-th elementary symmetric polynomial of a positive definite matrix $A$. Then for any positive definite $A, B, C$ we have
      \begin{equation*}
    E_k(A)E_k(A+B+C) \le E_k(A+B)E_k(A+C).
  \end{equation*}
      This log-submodularity, immediately implies the log-submodularity of $f_k(S) := E_k(A+\sum\nolimits_{i\in S}v_iv_i^T)$.