Reverse Minkowski (and related) Determinant Inequalities

Question

For positive semidefinite matrices $A,B,C \in \mathbb{R}^{n\times n}$, the following inequalities are well known:

$$(\det(A+B))^{1/n} \geq (\det A)^{1/n} + (\det B)^{1/n} $$

and

$$\det(A+B+C) + \det(C) \geq \det(A+C) + \det(B+C).$$

The first is, of course, Minkowski's determinant inequality. I'm not sure whether the second has a name, but it can be found, e.g., here and here.

I would like to know whether anyone knows of any reverse forms of these inequalities. To be more specific, I note that the Minkowski determinant inequality can be viewed as a consequence of the Entropy Power Inequality applied to Gaussian random vectors with covariance matrices $A$ and $B$. Applying similar reasoning to the entropy power inequality appearing in Theorem 4 of this paper, we may conclude that the following reverse counterparts to the above inequalities hold:

$$(\det(A+B))^{1/n} \leq (\det A)^{1/n}\left(\frac{\frac{1}{n}\operatorname{Tr}(B^{-1})}{(\det B^{-1})^{1/n}} \right) + (\det B)^{1/n}\left(\frac{\frac{1}{n}\operatorname{Tr}(A^{-1})}{(\det A^{-1})^{1/n}} \right),~~~~~(\star) $$

and

$$(\det(A+B+C) \det(C) )^{1/n} + (\det(A) \det(B) )^{1/n} \leq (\det(A+C) \det(B+C) )^{1/n} .~~~(\star\star)$$

The first may be obtained from the second by taking $C = \varepsilon I$ and letting $\varepsilon \downarrow 0$. I view $(\star)$ as a reverse counterpart to Minkowski's determinant inequality, since the ratio

$$\frac{\frac{1}{n}\operatorname{Tr}(B^{-1})}{(\det B^{-1})^{1/n}}\geq 1$$

by the AM-GM inequality applied to the eigenvalues of $B^{-1}$ (and similarly for the ratio involving $A^{-1}$), with near equality if the matrix $B$ has eigenvalues all roughly the same.

Has anyone seen these inequalities before, or does anyone have a direct proof of $(\star\star)$?

Answer 1

Inequality ($\star\star$) essentially follows from the original Minkowski plus an implication of Lidkskii's inequality (Fiedler's inequality, noted below). $\newcommand{\da}{\downarrow} \newcommand{\ua}{\uparrow}$

Assume $C$ is strictly positive definite (otherwise, $\det C=0$ rendering ($\star\star$) trivial), and let $C^{1/2}$ denote the positive square root of $C$. Renaming the matrices $A \leftarrow C^{-1/2}A C^{-1/2}$ and $B \leftarrow C^{-1/2}B C^{-1/2}$, the original inequality reduces to showing that \begin{equation*} \det(A+B+I)^{1/n} + [\det A\det B]^{1/n} \le [\det(I+A)\det(I+B)]^{1/n}. \end{equation*}

Let $a^\da$ denote the eigenvalues of $A$ in decreasing order; similarly, define $b^\ua$ for $B$. Then, we have \begin{equation*} \det(I+A)\det(I+B) = \prod_i(1+a_i^\da)\prod_i(1+b_i^\ua) = \prod_i(1+a_i^\da + b_i^\ua + a_i^\da b_i^\ua), \end{equation*} whence by the Minkowski inequality we get \begin{equation*} [\det(I+A)\det(I+B)]^{1/n} \ge \prod_i(1+a_i^\da+b_i^\ua)^{1/n} + \prod_i(a_i^\da b_i^\ua)^{1/n}. \end{equation*} Now from Fiedler's (1971) determinantal inequality we know that \begin{equation*} \prod_i(1+a_i^\da+b_i^\ua) \ge \det(I+A+B), \end{equation*} while clearly $\prod_i(a_i^\da b_i^\ua) = \det(A)\det(B)$. Thus, the inequality ($\star\star$) follows.