Is $|\bf A \bf B^{-1} + \bf I| \geq |\bf A (\bf B + \bf C)^{-1} + \bf I|$ true? $\bf A$, $\bf B$, and $\bf C$ are Hermitian matrices of the same order.
They satisfy $\bf A \succeq \bf 0$, $\bf B \succ \bf 0$, and $\bf C \succeq \bf 0$.
Then, is
\begin{equation}
|{\bf A} {\bf B}^{-1} + {\bf I}| \geq |{\bf A} ({\bf B} + {\bf C})^{-1} + {\bf I}| \tag{1}
\end{equation}
true?
For the simple scalar case, it is obvious that
\begin{equation}
\frac{a}{b} + 1 \geq \frac{a}{b+c} + 1. \tag{2}
\end{equation}
I verified (1) in Matlab and feeled that it is true. But I could not prove it. I would be very appreciated if someone could provide some relevant hints or references to prove (1).
 A: Determinant inequality and positive definite matrix answers your question. Equivalently we should show $|B^{-1}+A^{-1}|\ge |(B+C)^{-1}+A^{-1}|$(Since we can approximate a semi-positive definite matrix by a strictly positive definite one, we can thus without loss of generality assume $A$ is also strictly positive definite so $A^{-1}$ exists). Then from this post, $B^{-1}-(B+C)^{-1}$ should be semi-positive definite. Then we regard $${B^{ - 1}} + {A^{ - 1}} = ({B^{ - 1}} - {(B + C)^{ - 1}}) + {(B + C)^{ - 1}} + {A^{ - 1}}$$ and applying the second conclusion in this post is OK.
A: We may use the continuity argument.
Let $\epsilon \ge 0$. Let
$$f(\epsilon) := |(A + \epsilon I)B^{-1} + I| - |(A + \epsilon I)(B + C)^{-1} + I|.$$
For any $\epsilon > 0$, we have
$$f(\epsilon) = |A + \epsilon I|
\cdot |B^{-1} + (A + \epsilon I)^{-1}|
- |A + \epsilon I| \cdot |(B + C)^{-1} + (A + \epsilon I)^{-1}| \ge 0$$
where we have used
i) $|A + \epsilon I| > 0$ and ii)
$|B^{-1} + (A + \epsilon I)^{-1}|
\ge |(B + C)^{-1} + (A + \epsilon I)^{-1}|$.
(See the remarks at the end.)
Note that $f(\epsilon)$ is a polynomial. Thus, $f(\epsilon)$ is continuous on $[0, \infty)$. Thus, $f(0) \ge 0$.
We are done.

Remarks:
We have
$$(B + C)\big(B^{-1} - (B + C)^{-1}\big)(B + C) 
= C + CB^{-1}C$$
which results in
$$B^{-1} - (B + C)^{-1} = (B + C)^{-1}(C + CB^{-1}C)(B + C)^{-1} \succeq 0.$$
Let $X = B^{-1} + (A + \epsilon I)^{-1}$
and $Y = (B + C)^{-1} + (A + \epsilon I)^{-1}$. We have $X \succeq Y \succ 0$.
We have
$$|X| = |Y^{1/2}(I + Y^{-1/2}(X - Y)Y^{-1/2})Y^{1/2}| = |Y|\cdot |I + Y^{-1/2}(X - Y)Y^{-1/2}| \ge |Y|$$
where we have used $Y^{-1/2}(X - Y)Y^{-1/2} \succeq 0$
and thus $|I + Y^{-1/2}(X - Y)Y^{-1/2}| \ge 1$.
