<s>[INST] Lower bound $\frac{\det(A+B+C)}{\det(A+C)}$ in terms of $\frac{\det(A+B)}{\det(A)}$ Assume matrix $A$ is symmetric and positive definite, and matrices B and C are symmetric and positive semi-definite. Originally I have ratio between determinants:
$$\frac{\det(A+B)}{\det(A)}$$
By adding another matrix C inside the determinant on both numerator and determinator, we obtain a new ratio between determinants:
$$\frac{\det(A+B+C)}{\det(A+C)}$$
From this question, the new ratio is proved to be upper bounded by the original one:
$$\frac{\det(A+B+C)}{\det(A+C)} \leq \frac{\det(A+B)}{\det(A)}$$
Now the question is, can we prove a lower bound in terms of the original ratio as well. For example:
$$ s(A,B,C) \frac{\det(A+B)}{\det(A)} \leq \frac{\det(A+B+C)}{\det(A+C)} \leq \frac{\det(A+B)}{\det(A)}$$
where $s(A,B,C) \in [0,1]$ is some scalar value that may depend on matrices $A, B, C$. Intuitively, if $C$ is a zero matrix, then $s(A,B,C)$ should be equal to 1, making the lower bound equal to upper bound.
My initial attemp is shown below:
\begin{aligned}
\frac{\det(A+B)}{\det(A)} &=\det(I+A^{-1}B)\\
&= \det(I+(A+C)^{-1}(A+C)A^{-1}B)\\
&= \det(I+(A+C)A^{-1}B(A+C)^{-1}) \quad (\text{[Weinstein–Aronszajn identity][2]})
\end{aligned}
I am wondering if the following inequality holds:
$$\det(I+(A+C)A^{-1}B(A+C)^{-1}) \leq \det((A+C)A^{-1})\det(I+B(A+C)^{-1})$$
If it holds, then we can show that:
$$\frac{\det(A+B)}{\det(A)}=\det(I+(A+C)A^{-1}B(A+C)^{-1}) \leq \frac{\det(A+C)}{\det(A)}\frac{\det(A+B+C)}{\det(A+C)}$$
Therefore,
$$\frac{\det(A)}{\det(A+C)}\frac{\det(A+B)}{\det(A)} \leq \frac{\det(A+B+C)}{\det(A+C)}$$
So $s(A,B,C)=\frac{\det(A)}{\det(A+C)}$
 [/INST] Turns out it is very easy to show the inequality holds.
\begin{aligned}
\frac{\det(A+B)}{\det(A)} &= \frac{det(A+C)}{det(A)}\frac{det(A+B)}{det(A+C)}\\
& \leq \frac{det(A+C)}{det(A)}\frac{det(A+B+C)}{det(A+C)}\\
\end{aligned}
</s>