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 $s(A,B,C)=\frac{\det(A)}{\det(A+C)}$
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}