Effect of adding a matrix to both numerator and denominator of a ratio between determinants of two matrices 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
$$\frac{\det(A+B)}{\det(A)}$$
which is obviously greater than or equal to 1.
How would this ratio change (increase or decrease) when I add another matrix $C$ inside the determinant on both numerator and determinator, as follows?
$$\frac{\det(A+B+C)}{\det(A+C)}$$
My intuition is that 
$$\frac{\det(A+B+C)}{\det(A+C)} \leq \frac{\det(A+B)}{\det(A)}$$ 
but I haven't been able to prove this. Any insight on this is appreciated!
 A: It's true. Let $S=A+C$. Then
\begin{aligned}
\frac{\det(A+B)}{\det(A)}
&=\det(I+A^{-1/2}BA^{-1/2})\\
&=\det(I+B^{1/2}A^{-1}B^{1/2})\quad(\text{because} \det(I+XY)=\det(I+YX))\\
&\ge\det(I+B^{1/2}S^{-1}B^{1/2})\quad(\text{because} B^{1/2}A^{-1}B^{1/2}\succeq B^{1/2}S^{-1}B^{1/2}\succeq0)\\
&=\det(I+S^{-1/2}BP^{-1/2})\\
&=\frac{\det(S+B)}{\det(S)}\\
&=\frac{\det(A+B+C)}{\det(A+C)}.
\end{aligned}
A: The below is missing a step and possibly incorrect.

Your intuition is correct. Throughout my proof below, positive (semi)-definite matrices are necessarily symmetric.
Denote $B' = A^{-1/2}BA^{-1/2}$ and $C' = A^{-1/2}CA^{-1/2}$. The statement that you are trying to prove can be rewritten as
$$
\frac{\det(I + B' + C')}{\det(I + C')} \leq \frac{\det(I + B')}{\det(I)} \implies
\det(I + B' + C') \leq \det(I + B')\det(I + C').
$$
We use $\leq$ to denote the Loewner ordering. That is, $A \leq B$ iff $B - A$ is positive semidefinite.  Now, we note that
$$
(I + C')^{1/2}(I + B')(I + C')^{1/2} = 
(I + C') + (I + C')^{1/2}(B')(I + C')^{1/2} \geq
I + C' + B'.
$$
It follows that 
$$
\det(I + B')\det(I + C') = \\
\det[(I + C')^{1/2}(I + B)(I + C')^{1/2}] \geq \\
\det[I + C' + B'],
$$
which is what we wanted.
