Assume that $B$ and $A$ are two positive definite matrices. Take $B^*$ a block diagonal matrix with block $B_{11}$ and $B_{22}$ of $B$. This means the following: $$ B=\left[\begin{array}{ll} B_{11}& B_{12}\\ B_{21}&B_{22} \end{array}\right], B^*=\left[\begin{array}{ll} B_{11}& 0\\ 0 &B_{22} \end{array}\right] $$ It can be seen that $2B^*\succ B$ and therefore: $$ \det(I+2AB^*)\geq \det(I+AB). $$ So I managed to prove that, however I have another conjecture that the following is true: $$ 2\det(I+AB^*)\geq \det(I+AB). $$ My question is about the proof of this conjecture.
A quick counterexample to your conjecture is \begin{equation*} A = \begin{pmatrix} 13 & 3 & 13 & 5\\ 3 & 4 & 3 & 4\\ 13 & 3 & 13 & 5\\ 5 & 4 & 5 & 10\\ \end{pmatrix},\quad B = \begin{pmatrix} 15 & 3 & 14 & 1\\ 3 & 18 & 6 & 3\\ 14 & 6 & 19 & 1\\ 1 & 3 & 1 & 1 \end{pmatrix} \end{equation*} For this choice, $2\det(I+AB^*) \approx 6\times 10^4$, while $\det(I+AB)\approx 8\times 10^4$. The $A$ above is semidefinite; if that displeases you, then just add a trivial $\epsilon I$ to it, the counterexample will still hold. 


From the comments, a true version is $$2^n\det(I+AB^*)\geq \det(I+AB), \quad (1)$$ where $n$ is the dimension of $A, B$. The proof is easy, as $2^n\det(I+AB^*)\geq\det(I+2AB^*)$. Can $2$ be replaced by a smaller constant in (1)? 

