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.



On a determinant inequality of positive definite matrices
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.


