By a limiting argument we may assume that $C := A+B$ is invertible.  If we write

$$ D := C^{-1/4} A^{1/2} C^{-1/4} $$
and
$$ E := C^{-1/4} B^{1/2} C^{-1/4} $$

then $D,E$ are positive semi-definite with $D^2+E^2=1$, so in particular $D,E$ commute.  The inequality can now be written in terms of $C,D,E$ as
$$ \det( C^{1/4} D C^{3/2} D C^{1/4} + C^{1/4} E C^{3/2} E C^{1/4} ) \geq \det( C^2 )$$
which on multiplying on left and right by $C^{-1/4}$ and setting $F := C^{3/2}$ becomes
$$ \det( D F D + E F E ) \geq \det( F ).$$
Now observe that the matrix
$$ \begin{pmatrix} D & E \\ -E & D \end{pmatrix} \begin{pmatrix} F & 0 \\ 0 & F \end{pmatrix} \begin{pmatrix} D & -E \\ E & D \end{pmatrix} = \begin{pmatrix} DFD + EFE & EFD-DFE \\ DFE-EFD & DFD+EFE \end{pmatrix}$$
is positive semi-definite and has determinant $\det(F)^2$ (the first and last matrices on the LHS are orthogonal).  Passing to the block-diagonal matrix
$$ \begin{pmatrix} DFD + EFE & 0 \\ 0 & DFD+EFE \end{pmatrix},$$
which is still positive semi-definite, the eigenvalues here are majorized by the previous matrix (by the [Schur-Horn theorem][1]), and so (by the [Schur concavity][2] of the product function $(\lambda_1,\dots,\lambda_n) \mapsto \lambda_1 \dots \lambda_n$), the determinant of the latter matrix must be at least as large as the determinant of the former.  (This inequality can also be established using [Schur complements][3].)  Thus
$$ \det( DFD + EFE )^2 \geq \det(F)^2 $$
and the claim follows.


  [1]: https://en.wikipedia.org/wiki/Schur%E2%80%93Horn_theorem
  [2]: https://en.wikipedia.org/wiki/Schur-convex_function
  [3]: https://en.wikipedia.org/wiki/Schur_complement