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), and so (by the Schur concavity 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.) Thus $$ \det( DFD + EFE )^2 \geq \det(F)^2 $$ and the claim follows.