Let $C := A^{1/2} (A+B) A^{1/2} + B^{1/2} (A+B) B^{1/2}$; this is a positive semi-definite matrix with the same trace as $(A+B)^2$. We show that the eigenvalues of $C$ are majorised by the eigenvalues of $(A+B)^2$, that is to say that the sum of the top $k$ eigenvalues of $C$ is at most the sum of the top $k$ eigenvalues of $(A+B)^2$ for any $k$. By the Schur concavity of $(\lambda_1,\dots,\lambda_n) \mapsto \lambda_1 \dots \lambda_n$, this gives the claimed determinantal inequality.

The sum of the top $k$ eigenvalues of $C$ can be written as
$$ \hbox{tr}( C P_V )$$
where $V$ is the $k$-dimensional space spanned by the top $k$ eigenvectors of $C$. This can be rearranged as
$$ \hbox{tr}( (A+B) (A^{1/2} P_V A^{1/2} + B^{1/2} P_V B^{1/2}) ). \quad\quad (*)$$

We can conjugate $A+B$ to be a diagonal matrix $\hbox{diag}(\lambda_1,\dots,\lambda_n)$ with $\lambda_1 \geq \dots \geq \lambda_n \geq 0$. In particular we have $A+B \leq \lambda_k I + D$ in the sense of positive definite matrices, where $D := \hbox{diag}(\lambda_1-\lambda_k, \dots, \lambda_{k-1}-\lambda_k, 0, \dots, 0)$. Using the fact that $\hbox{tr}(XZ) \leq \hbox{tr}(YZ)$ whenever $X,Y,Z$ are positive semi-definite with $X \leq Y$, we can bound (*) by

$$ \hbox{tr}( (\lambda_k I + D) (A^{1/2} P_V A^{1/2} + B^{1/2} P_V B^{1/2}) )$$

which rearranges as

$$ \lambda_k \hbox{tr}( (A+B) P_V ) + \hbox{tr}( P_V (A^{1/2} D A^{1/2} + B^{1/2} D B^{1/2}) ).$$
Using $A+B \leq \lambda_k I + D$ for the first term and $P_V \leq I$ for the second term, this is bounded by
$$ \lambda_k^2 \hbox{tr}( P_V ) + \lambda_k \hbox{tr}( D P_V ) + \hbox{tr}( A^{1/2} D A^{1/2} + B^{1/2} D B^{1/2} ).$$
For the second term we use $P_V \leq 1$, and for the third term we use the cyclic property of trace to bound by
$$ \lambda_k^2 \hbox{tr}( P_V ) + \lambda_k \hbox{tr}( D ) + \hbox{tr}( (A+B) D ).$$
For the first term we write $\hbox{tr}(P_V) = k = \hbox{tr}(P_W)$, where $W$ is the span of the first $k$ basis vectors $e_1,\dots,e_k$. For the third term we use $A+B \leq \lambda_k I + D$ to bound the above by
$$ \lambda_k^2 \hbox{tr}( P_W ) + \lambda_k \hbox{tr}( D ) + \hbox{tr}( (\lambda_k I + D) D ).$$
Since $D = P_W D P_W$, we can collect terms to obtain
$$ \hbox{tr}( P_W (\lambda_k I + D)^2 P_W ).$$
But by construction, $P_W (\lambda_k I + D)^2 P_W = \hbox{diag}( \lambda_1^2, \dots, \lambda_k^2, 0, \dots, 0 )$, so we have bounded (*) by the sum of the top $k$ eigenvalues of $(A+B)^2$, as required.