TITLE: A curious determinantal inequality
QUESTION [35 upvotes]: In my study, I come across the following curious inequality, which I do not know a proof yet (so I am asking it here). 
Let $A, B$ be $n\times n$ (Hermitian) positive definite matrices.  It is very likely true that
$$\det \left(A^{\frac{1}{2}}(A+B)A^{\frac{1}{2}}+B^{\frac{1}{2}}(A+B)B^{\frac{1}{2}}\right) \ge \det(A+B)^2. $$
Here $A^{\frac{1}{2}}$ is the unique positive definite square root of $A$.
I am able to confirm the $3\times 3$ case.
Comments: Only recently did I notice that the majorization $\lambda\left(A^{\frac{1}{2}}(A+B)A^{\frac{1}{2}}+B^{\frac{1}{2}}(A+B)B^{\frac{1}{2}}\right) \prec \lambda(A+B)^2$ follows immediately by THEOREM 2
of [R.B. Bapat, V.S. Sunder, On majorization and Schur products, Linear Algebra Appl. 72 (1985)
107–117.] http://www.sciencedirect.com/science/article/pii/0024379585901478

REPLY [5 votes]: Here is a complementary approach without using majorization. The answer is partial because it has an open "TODO". I am writing it down here already in case someone wishes to complete the argument.

Let $A, B, X, Y > 0$. It is easy to show using Schur complements that
\begin{equation*}
\tag{$*$}
 AX^{-1}A + BY^{-1}B \ge (A+B)(X+Y)^{-1}(A+B).
\end{equation*}
From $(*)$ it follows that $\det(AX^{-1}A + BY^{-1}B)\det(X+Y)\ge \det(A+B)^2$. 
Let $C=A^{1/2}(A+B)A^{1/2}$ and $D=B^{1/2}(A+B)B^{1/2}$. If we can find (TODO) $X$ and $Y$ such that 
\begin{equation*}
  X \gets A(X+Y)^{1/2}C^{-1}(X+Y)^{1/2}A,\quad Y \gets B(X+Y)^{1/2}D^{-1}(X+Y)^{1/2}B,
\end{equation*}
then we will obtain $$(X+Y)^{1/2}(AX^{-1}A + BY^{-1}B)(X+Y)^{1/2} = C+D = A^{1/2}(A+B)A^{1/2} + B^{1/2}(A+B)B^{1/2}.$$ Combining this identity with the above inequality immediately implies the desired inequality.
Notice that in particular, if $A$ and $B$ commute, then $X=A(A+B)^{-1}$ and $Y=B(A+B)^{-1}$ is a solution.