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>



A curious determinantal inequality
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.


