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.

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