Say $A$ and $B$ are symmetric, positive definite matrices. I've proved that 


$$\det(A+B) \ge \det(A) + \det(B)$$ 


in the case that $A$ and $B$ are two dimensional. Is this true in general for $n$-dimensional matrices? Is the following even true?


$$\det(A+B) \ge \det(A)$$ 


This would also be enough. Thanks.



Determinant of sum of positive definite matrices
The inequality 
$$\det(A+B)\geq \det A +\det B$$
is implied by the Minkowski determinant theorem
$$(\det(A+B))^{1/n}\geq (\det A)^{1/n}+(\det B)^{1/n}$$
which holds true for any non-negative $n\times n$ Hermitian matrices $A$ and $B$. The latter inequality is equivalent to the fact that the function $A\mapsto(\det A )^{1/n}$ is concave on the set of $n\times n$ non-negative Hermitian matrices (see e.g., [*A Survey of Matrix Theory and Matrix Inequalities*](http://books.google.com/books?id=hLHKwSNqLOcC&printsec=frontcover&dq=marcus++matrix&hl=en&ei=6hXVTfi-M4Gg-wag_5TvCw&sa=X&oi=book_result&ct=result&resnum=1&ved=0CDYQ6AEwAA#v=onepage&q&f=false) by Marcus and Minc, Dover, 1992, P. 115 and also the previous [MO thread](https://mathoverflow.net/questions/42594/concavity-of-det1-n-over-hpd-n)). 


