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 by Marcus and Minc, Dover, 1992, P. 115 and also the previous MO thread).

We have $((A+B)x,x)\ge (Ax,x)$. It then follows from the variational characterization of eigenvalues (min-max theorem) that the eigenvalues of $A+B$ are greater than or equal to those of $A$. This implies $det(A+B)\ge det(A)$.

Yet another way to see this is to note that $A = \overline{Q}^{t}Q$ for some invertible matrix $Q$. Then ${\rm det}(A+B) = |{\rm det}(Q)|^{2}{\rm det}{( I + (\overline{Q}^{-1}})^{t}BQ^{-1})$.` Now $(\overline{Q}^{-1})^{t}BQ^{-1}$ is Hermitian, and positive definite. It suffices to prove that if $X$ is positive definite and Hermitian, then ${\rm det}(I+X) \geq (1 + {\rm det}X)$. We may conjugate $X$ by a unitary matrix $U$ and assume that $X$ is diagonal. Let the eigenvalues of $X$ be $\lambda_{1},\ldots, \lambda_{n}$, (allowing repetitions). Then ${\rm det}(I+X) = \prod_{i=1}^{n}(1 + \lambda_{i}) \geq 1 + \prod_{i=1}^{n} \lambda_{i} = 1 + {\rm det}X.$ Such an argument appears in some proofs by R. Brauer, though I do not know whether it originates with him.

Later edit: Incidentally, I think that with the arithmetic-geometric mean inequality and a slightly more careful analysis, you can see by this approach that for $X$ as above, you do have ${\rm det}(I+X) \geq (1 +({\rm det}X)^{1/n})^{n}$ (a special case of the inequality of Minkowski mentioned in the accepted answer, but enough to prove the general case by an argument similar to that above). For set $d = {\rm det}X$. Let $s_{m}(\lambda_{1},\ldots ,\lambda_{n})$ denote the $m$-th elementary symmetric function evaluated at the eigenvalues. Using the arithmetic-geometric mean inequality yields that $s_{m}(\lambda_{1},\ldots ,\lambda_{n}) \geq \left( \begin{array}{clcr} n\\m \end{array} \right)d^{m/n}$, so we obtain ${\rm det}(I+X) \geq (1+d^{1/n})^{n}.$