Determinant of sum of positive definite matrices 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.
 A: 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). 
A: 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 $det(A+B)\ge det(A)$.
A: Here is yet another overkill, but hopefully not too bad a way to prove this inequality.
We have the following proof sketch.
$$\begin{eqnarray}
x^T(A+B)x &\ge& x^TAx\quad\forall x\\\\
-x^T(A+B)x &\le& -x^TAx\\\\
\exp(-x^T(A+B)x) &\le& \exp(-x^TAx)\\\\
\int\exp(-x^T(A+B)x)dx &\le& \int\exp(-x^TAx)dx\\\\
\frac{1}{\sqrt{\det(A+B)}} &\le& \frac{1}{\sqrt{\det(A)}}\\\\
\det(A+B) &\ge& \det(A)
\end{eqnarray}
$$
The only fancy thing that happened is in the second last line, where I used the formula for the Gaussian integral (see multivariate section)

Update. To expand upon my comment below, to note that the above idea actually with a little bit more care actually yields a proof of the Minkowski determinant inequality, by equivalently establishing log-concavity of the determinant. The key point to observe is
\begin{eqnarray}
\exp(-x^T((1-\lambda)A+\lambda)x) &=& [\exp(-x^TAx)]^{1-\lambda}[\exp(-x^TBx)]^\lambda\\\\
\int\exp(-x^T((1-\lambda)A+\lambda)x)dx &=& \int [\exp(-x^TAx)]^{1-\lambda}[\exp(-x^TBx)]^\lambda\ dx\\\\
&\stackrel{\text{Hölder}}{\le}& \left(\int\exp(-x^TAx)dx \right)^{1-\lambda}\left(\int \exp(-x^TBx)dx \right)^\lambda.
\end{eqnarray}
Now invoke the Gaussian integral as above to conclude
\begin{equation*}
\det((1-\lambda)A+\lambda B) \ge \det(A)^{1-\lambda}\det(B)^\lambda,
\end{equation*}
from which we can easily conclude $\det(A+B)^{1/n} \ge \det(A)^{1/n}+\det(B)^{1/n}$.
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 ${\rm det}(I+X) \geq (1+d^{1/n})^{n}.$
A: Let me add some more. If $A, B, C$ are positive semidefinite, then $$\det (A+B+C)+\det C\ge \det (A+C)+\det (B+C). \quad  (\star)$$
When $C=0$, this reduces to OP's question. 
A remarkable extension of ($\star$) were recently obtained by V. Paksoy, R. Turkmen, F. Zhang [ Electron. J. Linear Algebra 27 (2014)
332-341], which says that the determinant functional can be replaced by any generalized matrix function. 
A: The determinant of a positive definite matrix $G$ is proportional to $(1/\hbox{Volume}(\mathcal B(G)))^2$
where $\mathcal B(G)$ denotes the unit ball with respect to the metric defined by $G$. If
$A$ and $B$ are positive definite then the volume of $\mathcal B(A+B)$ is smaller than the volume of $\mathcal B(A)$ or $\mathcal B(B)$.
A: Just for those who might want to know, I think the min-max theorem mentioned by @Michael Renardy is Courant-Fischer: Supporse a real symmetric matrix A's eigen values are $\lambda_1 \leq \lambda_2 \leq ... \leq \lambda_n $. then k-th eigen value: $$ \lambda_k = \underset{V_k}{\text{min}} \space max\{x^TAx | x \in V_k, ||x||_2 = 1\}$$
$V_k$ is any k-dimentional subspace of $R^n$, $1 \leq k \leq n$ .
