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)][1]

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**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 $\det(A+B)^{1/n} \ge \det(A)^{1/n}+\det(B)^{1/n}$.
  [1]: http://en.wikipedia.org/wiki/Gaussian_integral