TITLE: How to prove this matrix inequality $\det(A+B)\ge 2^n\sqrt{\det(A)\det(B)}$
QUESTION [7 upvotes]: Question:
Let $A_{n\times n}$ and $B_{n\times n}$ be positive Hermitian matrices.

Show that
  $$\det(A+B)\ge 2^n\sqrt{\det(A)\det(B)}.$$

I know that
$$\det(A+B)\ge \det(A)+\det(B)$$
But my problem is that I can't,(maybe this is an old reslut,and also I can't find it),
Thank you very much!

REPLY [8 votes]: This seems to be a bit late for an answer, but the following theorem is proven in "Linear Algebra" by Lax in chapter 10.
Let $A$ and $B$ be self-adjoint, positive, $n \times n$ matrices.  Then for all $0<t<1,$
\begin{align}
\det(tA + (1-t)B) \geq (\det A)^{t}(\det B)^{1-t}.
\end{align}
Your answer follows with $t = \frac{1}{2}$.