# How to prove this matrix inequality $\det(A+B)\ge 2^n\sqrt{\det(A)\det(B)}$

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!

This is a corollary of Minkowski's Determinant Theorem: $\det(A+B)^\frac{1}{n}\geq \det(A)^\frac{1}{n}+\det(B)^\frac{1}{n}.$ Apply AM-GM inequality to the right-hand side.
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 $t = \frac{1}{2}$.