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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!

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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.

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  • $\begingroup$ this can AM-GM inequality? Thank you $\endgroup$ – china math Dec 5 '13 at 11:02
  • $\begingroup$ Sorry, I am not sure what you are asking? $\endgroup$ – Casteels Dec 5 '13 at 11:03
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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 $t = \frac{1}{2}$.

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  • $\begingroup$ It's never too late for a good answer! $\endgroup$ – Casteels Apr 15 '14 at 7:42
  • $\begingroup$ Can you send me pdf document or links ? Thanks you very much! $\endgroup$ – Road Human Apr 6 '15 at 16:07

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