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Let $a_1,a_2,\cdots,a_n$ be positive real numbers such that $a_1+a_2+\cdots+a_n=1$ and let $A_1,A_2,\cdots,A_n$ be positive definite complex matrices. Prove that the following inequality holds: $$\det(a_1A_1+\cdots+a_nA_n) \ge (\det A_1)^{a_1}\cdots(\det A_n)^{a_n}$$

My progress: I tried to use weighted AM-GM inequality and tried to prove that(is this even true?) $$\det(a_1A_1+\cdots+a_nA_n) \ge a_1\det A_1+\cdots+a_n\det A_n$$ Now this looks little bit like Jensen's inequality but I don't know how to proceed. Does anyone know of a solution that doesn't involve use of any powerful matrix inequality like Minkowski?

Takamoto Yuji
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  • Yes, that second inequality is true. Start with $n=2$ and use a square root of one of the two matrices $A+B=A^{-1/2}(I+A^{-1/2}BA^{1/2})A^{1/2}$. Take determinants. $\det(A+B)=\det(A)\det(I+A^{-1/2}BA^{1/2})$ Here the matrix in the second factor is also positive def. So, this is a product of the eigenvalues which are those of $A^{-1/2}BA^{1/2}$ increased by $1$. That is larger than $1+\det(A^{-1/2}BA^{1/2})$. – NDB Jul 12 '23 at 16:28
  • Oh, I meant with the $a_i$'s inside the $\det$. – NDB Jul 12 '23 at 16:33
  • @NDB could you explain why the second factor is also positive definite? – Takamoto Yuji Jul 12 '23 at 16:42
  • Also, it looks to me as if the coefficients inside will come out with the size of the matrix as the exponent? – Takamoto Yuji Jul 12 '23 at 16:53
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    It suffices to prove the inequality for $n=2$. Once this is proved, $f=\log\det$ is shown to be convex on the set of all positive definite matrices. Hence $f(\sum_ia_iA_i)\ge\sum_ia_if(A_i)$ for any convex combination $\sum_ia_iA_i$. – user1551 Jul 12 '23 at 17:10
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    Compare https://math.stackexchange.com/q/1192329/42969: $f(A) = \log( \det(A))$ is a concave function, and that implies your first inequality. – Martin R Jul 13 '23 at 07:59

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