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If $A, B$ and $C$ are $n\times n$ positive semidefinite matrices. How to show that $$\det(A + B) + \det(A + C)\le \det A + \det(A + B + C)?$$

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Using the fact that $A$ is symmetric positive definite, we can find $R$ symmetric such that $R^2=A$. Therefore we are reduced to the case $A=I$. We have to show that $$\sum_{J\subset [n]}(\det B^{(J)}+\det C^{(J)})\leq 1+\sum_{J\subset [n]}\det(B+C)^{(J)},$$ where $A^{(J)}$ means the matrix $A$ without the lines and columns of index in $J$. Indeed, we can show that for a matrix $A$, $\det(I+A)$ is a sum of $2^n$ determinants, when columns are from $I$ or $A$. Then for each determinant of this type, expand with respect to the columns which belong to $I$.

This topic show that $\det(M_1+M_2)\geq \det M_1+\det M_2$ where $M_1$ and $M_2$ are two positive define matrices gives the conclusion.

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  • $\begingroup$ So I know it is equivalent to $\det(I+B)+\det(I+C)\le I+ \det(I+B+C)$, but I don't understand the remaining part of your argument. $\endgroup$
    – Fischer
    Sep 23 '12 at 15:42
  • $\begingroup$ I have another approach to prove the determinantal inequality. $\endgroup$
    – Fischer
    Sep 23 '12 at 15:58
  • $\begingroup$ I've added the details. $\endgroup$ Sep 23 '12 at 18:26
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Positive definite matrices have unique positive definite square roots. Therefore, by pulling out $\det(A+B)$ (as $\left[\det\left((A+B)^{1/2}\right)\right]^2$), we may assume that $A+B=I$ and $\rho(A)<1$. So, the stated inequality reduces to $$1 + \det(A + C)\le \det A + \det(I + C),$$ which is true because \begin{align*} &1 + \det(A+C)\\ =&1 + \prod_{i=1}^n \lambda_i(A+C)\\ \le&1 + \prod_{i=1}^n \left(\lambda_i(A) + \lambda_i(C)\right)\\ =&1 + \det(A) + \det(C) + \sum_{\begin{array}{c}B_1,\ldots,B_n\in\{A,C\}\\ B_1,\ldots,B_n \text{ are not all equal}\end{array}} \prod_{i=1}^n \lambda_i(B_i)\\ \le&1 + \det(A) + \det(C) + \sum_{\begin{array}{c}B_1,\ldots,B_n\in\{I,C\}\\ B_1,\ldots,B_n \text{ are not all equal}\end{array}} \prod_{i=1}^n \lambda_i(B_i)\\ =&\det(A)+\det(I+C). \end{align*}

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