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The inequality $$\det(A+B)\geq \det A +\det B$$ is implied by the Minkowski determinant theorem $$(\det(A+B))^{1/n}\geq (\det A)^{1/n}+(\det B)^{1/n}$$ which holds true for any non-negative $n\times n$ Hermitian matrices $A$ and $B$. The latter inequality is equivalent to the fact that the function $A\mapsto(\det A )^{1/n}$ is concave on the set of $n\times n$ non-negative Hermitian matrices (see e.g., A Survey of Matrix Theory and Matrix Inequalities by Marcus and Minc, Dover, 1992, P. 115 and also the previous MO threadMO thread).

The inequality $$\det(A+B)\geq \det A +\det B$$ is implied by the Minkowski determinant theorem $$(\det(A+B))^{1/n}\geq (\det A)^{1/n}+(\det B)^{1/n}$$ which holds true for any non-negative $n\times n$ Hermitian matrices $A$ and $B$. The latter inequality is equivalent to the fact that the function $A\mapsto(\det A )^{1/n}$ is concave on the set of $n\times n$ non-negative Hermitian matrices (see e.g., A Survey of Matrix Theory and Matrix Inequalities by Marcus and Minc, Dover, 1992, P. 115 and also the previous MO thread).

The inequality $$\det(A+B)\geq \det A +\det B$$ is implied by the Minkowski determinant theorem $$(\det(A+B))^{1/n}\geq (\det A)^{1/n}+(\det B)^{1/n}$$ which holds true for any non-negative $n\times n$ Hermitian matrices $A$ and $B$. The latter inequality is equivalent to the fact that the function $A\mapsto(\det A )^{1/n}$ is concave on the set of $n\times n$ non-negative Hermitian matrices (see e.g., A Survey of Matrix Theory and Matrix Inequalities by Marcus and Minc, Dover, 1992, P. 115 and also the previous MO thread).

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Andrey Rekalo
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The inequality $$\det(A+B)\geq \det A +\det B$$ is implied by the Minkowski determinant theorem $$(\det(A+B))^{1/n}\geq (\det A)^{1/n}+(\det B)^{1/n}$$ which holds true for any non-negative $n\times n$ Hermitian matrices $A$ and $B$. The latter inequality is equivalent to the fact that the function $A\mapsto(\det A )^{1/n}$ is concave on the set of $n\times n$ non-negative Hermitian matrices (see e.g., A Survey of Matrix Theory and Matrix Inequalities by Marcus and Minc, Dover, 1992, P. 115 and also the previous MO thread).

The inequality $$\det(A+B)\geq \det A +\det B$$ is implied by the Minkowski determinant theorem $$(\det(A+B))^{1/n}\geq (\det A)^{1/n}+(\det B)^{1/n}$$ which holds true for any non-negative $n\times n$ Hermitian matrices $A$ and $B$. The latter inequality is equivalent to the fact that the function $A\mapsto(\det A )^{1/n}$ is concave on the set of $n\times n$ non-negative Hermitian matrices (see e.g. A Survey of Matrix Theory and Matrix Inequalities by Marcus and Minc, Dover, 1992, P. 115).

The inequality $$\det(A+B)\geq \det A +\det B$$ is implied by the Minkowski determinant theorem $$(\det(A+B))^{1/n}\geq (\det A)^{1/n}+(\det B)^{1/n}$$ which holds true for any non-negative $n\times n$ Hermitian matrices $A$ and $B$. The latter inequality is equivalent to the fact that the function $A\mapsto(\det A )^{1/n}$ is concave on the set of $n\times n$ non-negative Hermitian matrices (see e.g., A Survey of Matrix Theory and Matrix Inequalities by Marcus and Minc, Dover, 1992, P. 115 and also the previous MO thread).

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Andrey Rekalo
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The inequality $$\det(A+B)\geq \det A +\det B$$ is implied by Minkowski's inequalitythe Minkowski determinant theorem $$(\det(A+B))^{1/n}\geq (\det A)^{1/n}+(\det B)^{1/n}$$ which holds true for any non-negative $n\times n$ Hermitian matrices $A$ and $B$. The latter inequality is equivalent to the fact that the function $A\mapsto(\det A )^{1/n}$ is concave on the set of $n\times n$ non-negative Hermitian matrices (see e.g. A Survey of Matrix Theory and Matrix Inequalities by Marcus and Minc, Dover, 1992, P. 115).

The inequality $$\det(A+B)\geq \det A +\det B$$ is implied by Minkowski's inequality $$(\det(A+B))^{1/n}\geq (\det A)^{1/n}+(\det B)^{1/n}$$ which holds true for any non-negative $n\times n$ Hermitian matrices $A$ and $B$.

The inequality $$\det(A+B)\geq \det A +\det B$$ is implied by the Minkowski determinant theorem $$(\det(A+B))^{1/n}\geq (\det A)^{1/n}+(\det B)^{1/n}$$ which holds true for any non-negative $n\times n$ Hermitian matrices $A$ and $B$. The latter inequality is equivalent to the fact that the function $A\mapsto(\det A )^{1/n}$ is concave on the set of $n\times n$ non-negative Hermitian matrices (see e.g. A Survey of Matrix Theory and Matrix Inequalities by Marcus and Minc, Dover, 1992, P. 115).

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Andrey Rekalo
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