While reading a paper [*An Arithmetic Proof of John’s Ellipsoid Theorem*](http://dmg.tuwien.ac.at/schuster/john1.pdf) by Gruber and Schuster, I have a question on their proof.

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Consider an $n\times n$ real symmetric and [positive definite](http://en.wikipedia.org/wiki/Positive-definite_matrix) matrix $\mathbf A$. 

1. As this kind of matrix is symmetric, its $n(n+1)/2$ upper diagonal terms are enough to represent it. Hence, we can consider such a matrix as a point in $\mathbb R^{n(n+1)/2}$.
2. A [conical combination](http://en.wikipedia.org/wiki/Conical_combination) of two positive definite matrices is also positive definite. Hence, the set of all symmetric positive definite matrices forms an open convex cone $\mathcal P\in\mathbb R^{n(n+1)/2}$ with apex on the origin.

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Now they claim the following theorem without proving it.
>  ***Theorem:*** The set $\ \mathcal D = \{\mathbf A \in \mathcal P: \det \mathbf A \geq 1\}$ is a closed, smooth, strictly convex set in $\mathcal P$ with non-empty interior.

They just gave some hint that we can use [Implicit Function Theorem](http://en.wikipedia.org/wiki/Implicit_function_theorem) and Minkowski's Determinant Inequality which states that
> For two $n\times n$ positive semidefinite [Hermitian](http://en.wikipedia.org/wiki/Hermitian_matrix) matrices $\mathbf X$ and $\mathbf Y$,
$$\det (\mathbf X + \mathbf Y)^{1/n}\geq \det(\mathbf X)^{1/n} + \det(\mathbf Y)^{1/n} $$

Any hint or suggestion on how to prove the above theorem about the set $\mathcal D$?