I came across this problem and I'm not sure if the proof I have found is OK. I may be overcomplicating things. Any suggestions or confirmation of correctness would be appreciated.
The Problem:
Let $A$ be an $n \times n$ Hermitian matrix such that $\det A_k > 0 $ for all $k = 1, 2, . . . , n − 1$ and $\det A \geq 0$. Prove that $A$ is positive semidefinite.
My shot:
Consider the matrix $A_{n-1}$. By Silvester's Criterion of Positivity, we know $A_{n-1}$ is positive definite $(A_{n-1}>0)$ and therefore it can be diagonalized by congruence using simple row operations (making sure to conserve the determinant) into a diagonal matrix $A_{n-1}'$ with positive entries.
We now consider the matrix $A'$, defined as the matrix $A$ but with the submatrix $A_{n-1}$ diagonalized as explained above:
$$A'=\begin{pmatrix} A_{n-1}' & *\\* & a_{n,\,n}\end{pmatrix}$$
$A'$ has the same determinant as $A$, so $\det A'\geq0$. We can now further diagonalize the matrix $A'$, keeping the determinant, to obtain:
$$D=\begin{pmatrix} A_{n-1}' & 0\\0 & a_{n,\,n}'\end{pmatrix},\:\det D\geq0$$
We compute the determinant of $D$:
$$\det D=\begin{vmatrix} A_{n-1}' & 0\\0 & a_{n,\,n}'\end{vmatrix}= a_{n,\,n}'\cdot\det A_{n-1}' \geq 0$$
As we have $\det A_{n-1}'>0$, we know that $a_{n,\,n}' \geq0$. Therefore, we have diagonalized the matrix A into a diagonal matrix with all its entries satisfying $a_{i,\,i}' \geq 0$, $\forall i=1,\;2,\;...,\;n$.
As stated by Sylvester's Law of Inertia, the number of positive, zero and negative entries of the diagonal matrix depends only on $A$ and not on the specific diagonalization that we use. As the matrix $A$ is Hermitian, one of its possible diagonal forms will have the eigenvalues of $A$ in its diagonal entries. Therefore, by Sylvester's Law of Inertia, the eigenvalues $\lambda_i$ of $A$ will also satisfy $\lambda_i\geq0$, $\forall i=1,\;2,\;...,\;n$. This means that the matrix $A$ is positive semidefinite ($A\geq0$).
