How would you show that the set $$P=\bigg\{\begin{pmatrix} x\\y\\z \end{pmatrix} \in \mathbb{R}^3 : \det \begin{pmatrix} x&z\\z&y \end{pmatrix} \geq 0 \text{ , } x\geq 0, \text{ }y\geq 0\bigg\}$$ is convex?
My attempt is as follows (using the definition of positve semidefinite):
$v^t\begin{pmatrix} x&z\\z&y \end{pmatrix} v \geq 0 :\forall v \in \mathbb{R}^3$
Then the following should hold as well:
$v^t\bigg((1-t)\begin{pmatrix} x_1&z_1\\z_1&y_1 \end{pmatrix}+t\begin{pmatrix} x_2&z_2\\z_2&y_2 \end{pmatrix}\bigg) v \geq 0 :t\in[0,1] \text{ , }\forall v \in \mathbb{R}^3$
As in my mind, the scaling of a PSD matrix with a $t\in[0,1]$ should return a another PSD matrix. Or am I wrong in my thinking? I'm an undergraduate, apologies for my possibly gross ignorance.
The determinant being nonnegative does not imply that a $2 \times 2$ matrix is PSD. You also need the diagonal elements to be nonnegative. Your set $P$ is not convex. For example, $(10, 10, 1)^T$ and $(-10, -10, 1)^T$ are in $P$ but $\frac{1}{2} (10, 10, 1)^T + \frac{1}{2} (-10, -10, 1)^T$ is not.