I was trying to do it by induction.
First let $$ A = \begin{bmatrix} a & b \\ \overline{b} & c \end{bmatrix}.$$
Since $A$ is PSD, $det(A) = ac - |b|^2 \geq 0$. So $ac \geq |b|^2$.
Now assume the hypothesis holds for all PSD matrices of size $(n-1)\times(n-1),$ and let $A$ be an arbitrary PSD matrix of size $n\times n$.
Since $A$ is PSD, it is also hermitian. So we can partition it as $$ A = \begin{bmatrix} B & y \\ y^* & a \end{bmatrix},$$
where $B$ is of size $(n-1)\times(n-1)$, $y \in \mathbb{C}^{n-1}$, and $a\in\mathbb{R}$
Consider $a_{i,i}, a_{j,j}$ and $a_{i,j}$.
If $i,j \leq n-1$, then these entries correspond to the sub matrix, $B$. $B$ is also PSD, and so the claim is true via our induction assumption. Thus we only need consider the case where $i<j = n$.
And here's where I'm stuck. if $j = n$, then the inequality we're aiming for is $a_{i,i}a_{n,n} > |a_{i,n}|^2$.
Clearly $a_{n,n} = a$, and $a_{i,n}$ is one of the entires from $y$. But it's not clear to me how these will get me the remainder of what I need. I thought about doing the determinant again, but I don't think I can get it to workout.
Any thoughts would be greatly appreciated.
Thanks in advance.
No need for induction.
I am assuming that $A$ is Hermitian.
Let $P=\begin{bmatrix} e_i & e_j \end{bmatrix}$, then $P^T AP \ge 0$. Note that $P^TAP = \begin{bmatrix} A_{ii} & A_{ij} \\ \overline{A_{ij}} & A_{jj} \end{bmatrix}$, and since $\det (P^TAP) \ge 0$, you have the desired result.