Product of Diagonal Elements of Positive Semidefinite Matrix

Question

Let $A$ be an $n \times n$ positive semidefinite matrix. Show that for any $i, j$ with $1 \leq i, j \leq n$ and $i \neq j$, we have $A_{ii}A_{jj} \geq A^2_{ij}$.

I feel like I'm heading down the wrong way with this: Let $e_l \in \mathbb{R}^n$ be a vector of all zeros except for the $l^{th}$ position, where it is 1. Then by $A \succcurlyeq 0$ we know that $e_i^TAe_i = A_{ii} \geq 0$ and $e_j^TAe_j = A_{jj} \geq 0$, for some given $i \neq j$. I cannot seem to figure out how to combine these components and invoke something that yields the missing $A_{ij}^2$ term right now (I have tried to do this through Cauchy-Schwarz and come up short).

Answer 1

Consider this: $(e_i+x e_j)^T A (e_i + x e_j) \geq 0$. It is a quadratic polynomial with respect to $x$. Now its discriminant $D$ is equal or less than $0$. Check that $D/4 = A_{ij}^2 - A_{ii}A_{jj}$.

Answer 2

For all $x,y \in \mathbb{R}$, $(xe_i+ye_j)^\top A (xe_i+ye_j) = A_{ii}x^2+2A_{ij}xy+A_{jj}y^2$ must be non-nonegative.

So you get $A_{ii}A_{jj} \geq A_{ij}^2$.