Off-diagonal entries of a symmetric positive semi-definite matrix: Prove $a_{ij}^2 \leq a_{ii}a_{jj}$ for $i \neq j$. 
If $A=(a_{ij})$ is positive semi-definite, prove that $a_{ij}^2 \leq a_{ii}a_{jj}$ for $i \neq j$.

I've only been provided with the following definition:

A symmetric matrix A of order $n$ is semi-definite if for all non-zero $x \in \mathbb{R}^n, x^tAx \geq 0$.

I have tried substituting a vector $x$ with non-zero entries (1 in this case) in the $i$ and $j$ position, and $0$ otherwise.
The resulting equation I get is
\begin{equation}
a_{ij}+a_{ji}+a_{ii}+a_{jj} \geq 0
\implies 2a_{ij}+a_{ii}+a_{jj} \geq 0
\end{equation}
 as $A$ is symmetric.
How do I manipulate this into the required form? ($a_{ij}^2 \leq a_{ii}a_{jj}$)
 A: Take $x=e_i+\lambda e_j$ where $e_i$ is the vector with $1$ at $i$-th entry and $0$ for the other co-ordinates. Then $x'Ax=a_{ii}+2\lambda a_{ij}+\lambda^2 a_{jj}\geq 0$ for any $\lambda\in\mathbb{R}.$ This implies the quadratic form $a_{ii}+2\lambda a_{ij}+\lambda^2 a_{jj}$ can have atmost one real root. So, we must have $4a_{ij}^2-4a_{ii}a_{jj}\leq 0.$ Hence the result.
A: The diagonal entries $a_{ii}$ are real and non-negative. As a consequence the trace, $tr(A) \ge 0$. Now, since every principal sub-matrix (in particular, 2-by-2) is positive definite,
$|a_{ij}|\leq {\sqrt {a_{ii}a_{jj}}}$
From here, obviously, follows what you want to prove.
A: You're on the right track, but just plugging in $1$ isn't enough in this case.  Let $e_1,\dots,e_n$ denote the standard basis vectors, so that $x_i e_i + x_j e_j$ is a vector with $x_i$ in the $i$th position and $x_j$ in the $j$th position.  Setting $x = x_i e_i + x_j e_j$, we compute
$$
x^TAx = a_{ii} x_{i}^2 + a_{jj} x_j^2 + 2 a_ia_j x_ix_j = \pmatrix{x_i & x_j}\pmatrix{a_{ii} & a_{ij}\\a_{ij} & a_{jj}}\pmatrix{x_i\\x_j}
$$
Now, it suffices to note that if $(\begin{smallmatrix}a_{ii} & a_{ij}\\a_{ij} & a_{jj}\end{smallmatrix})$ is positive semidefinite, then its determinant $a_{ii}a_{jj} - a_{ij}^2$ must be positive.
