If $A=(a_{ij})$ is positive semidefinite, prove that $a_{ij}^2 \leq a_{ii}a_{jj}$ for all $i \neq j$ If $A=(a_{ij})$ is positive semidefinite, prove that $a_{ij}^2 \leq a_{ii}a_{jj}$ for all $i \neq j$.
I don't even know how to get started, any hint is appreciated, thanks a lot.
 A: Hint: By Cauchy-Schwarz you have $(x^tAy)^2 \le (x^tAx)(y^t Ay)$ for any $x,y \in \mathbb R^d$.
You can of course try to mimic the proof of Cauchy-Schwarz and start with 
$$ 0 \le (e_i + se_j)^tA(e_i + se_j) = a_{ii} + 2sa_{ij} + s^2a_{jj} $$
Now let $s = -\frac{a_{ij}}{a_{jj}}$. You of course have to consider the case $a_{jj} = 0$ seperately.
A: You need $(x^tAx)\geq0$ for all $x$ and $x$. Now if you set $x=e_i+te_j$, you get a quadratic polynomial in $t$ which must be non-negative for all $t$. So this polynomial cannot have any non-repeated roots, as polynomials must be positive on one side of these and negative on the other.
So the discriminant of this polynomial (i.e. $b^2-4ac$ if your polynomial is $at^2+bt+c$) must be non-positive - when you work out what the coefficients are in terms of the matrix entries, this will be the inequality you want.
A: Another approach, since it has not been mentioned:
A matrix $A$ is positive semidefinite if and only if every principal submatrix of $A$ is positive semidefinite.  Thus, if $A$ is positive definite, then every principal submatrix must have a non-negative determinant.
Note that
$$
\pmatrix{
a_{ii}&a_{ij}\\
a_{ji}&a_{jj}
}
$$
is a principal submatrix of $A$, so its determinant must be non-negative.  Note that since $A$ is positive semidefinite, it is symmetric so that $a_{ji} = a_{ij}$.  The conclusion follows.
