Prove that the sum of two positive real numbers is equal or greater than the square root of their product. Trying to prove this:
$A$ and $B$ are positive real numbers.
$A + B \geq \sqrt{AB}$
This is what I wrote:
Proof by Contradiction
$A + B < \sqrt{AB}$
$(A + B)^2 < AB$
$A^2 + AB + AB + B^2 < AB$
$A^2 + AB + B^2 < 0$
Inconsistent with $A>0$ and $B>0$.
Did I do this correctly? 

Yes, it's correct.
You could have proceeded directly, obtaining
$$
A^2+AB+B^2\geq0
$$
which is true when $A,B>0$.
In fact one even has
$$
\frac{A+B}{2}\geq\sqrt{AB}
$$
and you might want to take a look at the AM-GM inequality.
