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? 

It looks good, but contradiction isn't needed:
$AB\leq A^2+2AB+B^2 \implies \sqrt{AB} \leq A+B$.
