Question: 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?

Answer: 
**Hints:** $$1+\omega+\cdots+\omega^6 = 0, \qquad \alpha + \beta = -p, \quad \alpha\beta = q.$$
Remember that $p$ is the negative sum of the roots and $q$ is their product. The roots are equally spaced on the unit circle starting with $\omega^0=1$