Show that $\frac{a+b}{2} \ge \sqrt{ab}$ for $0 \lt a \le b$ I have to prove that
$$\frac{a+b}{2} \ge \sqrt{ab} \quad \text{for} \quad 0 \lt a \le b$$
The main issue I am having is determining when the proof is complete (mind you, this is my first time). So I did the following steps: 
$$\begin{align}
\frac{a+b}{2} &\ge \sqrt{ab} \\
\left(\frac{a+b}{2}\right)^2 &\ge \left(\sqrt{ab}\right)^2 \\
\frac{a^2+2ab+b^2}{4} &\ge ab \\
a^2+2ab+b^2 &\ge 4ab \\
a^2-2ab+b^2 &\ge 0\\
(a-b)^2 &\ge 0 \\
\end{align}$$
Now this is where I stopped because if I square root each side, I will be left with $a-b \ge 0$ or in other words, $a \ge b$ which doesn't make a whole lot of sense to me. So ultimately the question is: how do I know when I'm done? and is what I did above correct?
Thanks!

Actually, you don't need to square both sides, since $\frac{a+b}{2} \ge \sqrt{ab} \Leftrightarrow \frac{a+b-2\sqrt{ab}}{2} \ge 0
\Leftrightarrow \frac{(\sqrt a - \sqrt b)^2}{2} \ge 0$. And for any real number x, $x^2 \ge 0$ is always true, so $\frac{(\sqrt a - \sqrt b)^2}{2} \ge 0$ is true.
As for your last equation, $(a-b)^2 \ge 0$, for any real number $a, b$, it is always a truth.
