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:

$$\begin{align} \frac{a+b}{2} &\ge \sqrt{ab} \\ \left(\frac{a+b}{2}\right)^2 &\ge (\sqrt{ab})^2 \\ \frac{a^2+2ab+b^2}{4} &\ge ab \\ a^2+2ab+b^2 &\ge 4ab \\ a^2-2ab+b^2 &\ge 0\\ (a-b)^2 &\ge 0 \\ \end{align}$$

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!

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!