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!
Best Answer
You're working backward, but all of your steps are actually equivalent, so that's okay. When you take the square root, though, you get $|a-b|\ge 0.$ This is true, so you're fine.
I would start with a true statement like $|a-b|\ge 0,$ and proceed through these steps in reverse, to prove the desired inequality. Or, more simply, note that $(a-b)^2\ge 0$ for all real $a,b$, so we can get there even more quickly starting from that point.