Question about the proof in "Understanding Analysis" by Stephen Abbot My question is rather simple and involves a step that is taken when proving that there is a number squared that equals two. He uses the least upper bound theorem. The first few steps make sense, but how he gets between these two highlighted steps confuses me. I do not know how he derives that second inequality that I have highlighted.

Thank you in advance for your help.

He's trying to prove that if $\alpha^2\lt 2$ then he can find a number that is larger than $\alpha$ but still less than $2$.
He is trying to find a positive number $n$ such that $\alpha$ so that the $\left(\alpha+\frac{1}{n}\right)^2\lt 2$
He then expands that to get $$\left(\alpha+\frac{1}{n}\right)^2=\alpha^2+\frac{2\alpha}{n}+\frac{1}{n^2}$$ and notices that if $\frac{1}{n}\lt 1$ then $\frac{1}{n^2}\lt \frac{1}{n}$ which gives us that $$\left(\alpha+\frac{1}{n}\right)^2=\alpha^2+\frac{2\alpha}{n}+\frac{1}{n^2}\lt\alpha^2+\frac{2\alpha}{n}+\frac{1}{n}=\alpha^2+\frac{2\alpha + 1}{n}$$
We want $$\alpha^2+\frac{2\alpha+1}{n}\lt 2$$ so we solve for $n$ to get $$\frac{2\alpha+1}{n}\lt 2-\alpha^2\implies\frac{2\alpha+1}{2-\alpha^2}\lt n\implies\frac{1}{n}\lt\frac{2-\alpha^2}{2\alpha+1}$$
He uses $n_0$ instead of $n$ when he solves for the specific value of $n$ to emphasize that $n$ is a variable and $n_0$ is the specific value that keeps $$\left(\alpha+\frac{1}{n_0}\right)^2\lt 2$$
