For large enough $\displaystyle n$ we have $\displaystyle 4n > n+1$. Admittedly, $\displaystyle n$ doesn't have to be very large for this to be true.
And (for all positive $\displaystyle n$) $\displaystyle \sqrt{4n}=2\sqrt{n}$.
EDIT:// posted the same as Archie. Archie's answer is equivalent to mine (and better)
No skipped steps.
Compare sqrt(n+1) with sqrt(4n) for natural numbers n. Which one is greater? Using that inequality, you can deduce the inequality of line 3. I'll leave it to you as an exercise. Try it. Trust me. It's beneficial in the long run if you prove it yourself.
3sqrt(n) is the sum of sqrt(4n) + sqrt(n) = 2sqrt(n) + sqrt(n). The process of having a fraction such as 1/3sqrt(n) is that if the sum is greater than this fraction, and the fraction is greater than 0, then clearly the sum is greater than 0.