I am not sure where to start with the question:
For any positive integer m, show that there exist integers a,b satisfying
$$\lvert{a}\rvert \le m \ \ \ \ \ \lvert{b}\rvert \le m \\ \ 0 < a+b \sqrt{2} \leq \frac{1+\sqrt{2}}{m+2}$$
I tried making a table for values of a, b which satisfy $\ \ 0 < a+b \sqrt{2} \leq \frac{1+\sqrt{2}}{m+2} \ \ $ but there are not any noticeable patterns. I thought to compare multiples of b$\sqrt{2}$ with whole numbers for the portion $a+b \sqrt{2}$ but my approach is probably too specific and feel that I am getting nowhere.
This was a problem under a chapter which discussed about the pigeonhole principle as a problem solving strategy but so far, I am having a hard time understanding the problem and it is unclear how I would apply this tactic. The only connection that I could make to the pigeonhole principle is that there are $m^{m}$ number of sums for $a+b \sqrt{2}$ when a and b have different signs and that $1+\sqrt{2}$ on a number line can be divided into m+2 sections.
I am wondering if I am focusing too much on the details of the problem and if my thought process is in the right or wrong direction. A hint on where to start would also be appreciated!
Your Pigeonhole approach works. There are $(m+1)^2$ values from 0 to $m(1+\sqrt 2)$, so splitting it into $m^2+2m$ regions and subtracting gives a value less than $(1+\sqrt{2})m/(m^2+2m)$.