Vtyjkyuk

I have to show that $$left|sqrt 2 - frac m nright| ge frac 1 (2sqrt 2 + 1)n^2$$ given $m,n$ integers with $m,n$ greater than or equal to 1. I have shown it for the case where $|sqrt 2 - frac m n | ge 1|$, but I need to show it when $|sqrt 2 - frac mn < 1|$. I am trying to use the minimum of $$n^2left|fracm^2n^2-2right|$$ but I am stuck. Does anybody have any advice?

We have $|m^2-2n^2|ge 1$ (as $m^2-2n^2$ is never zero) so $left|fracm^2n^2-2right|ge frac1n^2$. Now $|sqrt2-frac mn|,|sqrt2+frac mn|ge frac1n^2$. We also have (as $sqrt2-1le frac mn le sqrt2+1$) and by the triangle inequality that $|sqrt2+frac mn|le sqrt2+frac mnle 2sqrt2+1$. Hence
$$|sqrt2-frac mn| ge frac1n^2 frac1ge frac1(2sqrt2+1)n^2
$$
as required.

Hint(s): The best rational approximation for $sqrt2$ are given by the convergents of its continued fraction:
$$ sqrt2=[1;2,2,2,ldots]tag1$$
If $fracp_nq_n$ and $fracp_n+1q_n+1$ are two consecutive convergents, $sqrt2$ is between them and:
$$ left|fracp_nq_n-fracp_n+1q_n+1right|=frac1q_nq_n+1.tag2$$
The claim then follows by studying the rate of growth of Pell numbers.