Claim: If $m/n$ is a good approximation of $\sqrt{2}$ then $(m+2n)/(m+n)$ is better.
My attempt at the proof:
Let d be the distance between $\sqrt{2}$ and some estimate, s.
So we have $d=s-\sqrt{2}$
Define $d'=m/n-\sqrt{2}$ and $d''=(m+2n)/(m+n)-\sqrt{2}$
To prove the claim, show $d''<d'$
Substituting in for d' and d'' yields:
$\sqrt{2}<m/n$
This result doesn't make sense to me, and I was wondering whether there is an other way I could approach the proof or if I am missing something.
Hint: Compare $\left|\dfrac{m^2}{n^2}-2\right|$ with $\left|\dfrac{(m+2n)^2}{(m+n)^2}-2\right|$. We need to take absolute values, because if one approximation is too big, the other turns out to be too small, and vice-versa.
Bring the expressions to the denominators $n^2$ and $(m+n)^2$ respectively. So the first becomes $\left|\dfrac{m^2-2n^2}{n^2}\right|$.
Make sure to expand the squares in the second one. The second one will simplify an awful lot: I will leave the pleasure to you. The result will jump out.