Show that if m/n is a good approximation of $\sqrt{2}$ then $(m+2n)/(m+n)$ is better 
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.

Assume $\dfrac mn\ne\sqrt2;$ otherwise $\dfrac mn$ is $\sqrt2$, not an approximation.
Then $d'\ne0$ so we can compute $\dfrac {d''}{d'}=\dfrac{\dfrac{m+2n}{m+n}-\sqrt2}{\dfrac mn-\sqrt2}=
\dfrac n{m+n}\dfrac{m+2n-\sqrt2(m+n)}{m-\sqrt2n}$
$=\dfrac n{m+n}\dfrac{m-\sqrt2n-\sqrt2(m-\sqrt2n)}{m-\sqrt2n}=\dfrac {1}{1+\dfrac mn}\left(1-\sqrt2\right).$
We could assume $\dfrac mn\ge0$ (otherwise $\dfrac mn$ is not "a good approximation of $\sqrt2$"), 
and $-1<1-\sqrt2<0$ since $1<\sqrt2<2$.  
From here it is easy to see that 
 $|d''|<|d'|,$ and $d''$ and $d'$ have opposite signs.
