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.
 A: 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.
A: 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.
A: First Proof of Convergence
Although not explicitly asked for by the OP, a convergent sequence can be defined.
Let $F(x) = \frac{x+2}{x+1}$ and $p_0 \gt 0$ a rational number. Define the sequence
$$\tag 1 p_{k+1} = F(p_k), \quad \text{ for } k \ge 0$$
Now J. W. Tanner's answer tells us that $(p_k)$ is alternating about $\sqrt 2$ with each term getting closer.
To show convergence we can also assume that $p_0 \lt \sqrt 2$,  so that we have an increasing sub-sequence
$$\tag 2 {(p_{2k})}_{\, k \ge 0} \text{ with each } p_{2k} \lt \sqrt 2$$
But to skip the odd entries, we can construct the function
$$\tag 3 G(x) = F \circ F(x) = \frac{3x+4}{2x+3}$$
so that
$$\tag 4 p_{2(k+1)} = \frac{3p_{2k}+4}{2p_{2k}+3}$$
The increasing bounded sequence $\text{(2)}$ has a limit $\alpha \gt 0$ and the following is also true,
$$\tag 5  \lim_{k\to +\infty} p_{2k} - p_{2(k+1)} = \lim_{k\to +\infty} p_{2k} - \frac{3p_{2k}+4}{2p_{2k}+3} = 0$$
Using more basic properties of limits, we can also write
$$\tag 5  \alpha - \frac{3\alpha+4}{2\alpha+3} = 0$$
and simple algebra shows that $\alpha = \sqrt 2$.
Using the fact that each next term is better as we alternate around $\sqrt2$, we must conclude that the starting sequence $\text{(1)}$ also converges to it.

Second Proof of Convergence
Following J. W. Tanner's hints (see comments), there is a simpler way to prove convergence via the following identity,
$\tag 6 \text{For every } x \in \Bbb R \setminus \{-1\}, \quad F(x) - \sqrt 2 = (x-\sqrt2) \left(\frac{1}{1+x}\right)\left(1-\sqrt2\right)$
We can use $\text{(6)}$ since the terms of our sequence $(p_k)$ contain only positive numbers. Moreover, to show convergence we can assume that $|p_0 - \sqrt 2| \lt 1$. But then
$\tag 7 \text{For every } k \ge 0, \quad    |p_{k+1}-\sqrt2|<(\sqrt2 -1)^k$
and we have convergence.
