Calculus Spivak Chapter 2 problem 16(c) The question asks to prove that if $\frac mn \lt \sqrt{2}$, then there is another rational number $\frac {m'}{n'}$ with $\frac mn \lt \frac {m'}{n'} \lt \sqrt{2}$.
Intuitively, it's clear that such a number exists, but I don't understand the solution to this problem. It states: let $m_1 = m + 2n$ and $n_1 = m + n$, and choose $m' = m_1 + 2n_1 = 3m + 4n$, and $n' = m_1 + n_1 = 2m + 3n$.  
Apparently $\frac {(m + 2n)^2}{(m+n)^2} \gt 2$, but can someone explain why and how plugging in those equations for $m'$ and $n'$ ensures that $\frac {m'}{n'}$ lies between $ \frac mn$ and $\sqrt {2}$?

Actually we have the following estimates:
\begin{gather*}
\frac{m}{n}<\frac{2(m+n)}{m+2n}<\sqrt{2},
\end{gather*}
provided $m,n$ are positive integers with $\frac{m}{n}<\sqrt{2}.$ 
Verification:  Since $0<\frac{m}{n}<\sqrt{2},$ we have $\frac{n}{m}>\frac{1}{\sqrt{2}},$ and so
\begin{gather*}
\frac{2(m+n)}{m+2n}=\frac{m}{n}\cdot\frac{2\left(1+\frac{n}{m}\right)}{2+\frac{m}{n}}>\frac{m}{n}\cdot\frac{2\left(1+\frac{1}{\sqrt{2}}\right)}{2+\sqrt{2}}=\frac{m}{n}.
\end{gather*}
In the other direction, we have
\begin{align*}
&\frac{2(m+n)}{m+2n}=\frac{2}{\frac{m+2n}{m+n}}=\frac{2}{1+\frac{n}{m+n}}\\
=&\frac{2}{1+\frac{1}{\frac{m}{n}+1}}<\frac{2}{1+\frac{1}{\sqrt{2}+1}}=\frac{2}{1+(\sqrt{2}-1)}=\sqrt{2}.
\end{align*}
Therefore, we have 
\begin{gather*}
\frac{m}{n}<\frac{m'}{n'}=\frac{2(m+n)}{m+2n}<\sqrt{2},
\end{gather*}
if $0<m/n<\sqrt{2}.$
