Prove the $\frac {2r+5}{r+2}$ is always a better approximation of $\sqrt {5}$ than $r$. 
Prove the $\frac {2r+5}{r+2}$ is always a better approximation of $\sqrt {5}$ than $r$.
SOURCE : Inequalities (Page Number 4 ; Question Number 207)

I tried a lot of approaches, but without success.
I rewrote $\frac {2r+5}{r+2}$ as $2 + \frac {1}{r+2}$.
$\sqrt {5} \approx2.2360679775$
Equating $\frac {1}{r+2}$ and $0.2360679774$ , I get $r=2.23606797929$.
So, $\frac {2r+5}{r+2}$ is a still a better approximation than $r$.
How to proceed ?
Any hints/ideas/pointers ?

Observe if $r<\sqrt{5}$, then there exists $\epsilon>0$ such that
\begin{align}
r+\epsilon<\sqrt{5} \ \ \Leftrightarrow& \ \ (r+\epsilon)^2 <5\\
\Leftrightarrow& \ \ r^2-5+2\epsilon r+ \epsilon^2=P(\epsilon)<0 
\end{align}
which is true if we sketch $P(\epsilon)$ as a function of $\epsilon$. In particular, we see that the positive root of $P(\epsilon)$ is
\begin{align}
\epsilon = \frac{-2r+\sqrt{20}}{2} = -r+\sqrt{5}= \frac{-r^2+5}{r+\sqrt{5}}>\frac{-r^2+5}{r+2}=\epsilon_0>0
\end{align}
since $\sqrt{5}>2$. Hence
\begin{align}
r<r+\epsilon_0 = r+\frac{-r^2+5}{r+2} = \frac{r^2+2r-r^2+5}{r+2} = \frac{2r+5}{r+2}.
\end{align}
Remark: The idea is to find $\epsilon_0>0$ rational such that $(r+\epsilon_0)^2<5$. 
