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 ?

$$\frac{2r+5}{r+2} - \sqrt{5} = \frac{2r+5-\sqrt{5}r-2\sqrt{5}}{r+2} = \frac{(2-\sqrt{5})r - (2-\sqrt{5})\sqrt{5}}{r+2} = \frac{2-\sqrt{5}}{r+2}(r-\sqrt{5})$$
That means as long as $$\left|\frac{2-\sqrt{5}}{r+2}\right| < 1$$ holds, the proposition is true. If we assume $r > 0$ as reasonable for an approximation of a square root of a positve integer, then the proposition always holds. 
