<s>[INST] 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 ?
 [/INST] Rather than doing manipulations with $\sqrt{5}$ it is better to use just the rationals. We can see that $$\left(\frac{2r+5}{r+2}\right)^{2}-5=\frac{4r^{2}+20r+25-5r^{2}-20r-20}{(r+2)^{2}}=\frac{5-r^{2}}{(r+2)^{2}}$$ Since $r>0$ it follows that $(2r+5)/(r+2)$ is a better approximation to $\sqrt{5}$ than $r$, but in a different direction.
The real benefit comes from iterating this procedure two times to get $$\dfrac{2\cdot\dfrac{2r+5}{r+2}+5}{\dfrac{2r+5}{r+2}+2}=\frac{9r +20}{4r+9}$$ so that $(9r+20)/(4r+9)$ is a better approximation to $\sqrt{5}$ and that too in same direction.
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