Double inequality involving gaussian and erf functions I want to establish the following inequality for $x>0$:
$$\phi(x) \left( \frac{1}{x} - \frac{1}{x^3}\right)\leq 1- \Phi(x) \leq \phi(x) \frac{1}{x}$$ with  $\phi(x)=\frac{1}{\sqrt{2 \pi}}e^{-\frac{1}{2} x^2} $ and $\Phi(x)= \int_{-\infty}^x \frac{1}{\sqrt{2 \pi}}e^{-\frac{1}{2} t^2} dt $
Can somebody give me a hint to show this inequality?
 A: This kind of (rather classical) inequalities are usually established using integration by parts. I would like to present here a different proof based on a certain continued fraction.
Let
$$\varphi(x):=e^{x^2/2}\int_x^{+\infty}e^{-t^2/2}dt\tag{1}$$
It is easy to prove that 
$$\varphi(x)=\dfrac{1-\Phi(x)}{\phi(x)}\tag{2}$$
Thus, we have to establish that :
$$\text{for all} \ x>0 : \ \  \left( \frac{1}{x} - \frac{1}{x^3}\right) \leq \varphi(x) \leq \dfrac{1}{x}\tag{3}$$
(3) is going to be a rather easy consequence of the following  "beautiful" continued fraction decomposition, valid for any $x>0$ :
$$\varphi(x)=\cfrac{1}{x +\cfrac{1}{x +\cfrac{2}{x +\cfrac{3}{\ddots  x + \cfrac{n}{x+\cdots}}}}}\tag{4}$$
Reference for (4) : this paper about inequalities dealing with "erf" function. 
Indeed, due to the positivity of $x$, this decomposition allows to "bracket" $\varphi(x)$ between its two first "convergents" :
$$\text{for all} \ x>0 : \ \ \dfrac{1}{x+\frac{1}{x}} \leq \varphi(x) \leq \dfrac{1}{x}\tag{5}$$
Besides : 
$$\text{for any} \ x>0, \ \ \dfrac{1}{x}-\dfrac{1}{x^3}<\dfrac{1}{x+\frac{1}{x}}\tag{6}$$
(due to the fact that $x^2-1<x^4$ for any $x$).
Combining (5) and (6) we get the awaited result (3).
Remarks : 
1) One has remarked that (5) is a stronger bracketing than (3).
2) Oddly, https://en.wikipedia.org/wiki/Error_function#Continued_fraction_expansion doesn't give a continued fraction similar to (4).
3) The author of the cited publication, Omran Kouba, has an interesting personal site : https://www.sites.google.com/site/koubamath/home/reaserch-papers
