How to prove that $\text{erf}\left(\frac{x}{\sqrt{2}}\right)\geq\left(1-\frac{1}{x^2}\right)$ for $x>0$? Having in mind that the error function is a function such that:
$$\text{erf}(x)=\displaystyle\int_0^{x}\frac{2}{\sqrt{\pi}}e^{-u^2}du$$
Graphically I can see that for $x>0$
$$\text{erf}\left(\frac{x}{\sqrt{2}}\right)\geq\left(1-\frac{1}{x^2}\right)$$
But how can that be proven mathematically?
Is there some particular result (something I should already detect from what I am given) to use or is the "differentiating on both sides approach" to be used? I have tried to differentiate on both sides wrt $x$ but this doesn't help me permorm further steps and I get stuck again.
Or is the fact that both r.h.s. and l.h.s. are non-decreasing, but at a different rate (second derivative), influential? That is, r.h.s. seems to grow at a faster rate but anyway not sufficient to 'go above' l.h.s.
 A: We want to prove that for the complementary error function $\operatorname{erfc}(x):=1-\operatorname{erf}(x)$,
$$
\operatorname{erfc}(x) \le \frac{1}{{2x^2 }}
$$
whenever $x>0$. Integration by parts gives
$$
\operatorname{erfc}(x) = \frac{2}{{\sqrt \pi  }}\int_x^{ + \infty } {e^{ - t^2 } dt}  = \frac{{e^{ - x^2 } }}{{x\sqrt \pi  }} - \frac{1}{{\sqrt \pi  }}\int_x^{+\infty}  {\frac{{e^{ - t^2 } }}{{t^2 }}dt}  < \frac{{e^{ - x^2 } }}{{x\sqrt \pi  }}
$$
for all $x>0$. Thus, it is enough to show that
$$
\frac{{e^{ - x^2 } }}{{x\sqrt \pi  }} \le \frac{1}{{2x^2 }} \Leftrightarrow xe^{ - x^2 }  \le \frac{{\sqrt \pi  }}{2}
$$
for all $x>0$. The function $x \mapsto xe^{ - x^2 }$, if defined for positive $x$, has a global maximum at $x=\frac{\sqrt{2}}{2}$ with value $0.4288819424\ldots< \frac{{\sqrt \pi  }}{2}=0.8862269254\ldots$. This completes the proof.
A: Disclaimer: This is not a complete answer, but I believe the methods could perhaps be tweaked to give the desired inequality.
I assume that you are interested in the case where $x \geq 0$. 
The tail-CDF of a standard normal distribution $Z$ is given by 
$$\frac{1}{2} - \frac12 \text{erf} \left( \frac{x}{\sqrt 2} \right)$$
and represents the probability that $Z$ is greater than $x$. By Chebyshev's Inequalities for higher moments,
$$\frac{1}{2} - \frac12 \text{erf} \left( \frac{x}{\sqrt 2} \right) =\mathbb{P}(Z > x) \leq \frac{\mathbb{E}[(Z - \mathbb{E}[Z])^4]}{x^4} = \frac{3}{x^4}$$
Here we have used the fact that the fourth moment of the standard normal is $3$. This inequality implies that
$$\text{erf}\left(\frac{x}{\sqrt 2}\right) \geq 1 - \frac{6}{x^4} > 1 - \frac{1}{x^2}$$
whenever $x \geq 2.5$. It seems that the harder part of the problem is to show that for small $x$ the inequality holds. 
Note that direct application of Chebyshev's Inequality on the standard normal distribution gives the inequality
$$\text{erf}\left(\frac{x}{\sqrt 2}\right) \geq 1 - \frac{2}{x^2}$$
which is not quite the inequality that we want.
A: Consider that we look at the minimum of function
$$f(x)=\text{erf}\left(\frac{x}{\sqrt{2}}\right)-\left(1-\frac{1}{x^2}\right)$$ We have
$$f'(x)=\sqrt{\frac{2}{\pi }} e^{-\frac{x^2}{2}}-\frac{2}{x^3}$$ If the derivative can cancel, then
$$\sqrt{\frac{2}{\pi }} e^{-\frac{x^2}{2}}=\frac{2}{x^3}\implies x^2-3 \log \left(x^2\right)+\log (2 \pi )=0$$ Let $y=x^2$
$$y-3 \log \left(y\right)+\log (2 \pi )=0\implies y=-3 W\left(-\frac{\sqrt[3]{2 \pi }}{3} \right)$$ and since
$$\frac{\sqrt[3]{2 \pi }}{3}  > \frac 1 e$$ $y$ is a complex number (then $x$ is) and there is no minimum. Since $f(1) > 0$, then the inequality holds for any $x>0$.
