I am trying to prove the following statement.
$$ 2\sqrt{2}\sigma - k \exp{\frac{k^2}{2\sigma^2}}\sqrt{\pi} \textit{Erfc}\Big[ \frac{k}{\sqrt{2}\sigma} \Big] >0 $$
My approach is as follows:
It is evidet Using L'Hôpital's rule yields:
$$ \lim_{k\to\infty} k \exp{\frac{k^2}{2\sigma^2}}\sqrt{\pi} \textit{Erfc}\Big[ \frac{k}{\sqrt{2}\sigma} \Big]= \sqrt{2}\sigma. $$
or, alternatively,
$$\lim_{k\to\infty} 2\sqrt{2}\sigma - k\exp{\frac{k^2}{2\sigma^2}}\sqrt{\pi} \textit{Erfc}\Big[ \frac{k}{\sqrt{2}\sigma} \Big] = \sqrt{2}\sigma >0.$$
As far as I concerned, it is now necessary to prove that $ k \exp{\frac{k^2}{2\sigma^2}}\sqrt{\pi} \textit{Erfc}\Big[ \frac{k}{\sqrt{2}\sigma} \Big]$ is a monotonically non-decreasing function. However, I am fail to do it so far.
Any help would be appreciated.
Assuming $\sigma >0$, let $k=\sqrt{2} \sigma t$ and you need to prove that $$2 >\sqrt{2 \pi }\, t\, e^{t^2}\,\text{erfc}(t)$$ Using the expansion for large values of $t$, we have $$t e^{t^2}\text{erfc}(t)=\frac{1}{\sqrt{\pi }}-\frac{1}{2 \sqrt{\pi } t^2}+\frac{3}{4 \sqrt{\pi } t^4}+O\left(\frac{1}{t^6}\right)$$ Making the rhs to be $$\sqrt{2}-\frac{1}{\sqrt{2} t^2}+\frac{3}{2 \sqrt{2} t^4}+O\left(\frac{1}{t^6}\right)$$ Then ... ?