I've seen the following lower bound for the complementary error function ([erfc](http://en.wikipedia.org/wiki/Erfc)) but I haven't been able to prove it. Does anyone know how to establish the following?


$$erfc(x) > \frac{ x \exp(-x^2) }{ \pi(1 + 2x^2) }$$



erfc lower bound
Durrett, *Probability: Theory and Examples*, 3rd edition, p. 6 gives


$$(x^{-1} - x^{-3}) e^{-x^2/2} \le \int\_x^\infty e^{-y^2/2} \: dy $$


The proof Durrett gives is from the observation that


$$ \int\_x^\infty (1-3y^{-4}) e^{-y^2/2} \: dy = \left( x^{-1} + x^{-3} \right) e^{-x^2/2} $$


which I suspect can be found by integration by parts, although I haven't written it out; in any case, differentiate it to check.


After this, some changes of variables give


$$ \left( {1 \over z} - {1 \over 2z^3} \right) e^{-z^2}/\sqrt{\pi} \le erfc(z). $$


Finally, $z/(1+2z^2) < 1/z-1/(2z^3)$ for $z > 2^{-1/4}$, giving your bound for $z > 2^{-1/4}$ if $\pi$ is replaced with $\sqrt{\pi}$.


Obviously this is a hack trying to get your proposed bound in the form of the bound I already knew, but hopefully it helps.


