erfc lower bound I've seen the following lower bound for the complementary error function (erfc) but I haven't been able to prove it.  Does anyone know how to establish the following?
$$erfc(x) > \frac{ x \exp(-x^2) }{ \pi(1 + 2x^2) }$$
 A: 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.
A: Paradoxically, it is quite easier to prove stronger bounds.
Let $X$ be a random variable with gaussian distribution and density $$ f(x)=\frac{1}{\sqrt{2\pi}}\exp(-x^2/2).$$
Now let, for any $k\in\mathbb{R}^+$, 
$$A_k = \sqrt{2\pi}\;\exp(k^2/2)\;\mathbb{P}[X>k] = \sqrt{\frac{\pi}{2}}\;\exp(k^2/2)\;\operatorname{Erfc}\left(\frac{k}{\sqrt{2}}\right).$$
Since $\mathbb{E}\left[\left(X-\mathbb{E}[X]\right)^2\right]\geq 0$, $\mathbb{E}[X^2]\geq\mathbb{E}[X]^2$, and the same holds for the conditional expected values, under the hypothesis $X>k$. That gives, in terms of $A_k$:
$$ (1-k\; A_k)^2 \leq A_k\;\left(-k+(1+k^2)\; A_k\right), $$
that can be restated as:
$$(\heartsuit)\quad A_k^2+k\;A_k-1\geq 0.$$
From:
$$ A_k \geq \frac{2}{k+\sqrt{k^2+4}}, $$
we immediately have a lower bound for the $\operatorname{erfc}$ function:
$$ e^{k^2}\;\operatorname{erfc}(k) \geq \frac{2}{\sqrt{\pi}}\left(\frac{1}{k+\sqrt{k^2+2}}\right).$$
An interesting fact is that the "reverse inequality"
$$(\spadesuit)\quad (1-k\; A_k)^2 \geq \frac{2}{\pi}\;A_k\;\left(-k+(1+k^2)\; A_k\right), $$
equivalent to:
$$(\spadesuit_2)\quad\pi\left(\int_{k}^{+\infty}(x-k)\;e^{-x^2/2}\;dx\right)^2\geq 2\int_{k}^{+\infty}e^{-x^2/2}\;dx\int_{k}^{+\infty}(x-k)^2\;e^{-x^2/2}\;dx,$$
holds, too. Since:
$$A_k =\int_{0}^{+\infty}\exp\left(-x^2/2-kx\right)\,dx, $$
by using the Fubini-Tonelli theorem and switching to polar coordinates one can see that $(\spadesuit_2)$ is equivalent to:
$$\int_{0}^{+\infty}\rho^3 e^{-\rho^2/2}\int_{0}^{\pi/4}(\pi\cos(2\theta)-2)\; e^{-k\rho\sqrt{2}\cos\theta}\;d\theta \; d\rho\geq 0, $$
also equivalent to (by integrating by parts):
$$\int_{0}^{+\infty}\rho^3 e^{-\rho^2/2}\int_{0}^{\pi/4}(\pi\sin\theta\cos\theta-2\theta)\;k\rho\sqrt{2}\sin\theta\; e^{-k\rho\sqrt{2}\cos\theta}\;d\theta \; d\rho\geq 0, $$
that is trivial since over $[0,\pi/2]$, by the concavity of the sine function, we have $\sin\phi\geq\frac{2\phi}{\pi}$. Following the line of the previous proof, the $(\spadesuit)$-inequality can be used to have tight upper bound for the $e^{k^2}\;\operatorname{erfc}(k)$ - function. Moreover, from the $(\spadesuit_2)$-inequality the convexity of $\frac{1}{A_k}$ follows.
A: Here's an approach that will establish the inequality, but it doesn't provide any insight into where the inequality came from.  Let f(x) be the left side minus the right side, i.e.
$f(x) = erfc(x) - \frac{ x \exp(-x^2) }{ \pi(1 + 2x^2) }$
Clearly $f(x) > 0$ and $ \lim_{x\to\infty}$ $f(x) = 0.$ A calculation shows that $f'(x) < 0$ for all $x > 0$, and so $f(x)$ must be positive for all $x > 0$. See these notes for details.  The notes also state improved bounds but without proof.
