I am trying to prove that the following function:
$$g(x) = x - \sqrt{\frac{2}{\pi}} e^{-x^2/2} \frac{\operatorname{Erf}\left(\frac{x}{\sqrt{2}}\right)}{1-\operatorname{Erf}\left(\frac{x}{\sqrt{2}}\right)^2} $$
where $\operatorname{Erf}$ is the Error function, has a unique real root (i.e. $g(x)=0$ has a unique real solution) at $x=0$.
Visual inspection with WolframAlpha: http://wolfr.am/1iXZ5LB
One candidate way to prove this would be to prove that $g(x)>0$ for $x>0$ and $g(x)<0$ for $x<0$ (seemingly true by visual inspection), but I can't seem to find appropriate inequalities that help me establish this.
A second candidate way to prove this would be to prove that the derivative $g'(x)$ is always positive (WolframAlpha visual inspection: http://wolfr.am/KJhoFz), since this would establish that g(x) only crosses $x=0$ once. Again, I can't seem to figure out a way to prove this.
A third candidate way would be to use that $g(x)$ is odd (easy to see/show), and thus only has non-zero coefficients in its Taylor expansion at $x=0$, $g(0) = \sum_{n=0}^{\infty} \frac{g^{(n)}(0)}{n!} x^n$, for odd $n$. If $g^{(n)}(0)$ is non-negative for all odd $n$ then the function has only one real root, at $x=0$. But I don't see a way to prove that all the odd coefficients are non-negative.
I'm wondering if anyone sees any tricks involving Erf, or more general tricks, for analytically proving this fact (that I'm very convinced is true by visual inspection).
Another thought. The only two pieces of that function that change sign are $x$ and $\mathrm{Erf}(x/\sqrt{2})$. Both change sign at 0. So, it comes down to showing that
$$ x < \sqrt{\frac{2}{\pi}}\mathrm{e}^{-x^2/2} \frac{\mathrm{Erf}(\frac{x}{\sqrt{2}})}{1-\mathrm{Erf}(\frac{x}{\sqrt{2}})^2} $$
when $x < 0$
and
$$ x > \sqrt{\frac{2}{\pi}}\mathrm{e}^{-x^2/2} \frac{\mathrm{Erf}(\frac{x}{\sqrt{2}})}{1-\mathrm{Erf}(\frac{x}{\sqrt{2}})^2} $$
when $x > 0$