# Upper bound for standard normal $\frac{x \phi(x)}{2 \Phi(x) - 1}$

I am looking to upper bound the expression:

$$\frac{x \phi(x)}{2 \Phi(x) - 1}$$ for $$x \geq 0$$. Here $$\phi(x), \Phi(x)$$ denote the PDF/CDF (respectively) of a $$\mathcal{N}(0, 1)$$ random variable.

Based on numerical plots it appears to be upper bounded by $$\frac{1}{2}$$, with that bound achieved at $$x = 0$$. Could anyone please show how to obtain this using elementary methods and properties of the standard normal random variable?

One remark first: technically $$\frac{x\phi(x)}{2\Phi(x)-1}$$ is undefined at $$x=0$$, because plugging in zero leads to an expression of the form $$\frac{0}{0}$$. To resolve this problem, we note that by l'Hopital's Rule, $$\lim_{x\to 0}\frac{x\phi(x)}{2\Phi(x)-1}=\lim_{x\to 0}\frac{\phi(x)+x\phi'(x)}{2\phi(x)}=\frac{1}{2},$$ (the last equality by direct evaluation) and thus $$\frac{x\phi(x)}{2\Phi(x)-1}$$ has a continuous extension to all of $$\mathbb{R}$$. So when $$x=0$$ we interpret $$\frac{x\phi(x)}{2\Phi(x)-1}=\frac{1}{2}$$. Okay, with that technicality out of the way, to show that for $$x\geq 0$$ $$\frac{x\phi(x)}{2\Phi(x)-1}\leq \frac{1}{2}$$ is equivalent to showing that $$x\phi(x)\leq \Phi(x)-\frac{1}{2},$$ and it suffices to show that if $$f(x):=x\phi(x)$$ and $$g(x):=\Phi(x)-\frac{1}{2}$$, then $$f(0)=g(0)$$ and that $$f'(x)\leq g'(x)$$ for all $$x\geq 0$$. We have $$f'(x)=\phi(x)+x\phi'(x)\text{ and }g'(x)=\phi(x),$$ so showing $$f'(x)\leq g'(x)$$ reduces to showing $$x\phi'(x)\leq 0$$, but this clearly holds because by direct calculation $$x\phi'(x)=-x^2\phi(x)\leq 0$$. This finishes the proof. Please feel free to comment below if you aren't sure about any of the above; I omitted some details which I thought are easy to verify.
$$\frac{x \,\phi(x)}{2 \Phi(x) - 1}=\frac{x\,e^{-\frac{x^2}{2}} }{\sqrt{2 \pi }\, \text{erf}\left(\frac{x}{\sqrt{2}}\right)}$$
Using Padé approximants $$\frac{x \,\phi(x)}{2 \Phi(x) - 1} \leq \frac 12 \,\frac{1-\frac{2 }{9}x^2+\frac{1}{63}x^4 } {1+\frac{1}{9}x^2+\frac{8 }{945}x^4 }$$
Expanded as series $$\frac 12 \,\frac{1-\frac{2 }{9}x^2+\frac{1}{63}x^4 } {1+\frac{1}{9}x^2+\frac{8 }{945}x^4 }-\frac{x \,\phi(x)}{2 \Phi(x) - 1}=\frac{4 x^{10}}{654885}+O\left(x^{12}\right)$$