Ratio of CDF to PDF increasing? Let $\Phi(x)$ be a cumulative normal distribution function and $\phi(x)$ the associated probability density function.
Is the ratio $\frac{\Phi(x)}{\phi(x)}$ increasing in x?
Numerically it seems to be true. Is there any ways to prove it analytically?
Thanks

Maybe there is a simpler way, but here is a thought. Use the relations $ \Phi(x)' = \phi(x)$ and $\phi(x)'= -x\phi(x)$
$$\left(\frac{\Phi(x)}{\phi(x)}\right)' = \frac{\Phi(x)'\phi(x) - \Phi(x)\phi(x)'}{\phi(x)^2} = \frac{\phi^2(x) + x \Phi(x)\phi(x)}{\phi(x)^2} = 1 + x \frac{\Phi(x)}{\phi(x)}$$ 
If $\left(\frac{\Phi(x)}{\phi(x)}\right)' \geq 0$ we are done.
But note that since $x$ can take negative values, we can't be sure that $1 + x \frac{\Phi(x)}{\phi(x)} \geq 0$.
If $1 + x \frac{\Phi(x)}{\phi(x)} = 0$ then 
$$\left(\frac{\Phi(x)}{\phi(x)}\right)'' =  \left(1 + x \frac{\Phi(x)}{\phi(x)}\right)' = \frac{\Phi(x)}{\phi(x)} + x\left(1 + x \frac{\Phi(x)}{\phi(x)}\right) = \frac{\Phi(x)}{\phi(x)} \geq 0 $$ 
Therefore once we prove that $\lim_{x \to -\infty}\left(\frac{\Phi(x)}{\phi(x)}\right)' \geq 0$ then we will have proved that $\left(\frac{\Phi(x)}{\phi(x)}\right)' \geq 0$ holds for every $x\in \Bbb{R}$
Therefore we calculate:
$$\lim_{x \to -\infty}1 +  x \frac{\Phi(x)}{\phi(x)} = 1 + \lim_{x \to -\infty}x\frac{\int_{-\infty}^x e^{-u^2/2}\, du}{e^{-x^2/2}}  = 1 + \lim_{x \to -\infty}\int_{-\infty}^x xe^{-(u^2 - x^2)/2}\, du  = *$$
And 
\begin{align}
\int_{-\infty}^x xe^{-(u^2 - x^2)/2}\, du &= -\int_x^{\infty} xe^{-(u^2 - x^2)/2}\, du \\
&= -\int_0^{\infty} x e^{-((x+h)^2 - x^2)/2}\, dh  = \int_0^{\infty} -x e^{-h^2/2} e^{ - xh}\, dh \xrightarrow[x \to \infty]{} 0
\end{align}
Once $xe^{ - xh}\xrightarrow[x \to \infty]{} 0$ Now note that
$$ 1 + \lim_{x \to -\infty}\int_{-\infty}^x xe^{-(u^2 - x^2)/2}\, du = 1 - \lim_{x \to \infty}\int_{-\infty}^x xe^{-(u^2 - x^2)/2}\, du = 1 \geq 0$$
