Properties of normal distribution I've been trying to prove the following:
$$\phi'(x)(1-\Phi(x))+2\phi(x)^2>0$$
where $\phi$ is the standard Normal pdf and $\Phi$ its cdf. I've tried many simulations and I believe it's true in general, how would one go proving it formally?
 A: $$\phi(x) = \frac1{\sqrt{2\pi}}\exp\left( -\frac{x^2}{2}\right)$$
$$\phi'(x) = \frac{-x}{\sqrt{2\pi}}\exp\left( -\frac{x^2}{2}\right)=-x\phi(x)$$
Hence the problem is equivalent to 
$$-x(1-\Phi(x))+2\phi(x) >0$$
As discussed in the comment, when $x \le 0$, the problem is trivial and I will only focuses on when $x>0$.
We want to show that 
$$ (1-\Phi(x))< \frac{2\phi(x)}{x} $$
for $x>0$.
\begin{align}
1-\Phi(x) &= \int_{x}^\infty \phi(t) \, dt\\
&< \int_x^\infty \frac{t}{x} \phi(t) \, dt \\
&= \int_x^\infty \frac{-\phi'(t)}{x}  \, dt \\
&= \frac{\phi(x)}{x} \\
&< \frac{2\phi(x)}{x}
\end{align}
It seems that a stronger statement is true:
$$\phi'(x)(1-\Phi(x))+\phi(x)^2 >0$$
A: Consider that for a random variable with any density above $x$, $E(X|X>x) > x$
For a standard normal variate, you should be able to show that $E(X|X>x) = \phi(x)/[1-\Phi(x)]$
(e.g. for the integral in the numerator use the fact that $x\phi(x)=-\phi'(x)$)
Hence $\phi(x)/[1-\Phi(x)]>x$ or  $\phi(x)>x[1-\Phi(x)]$ which is stronger than your result. 
