Let $ Ф(x) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{x} e^{-t^2/2} \, dt $ be the cumulative distribution function of the standard normal distribution.
Numerical calculations suggest the following inequality is true:
$$ 1 – Ф(x) > \frac{2}{\pi} \log( e^{x^2/2} -1 ) e^{-x^2} $$
for $ x > 0 $. The question is how to prove it?
There are some known lower bounds on $ 1 – Ф(x) $ like
$$ 1 – Ф(x) > \frac{1}{\sqrt{2\pi}} \frac{x}{x^2+1} e^{-x^2/2} $$
and
$$ 1 – Ф(x) > \frac{1}{2\sqrt{2\pi}} (\sqrt{x^2+4}-x) e^{-x^2/2}. $$
In both cases the difference of left and right sides is monotonically decreasing, but in my inequality the difference isn't monotonic.
Best Answer
For sufficiently large $x$, the inequality is true, as $$1 - \Phi(x) \sim \dfrac{e^{-x^2/2}}{\sqrt{2\pi} x}$$
while your expression (let's call it $R(x)$) is $O(x^2 e^{-x^2})$. In principle one can get explicit bounds corresponding to these asymptotics and thereby determine a finite value $B$ such that $1 - \Phi(x) > R(x)$ for all $x \ge B$. You then want to look at the interval $[0,B]$. A finite number of computations using interval arithmetic should be able to handle this. Thus if $1 - \Phi(x_0) - R(x_0) \ge a$ and you have an estimate $-\Phi'(x) - R'(x) \le b$ in an interval $[a,c]$, then $1 - \Phi(x) - R(x) \ge a + b (x-x_0)$ in that interval.