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.