 Sharp two-sided Gaussian tail bound

For $G$ with the standard normal distribution and $t\ge 0$, $\mathbb{P}(|G|>t)=2(1-\Phi (t))\le e^{-t^2/2}$. The difference has derivative $e^{-t^2/2}(\sqrt{2/\pi }-t)$, starts at zero, and tends to zero, proving its nonnegativity.