Normal distribution tail probability inequality
For every $t\gt0$, $$P(X\gt t)=\int_0^\infty\frac1{\sqrt{2\pi}}\mathrm e^{-(x+t)^2/2}\mathrm dx,$$ and, for every $x\geqslant0$, $$\mathrm e^{-(x+t)^2/2}\leqslant\mathrm e^{-t^2/2}\mathrm e^{-x^2/2},$$ hence $$P(X\gt t)\leqslant\mathrm e^{-t^2/2}\int_0^\infty\frac1{\sqrt{2\pi}}\mathrm e^{-x^2/2}\mathrm dx=\mathrm e^{-t^2/2}\,P(X\gt0).$$