Show that $\int_{x}^{\infty}e^{-\frac{t^2}{2}}dt\geq e^{-\frac{t^2}{2}}(\frac{1}{x}-\frac{1}{x^3}) $ for all positive $x$
Does it require the mean value theorem, or the Taylor series expansion? It is easy to show that the above inequality is reversed if the $x^{-3}$ term is dropped, i.e. $$\int_{x}^{\infty}e^{-\frac{t^2}{2}}dt\leq e^{-\frac{t^2}{2}}\frac{1}{x} $$ for all positive $x$.
Hint: Start by using integration by parts, $u=\frac{1}{t}$ and $dv=te^{-t^2/2}\,dt$. After you do the calculation, how to continue may be clear.