Q:

Prove $\int_t^{\infty} e^{-x^2/2}\,dx > e^{-t^2/2}\left(\frac{1}{t} - \frac{1}{t^3}\right)$

How to formally prove the following inequality - 
$$\int_t^{\infty} e^{-x^2/2}\,dx > e^{-t^2/2}\left(\frac{1}{t} - \frac{1}{t^3}\right)$$

A:

For a better lower-bound you may use the following proof by @robjohn:
$$x\int_x^\infty e^{-t^2/2}\,\mathrm{d}t \le \int_x^\infty e^{-t^2/2}\,t\,\mathrm{d}t =e^{-x^2/2}$$ 
Integrate both sides of the preceding: 
$$ 
\begin{align} 
\int_s^\infty e^{-x^2/2}\,\mathrm{d}x 
&\ge\int_s^\infty x\int_x^\infty e^{-t^2/2}\,\mathrm{d}t\,\mathrm{d}x\\ 
&=\int_s^\infty\int_s^txe^{-t^2/2}\,\mathrm{d}x\,\mathrm{d}t\\ 
&=\int_s^\infty\frac12(t^2-s^2)e^{-t^2/2}\,\mathrm{d}t\\ 
\left(1+\frac12s^2\right)\int_s^\infty e^{-x^2/2}\,\mathrm{d}x 
&\ge\frac12\int_s^\infty t^2e^{-t^2/2}\,\mathrm{d}t\\ 
&=-\frac12\int_s^\infty t\,\mathrm{d}e^{-t^2/2}\\ 
&=\frac12se^{-s^2/2}+\frac12\int_s^\infty e^{-t^2/2}\,\mathrm{d}t\\ 
\left(s+\frac1s\right)\int_s^\infty e^{-x^2/2}\,\mathrm{d}x 
&\ge e^{-s^2/2} 
\end{align} 
$$
and note that $\displaystyle \left(s+\frac{1}{s}\right)^{-1} = \frac{s}{1+s^2} > \frac{1}{s} - \frac{1}{s^3}$ for $s > 0$,
which is equivalent to $\displaystyle s^4 > s^4 - 1$.

Expalanation for integartion by parts: 
\begin{align*}
&\int_x^{\infty} e^{-t^2/2} \mathrm dt\\
=& \int_x^{\infty} \color{red}{\frac{1}{t}} .\color{blue}{te^{-t^2/2}} \mathrm dt\\
=& \left[\color{red}{\frac{1}{t}} .\int\color{blue}{te^{-t^2/2}}\,dt\right]_x^{\infty} 
- \int_x^{\infty} \left( \color{red}{\frac{1}{t}} \right )' \left (\int\color{blue}{te^{-t^2/2}}\,dt\right)\mathrm dt\\
=& \frac{e^{-x^2/2}}{x} - \int_x^{\infty} \frac{e^{-t^2/2}}{t^2} \mathrm dt.
\end{align*}
Now, for the second integral that we obtained in the previous line we employ similar idea as done above:
\begin{align*}
&\int_x^{\infty} \frac{e^{-t^2/2}}{t^2} \mathrm dt\\
=& \int_x^{\infty} \color{red}{\frac{1}{t^3}} .\color{blue}{te^{-t^2/2}} \mathrm dt\\
=& \left[\color{red}{\frac{1}{t^3}} .\int\color{blue}{te^{-t^2/2}}\,dt\right]_x^{\infty} 
- \int_x^{\infty} \left( \color{red}{\frac{1}{t^3}} \right )' \left (\int\color{blue}{te^{-t^2/2}}\,dt\right)\mathrm dt\\
=& \frac{e^{-x^2/2}}{x^3} -3 \int_x^{\infty} \frac{e^{-t^2/2}}{t^4} \mathrm dt.
\end{align*}
Combining these together we have $$\int_x^{\infty} e^{-t^2/2} = e^{-x^2/2}\left(\frac{1}{x} - \frac{1}{x^3}\right)+3 \int_x^{\infty} \frac{e^{-t^2/2}}{t^4} \mathrm dt$$
Since, the integrand $\dfrac{e^{-t^2/2}}{t^4}$ is always positive, we get the desired inequality:
$$\int_x^{\infty} e^{-t^2/2} > e^{-x^2/2}\left(\frac{1}{x} - \frac{1}{x^3}\right)$$

