How to show that $\int_{t}^{\infty} e^{-\frac{x^2}{2}} dx \leq \frac{1}{t}e^{-\frac{t^2}{2}}$ where $1 \leq t$? How to show that

$$
\int_{t}^{\infty} e^{-\frac{x^2}{2}} dx \leq \frac{1}{t}e^{-\frac{t^2}{2}}
$$
  where $t \geq 1$.

 A: $$t\int_{t}^{\infty} e^{-\frac{x^2}{2}} dx =\int_{t}^{\infty}t e^{-\frac{x^2}{2}} dx \leq \int_{t}^{\infty} xe^{-\frac{x^2}{2}} dx =- e^{-\frac{x^2}{2}}\Big|_{t}^{\infty} =e^{-\frac{t^2}{2}}$$
where $x\geq t \geq 1$.
A: Hint: Inside the integrand, multiply and divide by $x$. Then as $x>t$, we have $1/x < 1/t$. Hope this helps. Interestingly this holds for any $ t>0$.
A: This is taken from my answer here:
But how to use this to show that $ \ \int_5^{\infty} e^{-x^2} dx \ $ is negligible as compared to $ \int_5^{\infty} e^{-5x} dx \ $?
$(\dfrac1{x}e^{-x^2/2})'
=\dfrac1{x}(e^{-x^2/2})'-\dfrac1{x^2}e^{-x^2/2}
=-e^{-x^2}-\dfrac1{x^2}e^{-x^2}
$
so
$\int e^{-x^2/2}dx
=-\dfrac1{x}e^{-x^2/2}-\int \dfrac1{x^2}e^{-x^2/2}dx
$
so
$\begin{array}\\
\int_a^{\infty} e^{-x^2/2}dx
&=-\dfrac1{x}e^{-x^2/2}|_a^{\infty}-\int_a^{\infty} \dfrac1{x^2}e^{-x^2/2}dx\\
&=\dfrac{e^{-a^2/2}}{a}-\int_a^{\infty} \dfrac1{x^2}e^{-x^2}dx\\
&<\dfrac{e^{-a^2/2}}{a}\\
\end{array}
$
You can also get a lower bound
as shown in that answer.
