<s>[INST] Why $\frac{1}{\sqrt{2\pi}} \int_x^\infty e^{-t^2/2} dt0$? Why does the following inequality hold for $x > 0$?
$$
Q(x)<e^{-x^2/2}, \tag{1}
$$
where
$$
Q(x) = \frac{1}{\sqrt{2\pi}} \int_x^\infty e^{-t^2/2} dt. \tag{2}
$$

The following is my attempt: Define
$$
\begin{align}
f(x)
& \triangleq e^{-x^2/2} - Q(x) \\
& = e^{-x^2/2} - \frac{1}{\sqrt{2\pi}} \int_x^\infty e^{-t^2/2} dt \\
& = e^{-x^2/2} - \left( 1 - \frac{1}{\sqrt{2\pi}} \int_{-\infty}^x e^{-t^2/2} dt \right) \\
& = e^{-x^2/2} - 1 + \frac{1}{\sqrt{2\pi}} \int_{-\infty}^x e^{-t^2/2} dt.
\tag{3}
\end{align}
$$
Then
$$
\begin{align}
f'(x)
& = -\frac{2x}{2} e^{-x^2/2} + \frac{1}{\sqrt{2\pi}} e^{-x^2/2} \\
& = \left(-x+\frac{1}{\sqrt{2\pi}} \right) e^{-x^2/2}. \tag{4}
\end{align}
$$
When $0<x<\frac{1}{\sqrt{2\pi}}$, $f'(x)>0 \Rightarrow f(x)$ is a monotonically increasing function.
When $x=\frac{1}{\sqrt{2\pi}}$, $f'(x)=0$.
When $x>\frac{1}{\sqrt{2\pi}}$, $f'(x)<0 \Rightarrow f(x)$ is a monotonically decreasing function.
So $f(x)$ has a local maximum at $x = \frac{1}{\sqrt{2\pi}}$.
$$
\lim_{x \rightarrow \infty} f(x) = 0 - 0 =0.
$$
$$
f(0) = e^{-0^2/2} - Q(0) = 1 - \frac{1}{2} = \frac{1}{2}.
$$
Then I don't know how to continue. Any comments and answers are welcome. Thanks in advance.
 [/INST] $$e^{-x^2/2}=\int^{\infty}_xe^{-t^2/2}t\,dt$$
Therefore 
$$Q(x)-e^{-x^2/2}=\frac{1}{\sqrt{2\pi}}\int^{\infty}_xe^{-t^2/2}\,dt-\int^{\infty}_xe^{-t^2/2}t\,dt=\int^{\infty}_x\Big(\frac{1}{\sqrt{2\pi}}-t\Big)e^{-t^2/2}\,dt$$
If $x\geqslant1/\sqrt{2\pi}$ then 
$$Q(x)-e^{-x^2/2}=\int^{\infty}_x\Big(\frac{1}{\sqrt{2\pi}}-t\Big)e^{-t^2/2}\,dt<0$$
otherwise split the integral into two parts
$$\int^{\infty}_x\Big(\frac{1}{\sqrt{2\pi}}-t\Big)e^{-t^2/2}\,dt=\int^{1/\sqrt{2\pi}}_x\Big(\frac{1}{\sqrt{2\pi}}-t\Big)e^{-t^2/2}\,dt+\int^{\infty}_{1/\sqrt{2\pi}}\Big(\frac{1}{\sqrt{2\pi}}-t\Big)e^{-t^2/2}\,dt$$
The first is positive and the second part negative. You can estimate each part
$$\int^{1/\sqrt{2\pi}}_x\Big(\frac{1}{\sqrt{2\pi}}-t\Big)e^{-t^2/2}\,dt\leqslant\int^{1/\sqrt{2\pi}}_0\Big(\frac{1}{\sqrt{2\pi}}-t\Big)e^{-t^2/2}\,dt<0.08$$
and 
$$\Big|\int^{\infty}_{1/\sqrt{2\pi}}\Big(\frac{1}{\sqrt{2\pi}}-t\Big)e^{-t^2/2}\,dt\Big|>0.5$$
The negative part outweights the positive part. So whenever $x>0$ the inequality holds. 
</s>