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.

Your idea is quite good: if
$$
f(x)=e^{-x^2/2}-\frac{1}{\sqrt{2\pi}}\int_x^{\infty} e^{-t^2/2}\,dt
$$
then $f(0)=1/2$ and $\lim_{x\to\infty}f(x)=0$. On the other hand,
$$
f'(x)=-xe^{-x^2/2}+\frac{1}{\sqrt{2\pi}}e^{-x^2/2}
$$
so $f$ is increasing in the interval $[0,1/\sqrt{2\pi}]$ and decreasing in the interval $[1/\sqrt{2\pi},\infty)$.
Hence $f(x)>0$ for every $x>0$.
When a function is decreasing over an interval $[a,b)$ (where $b$ can be $\infty$), then
$$
\lim_{x\to b}f(x)=\inf_{x\in[a,b)}f(x)
$$
Thus $\inf_{x\in[1/2\pi,\infty)}f(x)=0$. The infimum cannot be reached, or the function would be constant in some interval $[c,\infty)$, which it isn't, because the derivative is not zero over $(1/\sqrt{2\pi},\infty)$.
