If $x>0$ then the inequality in (1.9) must be true, because the left side is then slightly less than n(x), and the right side is slightly more than n(x).
Let's consider the derivatives of the expressions in (1.8).
$$\frac d{dx}(1-\Re(x)) = -\Re'(x) = -n(x)$$
Let's first find $n'(x)$.
We have:
$$n'(x)=\frac d{dx} \frac 1{\sqrt{2\pi}} e^{-\frac 12x^2} = \frac 1{\sqrt{2\pi}} e^{-\frac 12x^2} \cdot -x =-xn(x)$$
Then we have for instance:
$$\frac d{dx}(x^{-1}n(x)) = -x^{-2}n(x) + x^{-1}n'(x) = -x^{-2}n(x)+x^{-1}\cdot -x n(x) = -(1+x^{-2})n(x)$$
So we see that the derivatives of the expressions in (1.8) are indeed the negatives of the expressions in (1.9).
We have that $1-\Re(x)$ is in between 2 expressions, so its integration must also be between the integrations of the those 2 expressions.
Qed.
To prove the more general formula, we need to repeat these steps for the additional terms.
