Inequality for standard normal distribution Suppose $X\sim N(0,\,1)$ and $c, \, y\in\Bbb R$ with $c>0$. Prove that:
$$|P(X<y)-P(X<cy)| \leq \frac{|c-1|\max(1,1/c)}{\sqrt{2\pi e}} $$

WLOG, suppose $c>1$ (for $c<1$, we work with $\frac{1}{c}$ instead) and $y>0$ (for $y<0$, we work with $-X$).
The left hand side becomes
$$
\begin{align}
P(X<cy)-P(X<y) &= P(y<X<cy) \\
&= \int_y^{cy}\frac{e^{-\frac{-t^2}{2}}}{\sqrt{2\pi} }dt \\
&\le \int_y^{cy}\frac{e^{-\frac{-y^2}{2}}}{\sqrt{2\pi} }dt =(c-1)y\frac{e^{-\frac{-y^2}{2}}}{\sqrt{2\pi} }\\
\end{align}
$$
But we have, according to Wolfram Alpha here
$$ye^{-\frac{-y^2}{2}} < \frac{1}{\sqrt{e}}$$
Hence,The left hand side is less or equal than $(c-1)\frac{1}{\sqrt{2\pi e}}$  (QED).
