Finding an upper bound for a probability / "Name" of inequalty

Question

I have a normal distributed random variable $X$ with mean $\mu$ and variance $\sigma^2$. I want to prove that $$P(|X| > \delta)\leq \sqrt{\frac{2}{\pi}}\frac{\exp\left(-\frac{\delta^2}{2\sigma^2}\right)}{\frac{\delta}{\sigma}} + C\mu.$$ I am aware of Mill's inequalty, $$P(|X|>\delta)\leq\sqrt{\frac{2}{\pi}}\frac{\exp\left(-\frac{\delta^2}{2}\right)}{\delta}$$ for a standard normal random variable. This inequalty seems to be used here. And I get pretty close to it: $$\begin{align*}P(X>\delta) &= P\left(\frac{X-\mu}{\sigma}>\frac{\delta - \mu}{\sigma}\right)\\&\leq\frac{1}{\sqrt{2\pi\sigma^2}}\frac{\exp\left(-\frac{(\delta - \mu)^2}{2\sigma^2}\right)}{\frac{\delta - \mu}{\sigma}}\end{align*}$$ and the the other side follows by symmetry. But how do I get rid of the $\mu$ in the denominator and in the exponential? Or is it not a variant of Mill's inequalty and some other inequalty (if yes, how it is called?)