Here's a sketch and a link for how I prove it. Let
$$ f(x) = - \left( \frac{1}{x} - \frac{1}{x^3} + \frac{3}{x^5}\right) \phi(x) .$$

Now show (lemma): $\frac{df}{dx} = \left(1 + \frac{15}{x^6}\right)\phi(x)$.

(To prove this, use that $\frac{d\phi}{dx} = - x \phi(x)$, the product rule, and some cancellation.)

Now suppose $x > 0$:

\begin{align*}
1 - \Phi(x) &= \int_{t=x}^{\infty} \phi(x) dx \\
&\leq \int_{t=x}^{\infty} \left(1 + \frac{15}{x^6}\right) \phi(x) dx \\
&= \lim_{t\to\infty} f(t) - f(x) \\
&= - f(x) \\
&= \left(\frac{1}{x} - \frac{1}{x^3} + \frac{3}{x^5} \right) \phi(x) .
\end{align*}

Using the next term in the series gives $f(x) = -\left(\frac{1}{x} - \frac{1}{x^3} + \frac{3}{x^5} - \frac{15}{x^7}\right)\phi(x)$ and $\frac{df}{dx} = \left(1 - \frac{105}{x^8}\right)\phi(x)$. Notice because of the alternating positive/negative terms, $\frac{df}{dx}$ is now $\phi$ times something less than one, so plugging it into the same proof gives a **lower** bound on $1 - \Phi(x)$. I have a blog post on the general form of this (sorry to not point you to a more formal reference).