Example 15 (1996 National High School League Question) Find the range of real numbers $a$ such that for any $\theta \in[0$, $\left.\frac{\pi}{2}\right]$, it always holds that $(x+3+2 \sin \theta \cdot \cos \theta)^{2}+(x+a \sin \theta+a \cos \theta)^{2} \geqslant \frac{1}{8}$.
Let $y=x$, then we have
$$
\left\{\begin{array}{l}
x-y=0, \\
{[x-(-3-2 \sin \theta \cdot \cos \theta)]^{2}+[y-(-a \sin \theta-a \cos \theta)]^{2} \geqslant \frac{1}{8} .}
\end{array}\right.
$$

Thus, the original problem is transformed into considering the square of the distance from the point $(-3-2 \sin \theta \cdot \cos \theta, -a \sin \theta - a \cos \theta)$ to the line $y=x$ being greater than $\frac{1}{8}$ when finding the range of $a$ (where $\theta \in\left[0, \frac{\pi}{2}\right]$).
By the distance formula, we can find that $|-3-2 \sin \theta \cdot \cos \theta + a \sin \theta + a \cos \theta| \geqslant \frac{1}{2}$.
Let $\sin \theta + \cos \theta = t$, then $t = \sqrt{2} \sin \left(\theta + \frac{\pi}{4}\right) \in [1, \sqrt{2}]$.
The original inequality can be transformed into $\left|-3 - (t^2 - 1) + at\right| \geqslant \frac{1}{2}$,
Solving this inequality, we find that $a \geqslant \frac{7}{2}$ or $a \leqslant \sqrt{6}$.