Example 12 (Olympiad Training Question from "Intermediate Mathematics" Issue 2, 2003) Find the range of real numbers $a$ such that the inequality $\sin 2 \theta-(2 \sqrt{2}+\sqrt{2} a) \cdot \sin \left(\theta+\frac{\pi}{4}\right)-\frac{2 \sqrt{2}}{\cos \left(\theta-\frac{\pi}{4}\right)}>-3-2 a$ holds for all $\theta \in\left[0, \frac{\pi}{2}\right]$.
Let $x=\sin \theta+\cos \theta, \theta \in\left[0, \frac{\pi}{2}\right]$, then $x \in[1, \sqrt{2}]$.
Since $\sin 2 \theta=x^{2}-1, \sin \left(\theta+\frac{\pi}{4}\right)=\cos \left(\theta-\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2} x$,
the original inequality becomes $x^{2}-1-(2+a) x-\frac{4}{x}+3+2 a>0$,
which simplifies to $(x-2)\left(x+\frac{2}{x}-a\right)>0$.
Given $x \in[1, \sqrt{2}]$, then $x+\frac{2}{x}-ax+\frac{2}{x}, x \in[1, \sqrt{2}]$.
Let $f(x)=x+\frac{2}{x}$, we know that $f(x)$ is monotonically decreasing on $[1, \sqrt{2}]$, thus $f(x)_{\max }=f(1)=3$. Therefore, $a>f(x)_{\max }$, which means $a>3$.