Example 8 (The 1st China Southeast Mathematical Olympiad) Given the inequality
$$
\sqrt{2}(2 a+3) \cos \left(\theta-\frac{\pi}{4}\right)+\frac{6}{\sin \theta+\cos \theta}-2 \sin 2 \theta<3 a+6
$$

holds for all $\theta \in\left[0, \frac{\pi}{2}\right]$, find the range of values for $a$.
Let $\sin \theta + \cos \theta = x$, then $\cos \left(\theta - \frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} x$, $\sin 2\theta = x^2 - 1$, and $x \in [1, \sqrt{2}]$.
Thus, the original inequality can be transformed into $(2a + 3)x + \frac{6}{x} - 2(x^2 - 1) > 0$.
Rearranging gives $(2x - 3)\left(x + \frac{2}{x} - a\right) > 0$ for $x \in [1, \sqrt{2}]$.
Since $x \in [1, \sqrt{2}]$, we have $2x - 3 < 0$ and $\left(x + \frac{2}{x}\right)_{\max}(x \in [1, \sqrt{2}])$.
It is easy to see that $f(x) = x + \frac{2}{x}$ is decreasing on $[1, \sqrt{2}]$, so $\left(x + \frac{2}{x}\right)_{\max} = 3$ for $x \in [1, \sqrt{2}]$.

Therefore, $a > 3$.