6. Let the vectors $\boldsymbol{\alpha}=(x+3, x), \boldsymbol{\beta}=(2 \sin \theta \cos \theta, a \sin \theta+a \cos \theta)$ satisfy that for any $x \in \mathbb{R}$ and $\theta \in\left[0, \frac{\pi}{2}\right]$, we have $|\boldsymbol{\alpha}+\boldsymbol{\beta}| \geq \sqrt{2}$ always holds. Then the range of the real number $a$ is $\qquad$
6. $a \leq 1$ or $a \geq 5$

Analysis: Let $\sin \theta + \cos \theta = t$, then $t \in [1, \sqrt{2}]$, $2 \sin \theta \cos \theta = t^2 - 1$, $\vec{\alpha} + \vec{\beta} = (t^2 + x + 2, x + at)$,
since $(t^2 + x + 2)^2 + (x + at)^2 \geq \frac{1}{2}(t^2 + x + 2 - x - at)^2 = \frac{1}{2}(t^2 - at + 2)^2$,
therefore, $|\vec{\alpha} + \vec{\beta}| \geq \sqrt{2} \Leftrightarrow (t^2 + x + 2)^2 + (x + at)^2 \geq 2$ for any $x \in \mathbb{R}$ always holds $\Leftrightarrow \frac{1}{2}(t^2 - at + 2)^2 \geq 2 \Leftrightarrow t^2 - at \geq 0$ or $t^2 - at + 4 \leq 0 \Leftrightarrow a \leq t$ or $a \geq t + \frac{4}{t}$ for any $t \in [1, \sqrt{2}]$ always holds $\Leftrightarrow a \leq 1$ or $a \geq 5$.