Question 57, Find the range of real number $\mathrm{a}$, such that for any real number $\mathrm{x}$ and any $\theta \in\left[0, \frac{\pi}{2}\right]$, it always holds that $(\mathrm{x}+3+$ $2 \sin \theta \cos \theta)^{2}+(x+a \sin \theta+a \cos \theta)^{2} \geq \frac{1}{8}$.
Question 57, Solution: Note that for any real numbers $c$ and $d$, we have $(c+d)^{2} \geq 0 \Rightarrow c^{2}+d^{2} \geq \frac{(c-d)^{2}}{2}$, with equality holding if and only if $c+d=0$. Therefore,
$$
\begin{array}{l}
(x+3+2 \sin \theta \cos \theta)^{2}+(x+a \sin \theta+a \cos \theta)^{2} \\
\geq \frac{[(x+3+2 \sin \theta \cos \theta)-(x+a \sin \theta+a \cos \theta)]^{2}}{2} \\
=\frac{[(3+2 \sin \theta \cos \theta)-(a \sin \theta+a \cos \theta)]^{2}}{2}
\end{array}
$$

Equality holds if and only if
$$
\begin{array}{l}
x+3+2 \sin \theta \cos \theta+x+a \sin \theta+a \cos \theta=0 \\
\Leftrightarrow x=-\frac{3+2 \sin \theta \cos \theta+a \sin \theta+a \cos \theta}{2}
\end{array}
$$

Thus, we need to find the range of real numbers $\mathrm{a}$ such that:
$\frac{[(3+2 \sin \theta \cos \theta)-(a \sin \theta+a \cos \theta)]^{2}}{2} \geq \frac{1}{8}$, for all $\theta \in\left[0, \frac{\pi}{2}\right]$
$\Leftrightarrow \mathrm{a} \geq \frac{3+2 \sin \theta \cos \theta+\frac{1}{2}}{\sin \theta+\cos \theta}$ or $\mathrm{a} \leq \frac{3+2 \sin \theta \cos \theta-\frac{1}{2}}{\sin \theta+\cos \theta}$, for all $\theta \in\left[0, \frac{\pi}{2}\right]$.
Let $y=\sin \theta+\cos \theta$, then when $\theta \in\left[0, \frac{\pi}{2}\right]$, $y \in[1, \sqrt{2}]$, and $2 \sin \theta \cos \theta=y^{2}-1$, hence
$$
\begin{array}{l}
\max _{\theta \in\left[0, \frac{\pi}{2}\right]} \frac{3+2 \sin \theta \cos \theta+\frac{1}{2}}{\sin \theta+\cos \theta}=\max _{y \in[1, \sqrt{2}]} \frac{3+\left(y^{2}-1\right)+\frac{1}{2}}{y}=\max _{y \in[1, \sqrt{2}]} y+\frac{5}{2 y}=1+\frac{5}{2}=\frac{7}{2} \\
\min _{\theta \in\left[0, \frac{\pi}{2}\right]} \frac{3+2 \sin \theta \cos \theta-\frac{1}{2}}{\sin \theta+\cos \theta}=\min _{y \in[1, \sqrt{2}]} \frac{3+\left(y^{2}-1\right)-\frac{1}{2}}{y}=\min _{y \in[1, \sqrt{2}]} y+\frac{3}{2 y}=2 \sqrt{y \cdot \frac{3}{2 y}}=\sqrt{6}
\end{array}
$$

Therefore, the values of $\mathrm{a}$ that satisfy the condition are $\mathrm{a} \in(-\infty, \sqrt{6}] \cup\left[\frac{7}{2},+\infty\right]$.