Example 9 (17th "Hope Cup" Test Question) Given the function
$$
y=f(x)=\sqrt{1+\sqrt{1-x^{2}}}-\sqrt{1+x} .
$$
(1) Find the domain and range of $y$, and prove that $y$ is a monotonically decreasing function;
(2) Solve the inequality $\sqrt{1+\sqrt{1-x^{2}}}-\sqrt{1+x}>\frac{1}{2}$.
Solve (1) From $\left\{\begin{array}{l}1-x^{2} \geqslant 0, \\ 1+x \geqslant 0,\end{array}\right.$ we get $-1 \leqslant x \leqslant 1$,
so the domain of the function $y=f(x)$ is $[-1,1]$.
Since the function $y=f(x)$ is monotonically decreasing on $[-1,1]$ (to be proven),
the range of $y=f(x)$ is $[f(1), f(-1)]$, i.e., $[1-\sqrt{2}, 1]$.
The following proves that $y-f(x)$ is monotonically decreasing on $[-1,1]$.
When $x \in[-1,1]$,
let $x=\sin \theta, \theta \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right], \sin \theta$ is monotonically increasing, and $\frac{\theta}{2} \in\left[-\frac{\pi}{4}, \frac{\pi}{4}\right]$,
so
$$
\begin{aligned}
y & =\sqrt{1+\cos \theta}-\sqrt{1+\sin \theta}=\sqrt{2} \cos \frac{\theta}{2}-\left(\sin \frac{\theta}{2}+\cos \frac{\theta}{2}\right) \\
& =(\sqrt{2}-1) \cos \frac{\theta}{2}-\sin \frac{\theta}{2}=\sqrt{4-2 \sqrt{2}} \cos \left(\frac{\theta}{2}+\varphi\right),
\end{aligned}
$$

where $\varphi=\arctan \frac{1}{\sqrt{2}-1}=\arctan (\sqrt{2}+1)$.
Since $\varphi \in\left(\frac{\pi}{4}, \frac{\pi}{2}\right), \frac{\theta}{2} \in\left[-\frac{\pi}{4}, \frac{\pi}{4}\right]$,
so $\frac{\theta}{2}+\varphi \in\left(0, \frac{3 \pi}{4}\right) \subset[0, \pi]$.
Clearly, $y=\sqrt{4-2 \sqrt{2}} \cos \left(\frac{\theta}{2}+\varphi\right)$ is a monotonically decreasing function of $\theta$, and by the monotonicity of composite functions (an increasing function composed with a decreasing function), $y=f(x)$ is a monotonically decreasing function on $[-1,1]$.
(2) Let $f\left(x_{0}\right)=\frac{1}{2}$,
and since the function $y=f(x)$ is monotonically decreasing on $[-1,1]$,
the solution set of the inequality $\sqrt{1+\sqrt{1-x^{2}}}-\sqrt{1+x}>\frac{1}{2}$ is $-1 \leqslant x<x_{0}$.
Since when $x=0$, $y=\sqrt{2}-1<\frac{1}{2}$, so $x_{0} \in[-1,0)$.
Let $x_{0}=\sin \theta, \theta \in\left[-\frac{\pi}{2}, 0\right), \sqrt{1+\cos \theta}-\sqrt{1+\sin \theta}=\frac{1}{2}$,

squaring both sides gives $2+\cos \theta+\sin \theta-2 \sqrt{(1+\cos \theta)(1+\sin \theta)}=\frac{1}{4}$,
let $\sin \theta+\cos \theta=t\left(\theta \in\left[-\frac{\pi}{2}, 0\right)\right)$, then $\sin \theta \cos \theta=\frac{1}{2}\left(t^{2}-1\right), t \in[-1,1)$,
$2+t-2 \sqrt{1+t+\frac{1}{2}\left(t^{2}-1\right)}=\frac{1}{4}$,
$t^{2}+\frac{7}{2} t+\frac{49}{16}=4\left[1+t+\frac{1}{2}\left(t^{2}-1\right)\right]$,
$t^{2}+\frac{t}{2}-\frac{17}{16}=0$,
solving gives $t=\frac{-\frac{1}{2} \pm \sqrt{\frac{1}{4}+\frac{17}{4}}}{2}=\frac{-1 \pm 3 \sqrt{2}}{4}$.
Since $t \geqslant-1$, discard $t=\frac{-1-3 \sqrt{2}}{4}$.
From $\left\{\begin{array}{l}\sin \theta+\cos \theta=\frac{3 \sqrt{2}-1}{4}, \\ \sin ^{2} \theta+\cos ^{2} \theta=1, \\ -\frac{\pi}{2} \leqslant \theta<0,\end{array}\right.$
we get
$$
x_{0}=\sin \theta=\frac{3 \sqrt{2}-1-\sqrt{13+6 \sqrt{2}}}{8},
$$

the solution set of the original inequality is $\quad\left[-1, \frac{3 \sqrt{2}-1-\sqrt{13+6 \sqrt{2}}}{8}\right)$.