Example 9 (1) When $0 \leqslant x \leqslant 1$, find the range of the function
$f(x)=(\sqrt{1+x}+\sqrt{1+x}+2) \cdot\left(\sqrt{1-x^{2}}+1\right)$.
(2) Prove that when $0 \leqslant x \leqslant 1$, there exists a positive number $\beta$ such that the inequality $\sqrt{1+x}+\sqrt{1-x} \leqslant 2-\frac{x^{a}}{\beta}$ holds for the smallest positive number $\alpha=2$, and find the smallest positive number $\beta$ at this time.
Solution (1) From $1+x \geqslant 0$ and $1-x \geqslant 0$, we have $-1 \leqslant x \leqslant 1$, which also satisfies $1-x^{2} \geqslant 0$. Using the method of magnification and reduction, we have $00$, the inequality
$\sqrt{1+x}+\sqrt{1-x}-2 \leqslant-\frac{x^{\alpha}}{\beta}(x \in[0,1])$ does not hold.
Conversely, that is, $-\frac{2 x^{2}}{f(x)} \leqslant-\frac{x^{a}}{\beta}$,
which means $x^{2-\alpha} \geqslant \frac{f(x)}{2 \beta}$ holds.
Since $2-\alpha>0$, let $x \rightarrow 0$, we get $0 \geqslant \frac{f(0)}{2 \beta}$,
but $f(0)=8$, which is impossible.
This shows that $\alpha=2$ is the smallest positive number that satisfies the condition.
To find the smallest $\beta>0$ that makes the inequality $\sqrt{1+x}+\sqrt{1-x}-2 \leqslant-\frac{x^{\alpha}}{\beta}(x \in[0,1])$, i.e., $-\frac{2 x^{2}}{f(x)} \leqslant-\frac{x^{2}}{\beta}$ hold, is equivalent to finding $\beta=\max _{0 \leqslant x \leqslant 1} \frac{1}{2} f(x)$.
Because, by the Cauchy inequality, for non-negative real numbers $u, v$ we have $\sqrt{u}+\sqrt{v} \leqslant \sqrt{2(u+v)}$.
Let $u=1+x, v=1-x$, we get $\sqrt{1+x}+\sqrt{1-x} \leqslant 2$.
Thus, when $x \in[0,1]$, $\frac{1}{2} f(x)=\frac{1}{2}(\sqrt{1+x}+\sqrt{1-x}+2)\left(\sqrt{1-x^{2}}+1\right) \leqslant$ $2\left(\sqrt{1-x^{2}}+1\right) \leqslant 4$, and $\frac{1}{2} f(0)=4$. Therefore, the maximum value of the function $\frac{1}{2} f(x)$ on $[0,1]$ is 4, i.e., the smallest positive number $\beta$ that satisfies condition (2) is 4.