Example 1 Find the smallest positive number $\alpha$, such that there exists a positive number $\beta$, for which the inequality
$$
\sqrt{1+x}+\sqrt{1-x} \leqslant 2-\frac{x^{\alpha}}{\beta}
$$
holds for $0 \leqslant x \leqslant 1$.
For any $x \in [0,1]$, there is an identity
$$
\begin{array}{l}
(\sqrt{1+x}+\sqrt{1-x}-2)(\sqrt{1+x}+ \\
\sqrt{1-x}+2)\left(\sqrt{1-x^{2}}+1\right)=-2 x^{2}
\end{array}
$$

Since on the closed interval
$$
\begin{aligned}
0 & 0$, the following similar inequality holds:
$$
\sqrt{1+x}+\sqrt{1-x}-2 \leqslant-\frac{x^{\alpha}}{\beta}, x \in[0,1],
$$

i.e., $-\frac{x^{2}}{h(x)} \leqslant-\frac{x^{\alpha}}{\beta}$,
then $x^{2-\alpha} \geqslant \frac{h(x)}{\beta}(x \in[0,1])$.
Let $x \rightarrow 0$, we get $0 \geqslant \frac{h(0)}{\beta}$. But $h(0)=4$. The contradiction obtained indicates that $\alpha=2$ is the smallest number satisfying condition (1).
For the inequality
$$
\sqrt{1+x}+\sqrt{1-x}-2 \leqslant-\frac{x^{2}}{\beta}(x \in[0,1])
$$

i.e., $-\frac{x^{2}}{h(x)} \leqslant-\frac{x^{2}}{\beta}$ to hold, the smallest positive integer $\beta$ is the smallest positive integer $\beta$ that satisfies $h(x) \leqslant \beta(x \in[0,1])$. Therefore, $\beta=\max _{0 \leqslant x \leqslant 1} h(x)$.
For any $u, v \geqslant 0$, we have
$$
\sqrt{u}+\sqrt{v} \leqslant \sqrt{2(u+v)} \text {. }
$$

In (3), let $u=1+x, v=1-x$, we get
$$
\sqrt{1+x}+\sqrt{1-x} \leqslant 2 \text {. }
$$

Thus, we obtain again
$$
\begin{aligned}
h(x)= & \frac{1}{2}(\sqrt{1+x}+\sqrt{1-x}+2) \\
& \cdot\left(\sqrt{1-x^{2}}+1\right) \\
\leqslant & 2\left(\sqrt{1-x^{2}}+1\right) \leqslant 4 .
\end{aligned}
$$

On the other hand, it has already been pointed out that $h(0)=4$, therefore, $\max _{0 \leq x \leq 1} h(x)=4$.
Hence, the smallest positive number $\beta=4$ that satisfies (2).
In summary, the smallest positive number $\alpha=2$ that satisfies the conditions, and at this time, there must exist $\beta$ (in fact, $\beta \geqslant 4$) such that the original inequality holds.