3. 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 $0 \leqslant x \leqslant 1$.
3. For any $x \in [0,1]$, there is an identity
$(\sqrt{1+x}+\sqrt{1-x}-2)(\sqrt{1+x}+\sqrt{1-x}+2)\left(\sqrt{1-x^{2}}+1\right)=-2 x^{2}$ holds.
Since on the closed interval
$$
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00$, the following similar inequality holds:
$\sqrt{1+x}+\sqrt{1-x}-2 \leqslant-\frac{x^{2}}{\beta}, x \in[0,1]$,

If $-\frac{x^{2}}{h(x)} \leqslant-\frac{x^{a}}{\beta}$, 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 $\sqrt{1+x}+\sqrt{1-x}-2 \leqslant-\frac{x^{2}}{\beta}(x \in[0,1])$,

If $-\frac{x^{2}}{h(x)} \leqslant-\frac{x^{2}}{\beta}$ holds, the smallest positive integer $\beta$ is the smallest positive integer $\beta$ that satisfies $h(x) \leqslant \beta(x \in[0,1])$. Therefore, $\beta=$ $\max _{0 \leq 1 \leq 1} h(x)$. For any $u 、 v \geqslant 0$, we have $\sqrt{u}+\sqrt{v} \leqslant \sqrt{2(u+v)}$.
In (3), let $u=1+x, v=1-x$, we get $\sqrt{1+x}+\sqrt{1-x} \leqslant 2$. On the other hand, it has been pointed out that $h(0)=4$, thus, $\max _{0 \leqslant x \leqslant 1} h(x)=4$. Hence, the smallest positive number $\beta=4$ that satisfies (2).
In summary, the smallest positive number $\alpha=2$ that satisfies the condition, and at this time, there must exist $\beta$ (in fact, $\beta \geqslant 4$) such that the original inequality holds.