Three, (50 points) Try to find the smallest positive number $a$, such that there exists a positive number $b$, for which when $x \in [0,1]$, the inequality $\sqrt{1-x}+\sqrt{1+x} \leqslant 2 - b x^{a}$ always holds; for the obtained $a$, determine the largest positive number $b$ that satisfies the above inequality.
$$
\begin{array}{l}
\text { III. Solution: }-b x^{2} \geqslant \sqrt{1-x}+\sqrt{1+x}-2 \\
=\frac{(\sqrt{1-x}+\sqrt{1+x}-2)(\sqrt{1-x}+\sqrt{1+x}+2)}{\sqrt{1-x}+\sqrt{1+x+2}} \\
=\frac{2 \sqrt{1-x^{2}}-2}{\sqrt{1-x}+\sqrt{1+x}+2} \\
=\frac{-2 x^{2}}{(\sqrt{1-x}+\sqrt{1+x}+2)\left(\sqrt{1-x^{2}}+1\right)}, \\
\end{array}
$$

To make $\sqrt{1-x}+\sqrt{1+x} \leqslant 2-b x^{2}$,
it is required that $\frac{-2 x^{2}}{(\sqrt{1-x}+\sqrt{1+x}+2)\left(\sqrt{1-x^{2}}+1\right)}$
$$
\begin{aligned}
\leqslant & -b x^{a}, \\
& \Leftrightarrow b x^{a-2} \\
& \leqslant \frac{2}{(\sqrt{1-x}+\sqrt{1+x}+2)\left(\sqrt{1-x^{2}}+1\right)}. \\
& \text { Let } f(x)=(\sqrt{1-x}+\sqrt{1+x}+2)\left(\sqrt{1-x^{2}}+\right.
\end{aligned}
$$
$$
\text { 1), } x \in[0,1] \text {, }
$$

Let $x=\cos 2 \alpha, \alpha \in\left(0, \frac{\pi}{4}\right)$,
then $f(x)=(\sqrt{2} \sin \alpha+\sqrt{2} \cos \alpha+2)(1+\sin 2 \alpha)$
$$
=[\sqrt{2}(\sin \alpha+\cos \alpha)+2](\sin \alpha+\cos \alpha)^{2} \text {. }
$$

Let $t=\sin \alpha+\cos \alpha, t \in[1, \sqrt{2}]$,
then $f(x)=(\sqrt{2} t+2) t^{2}=\sqrt{2} t^{3}+2 t^{2}$ is an increasing function on $t \in[1, \sqrt{2}]$.
$$
\therefore f_{\max }=f(\sqrt{2})=8 \text {, }
$$

Thus, $b x^{a-2} \leqslant \frac{1}{4}$, hence $a=2, b=\frac{1}{4}$.
Next, we prove $a=2$,
Assume there exists $a<2$ and some positive number $b$ such that $\sqrt{1-x}+$
$$
\sqrt{1+x}-2=\frac{-2 x^{2}}{f(x)} \leqslant-b x^{2} \text {, }
$$

then $x^{a-2} \geqslant \frac{b}{2} f(x)$, let $x=0$, then $0 \geqslant 4 b$, which is a contradiction, hence $a=2$.

Finally, we prove $b=\frac{1}{4}$, by $\sqrt{1-x}+\sqrt{1+x}-2 \leqslant-b x^{2}, x$ $\in[0,1]$,
i.e., $b \leqslant \frac{2}{f(x)} \Rightarrow b \leqslant \min \frac{2}{f(x)}=\frac{1}{4}$.