Find the smallest $\alpha>0$ for which there is $\beta(\alpha)>0$ so that the following inequality holds $\forall x\in[0,1]$: $$\sqrt{1+x}+\sqrt{1-x}\le 2-\frac{x^\alpha}\beta$$ and find $\min\beta(a)$ for that particular $a$.
Source: Elezović, N., Odabrani zadatci elementarne matematike, Zagreb, 1992
EDIT: By $\beta(\alpha)$ I denoted a positive $\beta$ depends on $\alpha$. If $\beta>0$ exists for some $\alpha>0$, it doesn't have to be unique.
My thoughts:
Function $f(x)=\sqrt{1+x}+\sqrt{1-x}$ is concave on $(-1,1)$, and particulary on $(0,1)$, which follows from the fact that $$f''(x)=-\frac14\left(\frac1{\sqrt{(1+x)^3}}+\frac1{\sqrt{(1-x)^3}}\right)<0,\space\forall x\in(0,1).$$
Since $f$ is concave, each of its tangents is above the graph $\Gamma(f)$.
My first idea was to observe the polynomial $g(x)=f'(c)(x-c)+f(c),c\in (0,1),\deg g=1,$ which made me suspect that $\alpha\ge 1$.
I noticed that a function $f_2(x)=-\frac{x^\alpha}\beta$ is concave for $\alpha>1,$ convex for $0<\alpha<1$ and both convex and concave for $\alpha=1,$ when $f_2$ is linear. Then, $$\alpha\ge 1\implies f_3(x)=2-\frac{x^\alpha}\beta\text{ is also concave}.$$
and this polynomial $f_3$ and the function $f$ should have at most one intersection and at $x=0$.
However, I'm not able to justify my claim that this is the only possibility.
Also, $f'$ is strictly decreasing on an open interval, and hence, my attempt to express $\min\beta(\alpha)$ in terms of $f'$ in the second part of the task failed.
May I ask for advice on solving this task? Thank you very much in advance!