Finding the smallest $\alpha>0$ for which $\exists\beta(\alpha)>0$ so that $\sqrt{1+x}+\sqrt{1-x}\le 2-\frac{x^\alpha}\beta,\forall x\in[0,1]$.

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!

Solution 1:

Clearly, $2 - \sqrt{1 + x} - \sqrt{1 - x} > 0$ for all $x\in (0, 1]$.

We have, for all $x \in (0, 1]$,
\begin{align} \beta &\ge \frac{x^\alpha}{2 - \sqrt{1 + x} - \sqrt{1 - x}}\\ &= \frac{x^{\alpha - 2}}{4}(2 + \sqrt{1 + x} + \sqrt{1 - x}) (2 + 2\sqrt{1 - x^2})\\ &\triangleq f(x). \end{align} If $0 < \alpha < 2$, then $f(x) \to \infty$ as $x\to 0^{+}$.
If $\alpha \ge 2$, then $f(x) \le \frac{1}{4}(2 + \sqrt{2} + 1)(2 + 2)$ for all $x \in (0, 1]$.

Thus, $\alpha = 2$ is the smallest possible $\alpha$, with the corresponding smallest possible $\beta = 4$ (not difficult to obtain).

Solution 2:

For $0 \le x < 1$ we have from the binomial series $$ f(x) = (1+x)^{1/2} + (1-x)^{1/2} = \sum_{n=0}^\infty \binom{1/2}{n} (x^n + (-x)^n) \\ = 2 \sum_{k=0}^\infty \binom{1/2}{2k}x^{2k} \le 2 - \frac 14 x^2 $$ since $\binom{1/2}{2k}$ is negative for all $k \ge 1$. This shows that $\alpha = 2$ works, with $\beta = 4$ as the smallest possible $\beta$.

It also shows that $\alpha = 2$ is the smallest possible $\alpha$: For $0 < \alpha < 2$ and any $\beta > 0$ is $$ f(x) - \left( 2-\frac{x^\alpha}{\beta} \right) = \frac {1}{\beta} x^\alpha-\frac 14 x^2 + O(x^4) \\ = x^\alpha \left( \frac 1 \beta - \frac 14 x^{2-\alpha} + O(x^{4-\alpha})\right) $$ and that is strictly positive for sufficiently small $x$.