# 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.

• what do you mean B(a)... is B a function of a?
– Sid
Mar 15, 2021 at 13:08
• @Sid, I meant that $\beta$ depends on $\alpha$. Just like $\delta$ for some $\varepsilon$. (: Mar 15, 2021 at 13:20
• yep good question... i graphed it on desmos and it looks like there are some values of a and b where it holds. I'm curious to find the minimum as well!
– Sid
Mar 15, 2021 at 13:33

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).

• Nice, one of fastest way to get the solution is if we take the Taylor serious of $\sqrt{1+x}+\sqrt{1-x}$ which turns out to be $$2-\frac{x^2}{4}-\frac{5x^4}{64}-\frac{21 x^6}{512} - \cdots$$ Mar 15, 2021 at 18:52
• @AderinsolaJoshua: ... which happens to be what I did ... :) Mar 15, 2021 at 19:00
• @AderinsolaJoshua I just want to avoid series even limit, though this looks strange as we need calculus to define $x^\alpha$ for real number $\alpha$? Mar 16, 2021 at 0:58

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 $$x$$.