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!

  • 1
    $\begingroup$ what do you mean B(a)... is B a function of a? $\endgroup$
    – Sid
    Mar 15, 2021 at 13:08
  • $\begingroup$ @Sid, I meant that $\beta$ depends on $\alpha$. Just like $\delta$ for some $\varepsilon$. (: $\endgroup$ Mar 15, 2021 at 13:20
  • 1
    $\begingroup$ 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! $\endgroup$
    – Sid
    Mar 15, 2021 at 13:33

2 Answers 2


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).

  • $\begingroup$ 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 $$ 2-\frac{x^2}{4}-\frac{5x^4}{64}-\frac{21 x^6}{512} - \cdots $$ $\endgroup$ Mar 15, 2021 at 18:52
  • 2
    $\begingroup$ @AderinsolaJoshua: ... which happens to be what I did ... :) $\endgroup$
    – Martin R
    Mar 15, 2021 at 19:00
  • $\begingroup$ @AderinsolaJoshua I just want to avoid series even limit, though this looks strange as we need calculus to define $x^\alpha$ for real number $\alpha$? $\endgroup$
    – River Li
    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 $x$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.