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$$a>0.$$ $$\sin{a}\leq\frac{a}{\sqrt{1+\frac{a^2}{k}}} $$ Find the minimum of $k$. It’s obvious that we only need to prove the inequality holds when $a\in\left(0 , \frac{\pi}{2}\right)$.

I can prove that the inequality holds when$ a \in \left(0, \frac{\pi}{4}\right)$ and $k=4$. But I am not sure that $k=4$ is the minimum and also works when $a \in \left(\frac{\pi}{4}, \frac{\pi}{2}\right)$.

When $a \in \left(0, \frac{\pi}{4}\right)$:

$\sin{a} \leq a \leq \tan{a}$

Hence $\cos{a} \leq \frac{\sin{a}}{a} \leq \cos{\frac{a}{2}}$

$\cos^2{\frac{a}{2}}=1-\sin^2{\frac{a}{2}} \leq 1-\frac{a^2}{2}$

$\Rightarrow \cos^2{\frac{a}{2}} \leq \frac{1}{1+\frac{a^2}{4}}$

Hence $ \sin{a}\leq\frac{a}{\sqrt{1+\frac{a^2}{4}}}$

Any ideas?

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  • $\begingroup$ Welcome! Is $0\cdot\sin a=0$ right? $\endgroup$ – manooooh Feb 16 at 15:04
  • $\begingroup$ I’m not sure what you mean. Can you please show me more clearly? $\endgroup$ – qsa Feb 16 at 15:15
  • $\begingroup$ Yes, sorry. You wrote $0.\sin a$, which is equal to $0$ for all values of $a$. It seems that you did not want to write that expression, so I am asking if you wanted to write that. $\endgroup$ – manooooh Feb 16 at 15:17
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    $\begingroup$ k=3is the answer. Use a graphing tool and explore why! $\endgroup$ – John. P Feb 16 at 15:23
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    $\begingroup$ Isolate k and look at the resulting function of a. $\endgroup$ – marty cohen Feb 16 at 15:29
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Now that an answer has been accepted, I'll spell out the method I hinted at in a comment, leaving only a couple of details to fill in. (I haven't even checked the details myself, so it would be unwise to take my word for them!)

For all $a \in \left(0, \frac{\pi}{2}\right)$, the terms of the alternating series $$ \sin a = a - \frac{a^3}{6} + \frac{a^5}{120} - \cdots $$ decrease in absolute value. For instance, $\left(\frac{\pi}{2}\right)^2 < 6$. (Check the other terms.) Therefore: \begin{equation} \label{3115111:eq:1}\tag{1} a - \frac{a^3}{6} < \sin a < a - \frac{a^3}{6} + \frac{a^5}{120} \quad \left(0 < a < \frac{\pi}{2} \right). \end{equation}

The right hand side of the desired inequality also has a convergent series expansion, by the generalised binomial theorem: \begin{equation} \label{3115111:eq:2}\tag{2} a\left(1 + \frac{a^2}{k}\right)^{-1/2} = a - \frac{a^3}{2k} + \frac{3a^5}{8k^2} - \frac{5a^7}{16k^3} + \cdots \quad (0 < a < \sqrt{k}). \end{equation} One must check - I haven't! - that the terms of this series, too, alternate in sign and decrease in absolute value. Then: $$ a - \frac{a^3}{2k} < \frac{a}{\sqrt{1+\frac{a^2}{k}}} < a - \frac{a^3}{2k} + \frac{3a^5}{8k^2} \quad (0 < a < \sqrt{k}). $$ If $k < 3$, then for small enough $a$, we will have: $$ \frac{3a^5}{8k^2} < \frac{a^3}{2k} - \frac{a^3}{6}, $$ whence: $$ \frac{a}{\sqrt{1+\frac{a^2}{k}}} < \sin a, $$ i.e. the inequality is false for those values of $a$.

So, we must have $k \geqslant 3$.

From \eqref{3115111:eq:2}: $$ \frac{a}{\sqrt{1+\frac{a^2}{3}}} > a - \frac{a^3}{6} + \frac{a^5}{24} - \frac{5a^7}{432} \quad (0 < a < \sqrt3). $$ The desired inequality follows from this in conjunction with \eqref{3115111:eq:1}, so long as: $$ \frac{a^5}{24} - \frac{5a^7}{432} > \frac{a^5}{120}. $$ This reduces to: $$ a^2 < \frac{72}{25}, \text{ i.e. } a < \frac{6\sqrt2}{5}. $$ This holds - just! - for all $a \in \left(0, \frac{\pi}{2}\right)$. Therefore, the desired inequality is true when $k = 3$.

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  • $\begingroup$ Thank you for your answer sincerely! $\endgroup$ – qsa Feb 17 at 1:58
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My solution might not be that elegant, but using calculus is a way to analyze graphs. Let $f(x)=\sin x - \frac{x}{\sqrt{1+\frac{x^2}{k}}}$. Then, $$\frac{d}{dx}f(x)= \cos x- \frac{1}{(\frac{k+x^2}{k})^{\frac{3}{2}}}$$ We want $f'(x)\le 0$ when $x \ge 0$, and since $f(0)=f'(0)=0, f''(x)$ should be negative. $$f''(x)=\frac{3x}{k(\frac{k+x^2}{k})^{\frac{5}{2}}}-\sin x, f''(0)=0$$ In same way, $f'''(x)$should be negative around $x=0$. $$f'''(x)=\frac{3(k-4x^2)}{(k+x^2)^2(\frac{k+x^2}{k})^{\frac{3}{2}}}-\cos x , f'''(0)=\frac{3}{k}-1$$ Therefore $\frac{3}{k}-1\le0, k\ge3$

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  • $\begingroup$ Thank you for your answer sincerely! $\endgroup$ – qsa Feb 17 at 1:58

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