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I found that I could do this with the Mean Value theorem and also found a similar question here in the stackexchange and Socratic but couldn't solve the right hand side of the inequality.

I figured that by specifically choosing the function and the interval (combinations of the given question) we could prove that. But I am lost tin selecting the appropriate function.

Can you guys please suggest some ideas and also some example sums where I could check these kind of sums.

Thanks

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    $\begingroup$ You can't prove it, because it is not true. Take $x=1$. $\endgroup$
    – Hanul Jeon
    Jan 6 '21 at 5:39
  • $\begingroup$ @Hanul Jeon Oops sorry I wrote the question wrong $\endgroup$
    – Aghilan
    Jan 6 '21 at 5:40
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    $\begingroup$ Substitute $ x = \sin u$ $\endgroup$
    – N.S.JOHN
    Jan 6 '21 at 5:43
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    $\begingroup$ Is this only for x>0? Because the inequality breaks down if x<0. $\endgroup$ Jan 6 '21 at 6:00
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Let $f(x)=x-\sin^{-1} x \implies f'(x)=1-\frac{1}{\sqrt{1-x^2}}=\frac{\sqrt{1-x^2}-1}{\sqrt{1-x^2}}<0$ So $f(x)$ is a decreasing function. $$x\ge 0 \implies f(x)\le f(0) \implies f(x)\le 0$$ Next, we take $g(x)=\sin^{-1} x-\frac{x}{\sqrt{1-x^2}} \implies g'(x)=\frac{-x^2}{\sqrt{1-x^2}}<0.$ So $g(x)$ is decreasing function, then $$ x \ge 0 \implies g(x)\le g(0)=0 \implies \sin^{-1}x \le \frac{x}{\sqrt{1-x^2}}.$$

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  • $\begingroup$ @Z Ahmed I understand the second part but the first part doesn't prove $x< sin^{-1}x$ $\endgroup$
    – Aghilan
    Jan 6 '21 at 7:43
  • $\begingroup$ @Aghilan Oh! thanks now please see I have corrected it. $\endgroup$
    – Z Ahmed
    Jan 6 '21 at 8:12
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For $|x|<1$,

$$1<\frac1{\sqrt{1-x^2}}<\frac1{(1-x^2)^{3/2}}$$ seems obvious. Then by integration between $0$ and $x$ you get the claimed inequalities.

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  • $\begingroup$ thanks for the interesting answer $\endgroup$
    – Aghilan
    Jan 6 '21 at 9:27
  • $\begingroup$ @Aghilan: any analytical proof will be equivalent to this. For geometric proofs, you need to show equivalence of the geometric axioms and analytic ones. $\endgroup$
    – user65203
    Jan 6 '21 at 9:56
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This is pretty standard property of $\arcsin$ which is equivalent to the more famous inequality $\sin x <x<\tan x$ for $x\in(0, \pi/2)$.

A direct proof can be given by noting that if $x\in(0,1)$ and $t\in(0,x)$ then $$1<\frac{1}{\sqrt{1-t^2}}<\frac{1}{\sqrt{1-x^2}}$$ integrating the above with respect to $t$ on interval $[0,x]$ we get $$x<\arcsin x<\frac{x} {\sqrt{1-x^2}}$$ for all $x\in(0,1)$.

You can avoid integrals and achieve the same result via mean value theorem. If $x\in(0,1)$ then by mean value theorem we have $$\arcsin x=\arcsin x - \arcsin 0=\frac {x}{\sqrt{1-c^2}}$$ for some $c\in(0,x)$. But then we can obviously see that $$1<\frac{1}{\sqrt{1-c^2}}<\frac{1}{\sqrt{1-x^2}}$$ Multiplying by positive $x$ we get $$x<\arcsin x<\frac{x} {\sqrt{1-x^2}}$$

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  • $\begingroup$ I just have a small doubt. The c taken must satisfy for any value in the interval of (0,1) right? So we are taking it for some c in the interval (0,x) $\endgroup$
    – Aghilan
    Jan 6 '21 at 12:27
  • $\begingroup$ @Aghilan: yes $c\in(0,x)$ ie $0<c<x<1$. $\endgroup$
    – Paramanand Singh
    Jan 6 '21 at 12:45

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