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How to prove this inequality for any $x$ and $n$?

$$ \left|\arctan\frac 1{nx}\right| \leq \frac 1{nx} ;\, 0<x<+{\infty} $$

Is this bounded? But how that can help me in proving? I mean that I don't know the interval of boundedness..

Please tell me how to prove this inequality?

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Show that $\tan(u)\geqslant u$ for every $u$ in $[0,\pi/2)$. –  Did May 28 '12 at 16:46
    
Didier, thanks. How to show it? I thought that $$ \tan(u)< u $$ –  Kamil Hismatullin May 28 '12 at 17:14
    
@Didier, I'm sorry, I was wrong. And then, after showed? –  Kamil Hismatullin May 28 '12 at 17:29
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Once you know that $\tan(u)\geqslant u$ for every $u$ in $[0,\pi/2)$, deduce from this an inequality between $\arctan(v)$ and $v$, valid for every $v\geqslant0$. –  Did May 28 '12 at 17:34
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See also this question: Why $x<\tan{x}$ while $0<x<\frac{\pi}{2}$? –  Martin Sleziak May 31 '12 at 6:06
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1 Answer

up vote 7 down vote accepted

Let us look at $\arctan t$, say for $t \ge 0$. We would like to show that $\arctan t\le t$.

The standard approach is to let $f(t)=t-\arctan t$, and note that $$f'(t)=1-\frac{1}{1+t^2} \ge 0.$$

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