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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

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 $[0,\pi/2)$.2012-05-28
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    Didier, thanks. How to show it? I thought that $$ \tan(u)< u $$2012-05-28
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    @Didier, I'm sorry, I was wrong. And then, after showed?2012-05-28
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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 $v\geqslant0$.2012-05-28
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    See also this question: [Why $x<\tan{x}$ while $0?](http://math.stackexchange.com/questions/98998/why-x-tanx-while-0x-frac-pi2)2012-05-31

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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 $$f'(t)=1-\frac{1}{1+t^2} \ge 0.$$