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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?

  • 1
    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

1 Answers 1

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