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 October 28th, 2011, 04:05 PM #1 Member   Joined: Jun 2008 Posts: 45 Thanks: 0 inequality Could anybody give some hint how to prove $\forall n\in\mathbb{N}\qquad\frac{1}{(n+1)\sqrt{n}}<2(\arc tan\sqrt{n}-\arctan\sqrt{n-1})$ Thanks in advance.
 October 29th, 2011, 06:39 AM #2 Senior Member   Joined: Aug 2008 Posts: 113 Thanks: 0 Re: inequality Note that the inequality holds for $n= 1$ Next, consider the function: $f(x)= \frac{1}{(x+1)\sqrt{x}} - 2 \tan^{-1} \sqrt{x} - 2 \tan^{-1} \sqrt{x-1}$ You can easily show that $f'(x) < 0$ for $x > 1$ This completes the proof.
 October 29th, 2011, 11:40 AM #3 Member   Joined: Jun 2008 Posts: 45 Thanks: 0 Re: inequality Thanks. Though you have some mistakes, but first of all I can't understand why the derivative may be something to judge about inequality's correctness.
October 29th, 2011, 12:50 PM   #4
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 Originally Posted by JC Thanks. Though you have some [color=#FF0000]mistakes[/color], but first of all I can't understand why the derivative may be something to judge about inequalitys correctness.
Please point out the [color=#FF0000]mistakes[/color].

 October 29th, 2011, 01:18 PM #5 Member   Joined: Jun 2008 Posts: 45 Thanks: 0 Re: inequality I think you thought f(x) = 1 / ((x+1) sqrt(x)) - 2tan^(-1) sqrt(x) [color=#FF0000]+[/color] 2tan^(-1) sqrt(x-1). This gives f'(x) [color=#FF0000]>[/color] 0 for x > 1.
October 29th, 2011, 02:35 PM   #6
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 Originally Posted by dibyendu This completes the proof.
A slight exaggeration; it's important that $\lim_{x\to\infty}f(x)\,=\,0.$

October 29th, 2011, 04:32 PM   #7
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 Originally Posted by skipjack A slight exaggeration; it's important that $\lim_{x\to\infty}f(x)\,=\,0.$
Skipjack, after this comment everything is clear to me now. I solved it and understood.
Many thanks to dibyendu and skipjack.

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