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 JC October 28th, 2011 04:05 PM

inequality

Could anybody give some hint how to prove

$\forall n\in\mathbb{N}\qquad\frac{1}{(n+1)\sqrt{n}}<2(\arc tan\sqrt{n}-\arctan\sqrt{n-1})$

 dibyendu October 29th, 2011 06:39 AM

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.

 JC October 29th, 2011 11:40 AM

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.

 dibyendu October 29th, 2011 12:50 PM

Re: inequality

Quote:
 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.

 JC October 29th, 2011 01:18 PM

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.

 skipjack October 29th, 2011 02:35 PM

Quote:
 Originally Posted by dibyendu This completes the proof.
A slight exaggeration; it's important that $\lim_{x\to\infty}f(x)\,=\,0.$

 JC October 29th, 2011 04:32 PM

Re:

Quote:
 Originally Posted by skipjack A slight exaggeration; it's important that $\lim_{x\to\infty}f(x)\,=\,0.$
:D 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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