How to prove, for every $x>0$ $$\dfrac{1}{\ln^2\left(1+\dfrac{1}{x}\right)} \dfrac{1}{(x+1)x}-1>0.$$
The inequality $\dfrac{1}{\ln^2\left(1+\dfrac{1}{x}\right)} \dfrac{1}{(x+1)x}-1>0$
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calculus
inequality
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0Does $ln^2()$ mean square of the logarithm or take the logarithm twice? – 2012-02-08
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0Which operator is between first and second fraction ? – 2012-02-08
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0@pedja multiple – 2012-02-08
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0@Henry $(ln())^2$ – 2012-02-08
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0Define $f(x)$ as : $$f(x)=\frac{1}{x(x+1)}-(\ln(x+1)-\ln x)^2$$ then calculate $f'(x)$ and try to show that $f'(x)>0$ for all $x>0$ . – 2012-02-08
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0Have you heard about: $$ 1-\frac{1}{x} < \log x < x-1$$? – 2012-02-08
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0@Peter I think I will reconsider the problem – 2012-02-08
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0Alternately use: $$ \frac{1}{x+1} \leq \log \left(1+\frac{1}{x}\right) \leq \frac{1}{x}$$ – 2012-02-08
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0@gingerjin: the @-notification system doesn't work the way you think it does. [See this](http://meta.stackexchange.com/questions/43019/how-do-comment-replies-work). In particular, you cannot summon any user who has not already commented on the thread. – 2012-02-09
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0@Peter It seems that I can't work out the problem with the use of the inequality above – 2012-02-18