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I have to prove that $\ln(x) > 3*[(x-1)]/(x+1)$ for $x > 1$...

Using Lagrange's theoreme... I have no idea where to start because I dont even have an interval? Can you give me a hint?

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    Aren't you working over the interval $(1, \infty)$?2012-12-26
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    Yes but I need an interval like a2012-12-26
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    Are you sure the $3$ should be there?2012-12-26
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    yes.im pretty sure2012-12-26
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    it's written like that in the book I bought -.-2012-12-26
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    @djsakds As I showed in my answer, this does not hold. Your book must have a typo2012-12-26
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    What *exactly* do you call "Lagrange's Theorem" to within this context?2012-12-26
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    @DonAntonio I suppose he means the Mean Value Theorem2012-12-26
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    @Nameless, I'm almost 100% positive he did, yet I think it's a good idea to be more explicit when writing to an international audience. Thanks.2012-12-26
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    @djsakds, what is that book you bought?2012-12-26

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The inequality is not true for say $x=2$. Then you have $$\ln 2>3\frac{2-1}{2+1}=3\frac13=1$$ which is not true because $\ln 2<\ln e=1$

I suspect the correct inequality is $$\ln x>\frac{x-1}{x}\iff \frac{\ln x-\ln 1}{x-1}>\frac{1}{x}$$ If so consider the interval $(1,x)$