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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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    @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 $(1,x)$