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Prove $\frac{x}{x+1}<\ln(x+1)<x$ for $x>0$

What I have done is to look at the derivatives and interval of each function.

lets: $f_1=\frac{x}{x+1}$ , $f_2=\ln(x+1)$, $f_3=x$

the intervals of the functions are $f_1=(0,1)$ , $f_2=(0,\infty)$, $f_3=(0,\infty)$

now looking at the derivatives of the functions $f_1'=\frac{1}{(x+1)^2}$, $f_2'=\frac{1}{x+1}$ , $f_3'=1$

Is this enough to prove the inequalities? is there a way to use the mean value theorem or Rolle's theorem?

gebruiker
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gbox
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  • Comparing the derivatives is using the mean value theorem. – Martin R Mar 21 '16 at 16:26
  • consider the functions $ln(x+1)-\frac{x}{x+1}$ and $x-ln(x+1)$, and note that their derivatives are always positive. to show that they are positive at $x>0$, just put $x=0$ – Kerr Mar 21 '16 at 16:27

2 Answers2

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You have to show that $x<(x+1)\ln(x+1)$ for $x>0$. This equivalent to showing that $x<(x+1)\ln(x+1)$. Taking the derivatives of both sides, we find that $1<\ln(x+1)+1$ for all $x>0$. The rest of the argument is easy.

Mathematician 42
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$$\log(x+1) = \int_{0}^{x}\frac{dt}{t+1} > \int_{0}^{x}\frac{dt}{x+1} = \frac{x}{x+1}.$$

Jack D'Aurizio
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