Thursday, December 14, 2017

calculus - Showing $frac{x}{1+x}

I want to show that $$\frac{x}{1+x}<\log(1+x)0$ using the mean value theorem. I tried to prove the two inequalities separately.


$$\frac{x}{1+x}<\log(1+x) \Leftrightarrow \frac{x}{1+x} -\log(1+x) <0$$


Let $$f(x) = \frac{x}{1+x} -\log(1+x).$$ Since $$f(0)=0$$ and $$f'(x)= \frac{1}{(1+x)^2}-\frac{1}{1+x}<0$$ for all $x > 0$, $f(x)<0$ for all $x>0$. Is this correct so far?



I go on with the second part: Let $f(x) = \log(x+1)$. Choose $a=0$ and $x>0$ so that there is, according to the mean value theorem, an $x_0$ between $a$ and $x$ with


$f'(x_0)=\frac{f(x)-f(a)}{x-a} \Leftrightarrow \frac{1}{x_0+1}=\frac{ \log(x+1)}{x}$.


Since $$x_0>0 \Rightarrow \frac{1}{x_0+1}<1.$$ $$\Rightarrow 1 > \frac{1}{x_0+1}= \frac{ \log(x+1)}{x} \Rightarrow x> \log(x+1)$$

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