Inequalities via MVT

[want2learn]

Hello, I am not sure how to approach/prove this inequality using the Mean Value Thm. The problem states, "Prove x/(1+x) < log (1+x) < x, if -1< x< 0 or 0 < x".

I subtracted the log (1+x) from x/(1+x) to try and get the inequality to resemble the MVT f(b)-f(a)= (b-a)f'(x). I also let f(x)= log(1+x) a=-1 b=0. As a result, I got x/(1+x)-log(1+x) <0<x AND (x/(1+x)-log(1+x))/0-(-1)=1/(1+x). I'm not sure how to proceed from here; please any guidance will be greatly appreciated. Thanks.

 

[pka]

Hello, I am not sure how to approach/prove this inequality using the Mean Value Thm. The problem states, "Prove x/(1+x) < log (1+x) < x, if -1< x< 0 or 0 < x".

There is a well known inequality: \(\displaystyle 0 < a < b \Rightarrow \;\dfrac{1}{b} < \dfrac{{\log (b) - \log (a)}}{{b - a}} < \dfrac{1}{a}\).
The proof is done using the mean value theorem. You can let \(\displaystyle a = 1\;\& \;b = x + 1\).
That will do half of it for you.

For the part \(\displaystyle -1<x<0\), you will have to find a variation of that proof.

 

[want2learn]

Thank you for your guidance.

Greetings PKA,

Thank you very much for your assistance. I will try this approach. Have a nice day.