How to prove an inequality using the Mean value theorem I've been trying to prove that $\frac{b-a}{1+b}<\ln(\frac{1+b}{1+a})<\frac{b-a}{1+a}$ using the Mean value theorem. What I've tried is setting $f(x)=\ln x$ and using the Mean value theorem on the interval  $[1,\frac{1+b}{1+a}]$. I managed to prove that $\ln(\frac{1+b}{1+a})<\frac{b-a}{1+a}$ but not the other part, only that $\frac{1+a}{1+b}<\ln(\frac{1+b}{1+a})$. any help?
p.s: sorry if I have some mistakes in my terminology or so on, I'm not totally fluent in english.

Suppose you know $\displaystyle \ln\left(\frac{1+b}{1+a}\right)<\frac{b-a}{1+a}$.
If you rename $a$ and $b$ so that the number that was called $b$ is now called $a$ and vice-versa, then this says:
$$
\ln\left(\frac{1+a}{1+b}\right)<\frac{a-b}{1+b}.
$$
Multiplying both sides by $-1$ necessitates changing $\text{“}{<}\text{''}$ to $\text{“}{>}\text{''}$ and we get
$$
-\ln\left(\frac{1+a}{1+b}\right)>\frac{b-a}{1+b}.
$$
But $-\log\dfrac p q = \log\dfrac q p$, so that is the same as
$$
\ln\left(\frac{1+b}{1+a}\right)>\frac{b-a}{1+b}.
$$
