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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.

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  • $\begingroup$ Use MVT in the form $\frac{\ln (1+b) - \ln (1+a)}{(1+b) - (1+a)}$ .. $\endgroup$ – r9m Nov 25 '15 at 21:00
  • $\begingroup$ The two inequalities are practically identical; once you've proved one, you've proved the other. See my answer below for a simple explanation of this point. ${}\qquad{}$ $\endgroup$ – Michael Hardy Nov 25 '15 at 21:11
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You apply the mean value theorem to the function $f : x \mapsto \ln (1+x)$ on the interval $[a,b]$. Note that $f$ fulfills the hypothesis of the theorem : being continuous on $[a,b]$ and differentiable on $]a,b[$.

On this interval the function $f'$ is bounded below by $\frac{1}{1+b}$ and above by $\frac{1}{1+a}$, so that $$\frac{1}{1+b} \leq \frac{f(b)-f(a)}{b-a} \leq \frac{1}{1+a}.$$ Replacing $f$ with its explicit definition and $f'$ by what I let you calculate, and "multiplying the inequalities" by $b-a$ gives you the wanted result.

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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 $$ \ln\left(\frac{1+b}{1+a}\right)>\frac{b-a}{1+b}. $$

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We begin from the integral definition of the logarithm function

$$\log x=\int_1^x \frac{1}{u}\,du\tag 1$$

for $x>0$.

Note that $1/x$ is monotonically decreasing on $[1,x]$ for $x\ge 1$ and monotonically increasing on $[x,1]$ for $x\le 1$. Then, from $(1)$ and the Mean Value Theorem for Integrals, along with the Intermediate Value Theorem, we obtain the inequalities

$$\frac{x-1}{x}\le \log x\le x-1 \tag 2$$

Letting $x=(1+b)/(1+a)$ in $(2)$ reveals

$$\frac{b-a}{1+b}\le \log \left(\frac{1+b}{1+a}\right)\le \frac{b-a}{1+a}$$

thereby establishing the coveted inequality!

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  • $\begingroup$ Please let me know how I can improve my answer. I really want to give you the best answer I can. - Mark $\endgroup$ – Mark Viola Nov 27 '15 at 18:29
  • $\begingroup$ Please let me know how I can improve my answer. I really want to give you the best answer I can. - Mark $\endgroup$ – Mark Viola Dec 2 '15 at 0:18

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