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So I'm having trouble trying to solve this problem:

Let $a$ and $b$ be two real numbers such that $a \geqslant b \geqslant 1$. Using the Mean value Theorem prove that $$\ln\left(\frac{a}{b}\right)\leqslant a-b$$

I honestly don't even know how to start, so I was wondering if someone could please lend me a hand? Thank you very much.

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Take $f(x)=\ln x$ and apply Mean Value Theorem to $f(x)$ on the inerval $[b,a]$. Then there is $c\in(b,a)$ such that $f(a)-f(b)=f'(c)(a-b)$. But, $f'(c)=\frac{1}{c}$ and $1\le b< c< a$, so $\frac{1}{c}<1$. Resumming, $$\ln(a/b)=\ln a-\ln b=\frac{1}{c}(a-b)< a-b.$$ This arguments works if $a<b$. Otherwise, $\ln a-\ln b=0=a-b$ and you have equality.

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If $a=b$, we are done. Otherwise, by the MVT, there is some $c\in(b,a)$ such that $$ \frac{\log(a)-\log(b)}{a-b}=\frac{1}{c}\leq 1 $$ where the last inequality is because $c\geq b\geq 1$.

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  • $\begingroup$ Why c can be equal to b? $\endgroup$ – Guillem Figols Jun 20 at 17:25
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    $\begingroup$ @guillemfigols Oh you can write $c>b$ if you’d like. I just need $c\geq 1$. $\endgroup$ – yurnero Jun 20 at 18:21
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$I:= \displaystyle{\int_{b}^{a}} \dfrac{1}{x}dx= \log a- \log b$;

MVT of integration:

$I= \dfrac{1}{r} \displaystyle{\int_{b}^{a}} 1 dx =(1/r)(a-b)$, where $r \in [b,a]$.

Finally

$\log a -\log b=(1/r)(a-b) \le a-b$, since $r\ge1$.

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