Mean value theorem inequality problem.

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.

3 Answers

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 $$a. Otherwise, $$\ln a-\ln b=0=a-b$$ and you have equality.

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 $$c\geq b\geq 1$$.

• Why c can be equal to b? – Guillem Figols Jun 20 at 17:25
• @guillemfigols Oh you can write $c>b$ if you’d like. I just need $c\geq 1$. – yurnero Jun 20 at 18:21

$$I:= \displaystyle{\int_{b}^{a}} \dfrac{1}{x}dx= \log a- \log b$$;

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 $$r\ge1$$.