# Prove inequality using Lagrange's Mean Value Theorem [duplicate]

I'm trying to prove

$1 - \frac{a}{b} \leq \ln\frac{b}{a}\leq\frac{a}{b} - 1$ where $0 < a < b$ using Lagrange's Mean Value Theorem.

Applying the theorem to $\ln x$ results in: $$\exists\epsilon\in(a,b): \ln'(\epsilon)(b-a)=\ln b - \ln a$$ $$\frac{b}{\epsilon}-\frac{a}{\epsilon}=\ln \frac{b}{a}$$

This looks very similar to the target inequality (set $\epsilon=a$ and $\epsilon=b$), but I'm not sure how to get to it.

Edit: Looks like my question is an exact duplicate of Mean Value theorem problem?(inequality), The answer doesn't really explain how to get to the inequalities though.

## marked as duplicate by Martin R, Robert Z, Alessandro Codenotti, Clarinetist, FabianJan 5 '18 at 17:20

• Looks like a clear duplicate to me. What are you missing in math.stackexchange.com/a/618533/42969? – Martin R Jan 5 '18 at 14:40
• @MartinR How they get to the inequalities: why is $\frac{b-a}{b} < \ln b - \ln a$? – Todd Sewell Jan 5 '18 at 14:41
• $\ln b - \ln a = \frac{b-a}{c} > \frac{b-a}{b}$ because $a < c < b$. – Martin R Jan 5 '18 at 14:45
• @MartinR Yes, I understand it now, thanks to both you and the answers I got on this question. – Todd Sewell Jan 5 '18 at 14:47
• @ToddSewell If you are ok, you can accept the answer and set as solved. Thanks! – gimusi Jan 8 '18 at 22:40

Note that fro MVT

$$\ln b - \ln a=\ln \frac{b}a=\frac{b-a}{c} \quad c\in(a,b)$$

and varing $c$ between $a$ and $b$

$$1-\frac{a}{b}\leq\frac{b-a}{c}\leq \frac{b}{a}-1$$

thus

$$1-\frac{a}{b}\leq\ln \frac{b}a\leq \frac{b}{a}-1$$

use that $$\frac{\ln(b)-\ln(a)}{b-a}=\frac{1}{\xi}$$ and $$\xi \in (a,b)$$ it is not difficult to complete this, since we have $$0<a<b$$ we get that $$\frac{1}{\xi}<\frac{1}{a}$$ and $$\frac{1}{\xi}>\frac{1}{b}$$

• This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post. - From Review – Arnaud D. Jan 5 '18 at 17:36
• and why is it so like you say? – Dr. Sonnhard Graubner Jan 5 '18 at 17:39
• OP already has that equality written in the question. The question is how to find the inequalities from that equality, and your answer doesn't really adress that. – Arnaud D. Jan 5 '18 at 17:42
• now it does i think – Dr. Sonnhard Graubner Jan 5 '18 at 17:46