Question

using suitable function f(x), obtain the following inequality for a < b < 1 using mean value theorem

Answer 1

Consider f(x)= In x, so we apply the Lagrange Formula to it on the interval [a,b]

$\frac{f(b)-f(a)}{b-a}=f'(c)$ where c $\in(a, b)\\$

Hence

$\frac{Inb-Ina}{b-a}=\frac{1}{c}\\ In\frac{b}{a}=(b-a)\frac{1}{c}\\$

The right side of the equation is minimal when c = b and respectively takes a maximum at c = a, as a result, we get the inequality.

$\frac{b-a}{b}\le In\frac{b}{a}\le\frac{b-a}{a}$