Let's break down the problem using Lagrange's Mean Value Theorem (LMVT).

Given Problem:

Using Lagrange's mean value theorem, prove that:

$\frac{b-a}{b} < \ln\left(\frac{b}{a}\right) < \frac{b-a}{a}$

for $0 < a < b$. Hence, show that:

$\frac{1}{4} < \ln\left(\frac{4}{3}\right) < \frac{1}{3}$

Step 1: Apply Lagrange’s Mean Value Theorem

Lagrange's Mean Value Theorem states that if a function $f(x)$ is continuous on a closed interval $[a,b]$ and differentiable on the open interval $(a,b)$, then there exists a point $c \in (a, b)$ such that:

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

Here, consider the function $f(x) = \ln(x)$. This function is continuous and differentiable on $(0, \infty)$. Applying LMVT to $f(x) = \ln(x)$ over the interval $[a, b]$, we get:

$f'(c) = \frac{\ln(b) - \ln(a)}{b - a}$

Now, the derivative of $\ln(x)$ is:

$f'(x) = \frac{1}{x}$

Therefore, at some point $c$ in $(a, b)$, we have:

[ \frac{1}{c} = \frac