To prove the inequality

$\frac{a - b}{a} \leq \log \left( \frac{a}{b} \right) \leq \frac{a - b}{b}, \quad \text{for} \quad 0 < b \leq a,$

we can use the Mean Value Theorem (MVT). Here's a step-by-step proof:

Step 1: Define a function involving logarithms

Let $f(x) = \log(x)$, which is a continuous and differentiable function on the interval $(0, \infty)$. Now, apply the Mean Value Theorem to $f(x)$ on the interval $[b, a]$ where $0 < b \leq a$.

According to the MVT, there exists a point $c \in (b, a)$ such that

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

Step 2: Compute the derivative of $f(x)$

The derivative of $f(x) = \log(x)$ is

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

Thus, from the MVT,

$\frac{1}{c} = \frac{\log(a) - \log(b)}{a - b} = \frac{\log\left(\frac{a}{b}\right)}{a - b}.$

This gives us

$\log\left( \frac{a}{b} \right) = \frac{a - b}{c}, \quad \text{for some} \quad c \in (b, a).$

Step 3: Bound $\frac{1}{c}$

Since $c \in (b, a)$, we know that

$b < c < a.$

Taking the reciprocal of $c$, we get

$\frac{1}{a} < \frac{1}{c} < \frac{1}{b}.$

Step 4: Multiply by $a - b$

Multiplying the inequality $\frac{1}{a} < \frac{1}{c} < \frac{1}{b}$ by $a - b$ (which is positive because $a > b$) gives:

$\frac{a - b}{a} < \frac{a - b}{c} < \frac{a - b}{b}.$

Step 5: Conclusion

Since $\frac{a - b}{c} = \log\left( \frac{a}{b} \right)$, we conclude that

$\frac{a - b}{a} \leq \log\left( \frac{a}{b} \right) \leq \frac{a - b}{b}.$

Thus, the inequality is proven.

Do you want more details or have any questions?

Related questions:

1. How does the Mean Value Theorem work for other functions?
2. What is the geometric interpretation of the Mean Value Theorem?
3. Can we generalize this inequality to other logarithmic functions?
4. What happens if $a = b$ in this inequality?
5. How can we apply the Mean Value Theorem to prove other inequalities involving logarithms?

Tip:

The Mean Value Theorem is a powerful tool for proving inequalities, especially when dealing with continuous and differentiable functions.