Determine all pairs $(a, b)$ of positive real numbers with $a \neq 1$ such that
\[\log_a b < \log_{a+1} (b + 1).\]
To determine all pairs \((a, b)\) of positive real numbers with \(a \neq 1\) such that
\[
\log_a b < \log_{a+1} (b + 1),
\]
we will analyze the given inequality step-by-step.

1. **Rewrite the logarithmic inequality using the change of base formula:**
   \[
   \log_a b = \frac{\log b}{\log a} \quad \text{and} \quad \log_{a+1} (b + 1) = \frac{\log (b + 1)}{\log (a + 1)}.
   \]
   Thus, the inequality becomes:
   \[
   \frac{\log b}{\log a} < \frac{\log (b + 1)}{\log (a + 1)}.
   \]

2. **Consider the function \(f(x) = \frac{\log (x + 1)}{\log x}\):**
   We need to analyze the behavior of this function to understand the inequality. 

3. **Compute the derivative \(f'(x)\) to determine the monotonicity of \(f(x)\):**
   \[
   f(x) = \frac{\log (x + 1)}{\log x}.
   \]
   Using the quotient rule, we have:
   \[
   f'(x) = \frac{(\log x)' \cdot \log (x + 1) - (\log (x + 1))' \cdot \log x}{(\log x)^2}.
   \]
   Simplifying the derivatives:
   \[
   (\log x)' = \frac{1}{x} \quad \text{and} \quad (\log (x + 1))' = \frac{1}{x + 1}.
   \]
   Therefore,
   \[
   f'(x) = \frac{\frac{\log (x + 1)}{x} - \frac{\log x}{x + 1}}{(\log x)^2}.
   \]

4. **Analyze the sign of \(f'(x)\):**
   To determine whether \(f(x)\) is increasing or decreasing, we need to analyze the numerator:
   \[
   \frac{\log (x + 1)}{x} - \frac{\log x}{x + 1}.
   \]
   Since \(\log (x + 1) \geq \log x\) and \(\frac{1}{x} \geq \frac{1}{x + 1}\), it follows that:
   \[
   \frac{\log (x + 1)}{x} \geq \frac{\log x}{x + 1}.
   \]
   Hence, \(f'(x) \leq 0\), indicating that \(f(x)\) is a decreasing function.

5. **Evaluate \(f(x)\) at critical points:**
   - For \(x \leq 1\), \(f(x) \leq 0\).
   - For \(x \geq 1\), \(f(x) \geq 0\).

6. **Apply the decreasing property of \(f(x)\) to the inequality:**
   Since \(f(x)\) is decreasing, if \(b \geq 1\), then \(f(b) \geq f(a)\). If \(b \leq 1\), then \(f(b) \geq f(a)\).

7. **Determine the pairs \((a, b)\) that satisfy the inequality:**
   - If \(b = 1\), then \(a\) can be any positive real number except 1.
   - If \(a > b > 1\), the inequality holds.
   - If \(b > 1 > a\), the inequality holds.

Thus, the solutions are:
\[
(b = 1, a \in \mathbb{R}^+ \setminus \{1\}) \quad \text{and} \quad (a > b > 1) \quad \text{and} \quad (b > 1 > a).
\]

The final answer is \( \boxed{ (b = 1, a \in \mathbb{R}^+ \setminus \{1\}) \quad \text{and} \quad (a > b > 1) \quad \text{and} \quad (b > 1 > a) } \).