Prove that for all positive real numbers $x, y$ and $z$, the double inequality $$0 < \frac{1}{x + y + z + 1} -\frac{1}{(x + 1)(y + 1)(z + 1)} \le \frac18$$ holds. When does equality hold in the right inequality?

[i](Walther Janous)[/i]
1. **Initial Setup and Substitution:**
   Let \( x + y + z = 3u \). This substitution simplifies the expression and allows us to use the Arithmetic Mean-Geometric Mean (AM-GM) inequality effectively.

2. **Applying AM-GM Inequality:**
   By the AM-GM inequality, we have:
   \[
   \frac{x + y + z}{3} \geq \sqrt[3]{xyz}
   \]
   which implies:
   \[
   u \geq \sqrt[3]{\frac{xyz}{3}}
   \]

3. **Simplifying the Expression:**
   We need to show:
   \[
   0 < \frac{1}{x + y + z + 1} - \frac{1}{(x + 1)(y + 1)(z + 1)} \leq \frac{1}{8}
   \]
   Substituting \( x + y + z = 3u \), the expression becomes:
   \[
   \frac{1}{3u + 1} - \frac{1}{(x + 1)(y + 1)(z + 1)}
   \]

4. **Upper Bound Analysis:**
   To find the upper bound, we need to compare:
   \[
   \frac{1}{3u + 1} \quad \text{and} \quad \frac{1}{(u + 1)^3}
   \]
   Since \( (x + 1)(y + 1)(z + 1) \geq (u + 1)^3 \) by the AM-GM inequality, we have:
   \[
   \frac{1}{(x + 1)(y + 1)(z + 1)} \leq \frac{1}{(u + 1)^3}
   \]
   Therefore:
   \[
   \frac{1}{3u + 1} - \frac{1}{(u + 1)^3} \leq \frac{1}{3u + 1} - \frac{1}{(u + 1)^3}
   \]

5. **Bounding the Expression:**
   We need to show that:
   \[
   \frac{1}{3u + 1} - \frac{1}{(u + 1)^3} \leq \frac{1}{8}
   \]
   For \( u = 1 \), we have \( x = y = z = 1 \), and:
   \[
   \frac{1}{3 \cdot 1 + 1} - \frac{1}{(1 + 1)^3} = \frac{1}{4} - \frac{1}{8} = \frac{1}{8}
   \]
   This shows that the upper bound is achieved when \( x = y = z = 1 \).

6. **Lower Bound Analysis:**
   Since \( x, y, z \) are positive real numbers, \( x + y + z + 1 > 0 \) and \( (x + 1)(y + 1)(z + 1) > 0 \), ensuring:
   \[
   \frac{1}{x + y + z + 1} > 0 \quad \text{and} \quad \frac{1}{(x + 1)(y + 1)(z + 1)} > 0
   \]
   Thus:
   \[
   \frac{1}{x + y + z + 1} - \frac{1}{(x + 1)(y + 1)(z + 1)} > 0
   \]

Conclusion:
\[
0 < \frac{1}{x + y + z + 1} - \frac{1}{(x + 1)(y + 1)(z + 1)} \leq \frac{1}{8}
\]
Equality holds in the right inequality when \( x = y = z = 1 \).

The final answer is \( \boxed{ 0 < \frac{1}{x + y + z + 1} - \frac{1}{(x + 1)(y + 1)(z + 1)} \leq \frac{1}{8} } \)