Let $d(n)$ be the number of all positive divisors of a natural number $n \ge 2$.
Determine all natural numbers $n \ge 3$ such that  $d(n -1) + d(n) + d(n + 1) \le  8$.

[i]Proposed by Richard Henner[/i]
1. **Initial Assumption and Observation**:
   We are given that \( n \geq 3 \) and need to find all natural numbers \( n \geq 3 \) such that \( d(n-1) + d(n) + d(n+1) \leq 8 \). Here, \( d(n) \) denotes the number of positive divisors of \( n \).

2. **Prime Number Consideration**:
   Since \( n \geq 3 \), at least one of the numbers \( \{n-1, n, n+1\} \) must be a prime number. This is because if none of them were prime, they would all have more than 1 divisor, making it difficult to satisfy the inequality \( d(n-1) + d(n) + d(n+1) \leq 8 \).

3. **Case Analysis**:
   We will analyze the cases where there is exactly one prime and where there are exactly two primes in the set \( \{n-1, n, n+1\} \).

4. **Case 1: Exactly One Prime**:
   Suppose there is only one prime in \( \{n-1, n, n+1\} \). The other two numbers must be composite. However, if \( n-1 \) and \( n+1 \) are both composite, they must have at least 3 divisors each (since they are not prime). This would make \( d(n-1) + d(n) + d(n+1) \geq 1 + 3 + 3 = 7 \), which is close to the upper limit. We need to check specific values to see if this can work.

5. **Case 2: Exactly Two Primes**:
   Suppose there are exactly two primes in \( \{n-1, n, n+1\} \). This means one of the numbers is composite and the other two are prime. The composite number must have at least 2 divisors. Let's analyze this case further:
   - If \( n-1 \) and \( n+1 \) are primes, then \( n \) must be composite.
   - If \( n \) is composite, it can have 2, 3, or 4 divisors (since \( n \) must be greater than 2).

6. **Checking Specific Values**:
   - For \( n = 3 \):
     \[
     d(2) + d(3) + d(4) = 2 + 2 + 3 = 7
     \]
     This satisfies the inequality.
   - For \( n = 4 \):
     \[
     d(3) + d(4) + d(5) = 2 + 3 + 2 = 7
     \]
     This satisfies the inequality.
   - For \( n = 6 \):
     \[
     d(5) + d(6) + d(7) = 2 + 4 + 2 = 8
     \]
     This satisfies the inequality.

7. **Exclusion of Other Values**:
   - For \( n = 5 \):
     \[
     d(4) + d(5) + d(6) = 3 + 2 + 4 = 9
     \]
     This does not satisfy the inequality.
   - For \( n > 6 \), the number of divisors for \( n \), \( n-1 \), and \( n+1 \) will generally increase, making it difficult to satisfy the inequality \( d(n-1) + d(n) + d(n+1) \leq 8 \).

Conclusion:
The only values of \( n \geq 3 \) that satisfy the given condition are \( n = 3, 4, 6 \).

The final answer is \( \boxed{n = 3, 4, 6} \).