Let $a < b < c < d < e$ be positive integers. Prove that
$$\frac{1}{[a, b]} + \frac{1}{[b, c]} + \frac{1}{[c, d]} + \frac{2}{[d, e]} \le 1$$
where $[x, y]$ is the least common multiple of $x$ and $y$ (e.g., $[6, 10] = 30$). When does equality hold?

1. **Understanding the Problem:**
   We need to prove that for positive integers \(a < b < c < d < e\), the inequality 
   \[
   \frac{1}{[a, b]} + \frac{1}{[b, c]} + \frac{1}{[c, d]} + \frac{2}{[d, e]} \le 1
   \]
   holds, where \([x, y]\) denotes the least common multiple (LCM) of \(x\) and \(y\).

2. **Using Properties of LCM:**
   Recall that the LCM of two numbers \(x\) and \(y\) can be expressed as:
   \[
   [x, y] = \frac{xy}{(x, y)}
   \]
   where \((x, y)\) is the greatest common divisor (GCD) of \(x\) and \(y\).

3. **Bounding the LCM:**
   For \(x > y\), we have:
   \[
   [x, y] \geq 2y
   \]
   This is because the smallest multiple of \(y\) that is greater than \(y\) itself is \(2y\).

4. **Applying the Bound:**
   Using the above property, we can write:
   \[
   \frac{1}{[a, b]} \leq \frac{1}{2b}, \quad \frac{1}{[b, c]} \leq \frac{1}{2c}, \quad \frac{1}{[c, d]} \leq \frac{1}{2d}, \quad \frac{2}{[d, e]} \leq \frac{2}{2e} = \frac{1}{e}
   \]

5. **Summing the Inequalities:**
   Summing these inequalities, we get:
   \[
   \frac{1}{[a, b]} + \frac{1}{[b, c]} + \frac{1}{[c, d]} + \frac{2}{[d, e]} \leq \frac{1}{2b} + \frac{1}{2c} + \frac{1}{2d} + \frac{1}{e}
   \]

6. **Using the Harmonic Series:**
   Since \(a < b < c < d < e\), we know that:
   \[
   \frac{1}{2b} + \frac{1}{2c} + \frac{1}{2d} + \frac{1}{e} \leq \frac{1}{a} \leq 1
   \]
   This is because the sum of the reciprocals of increasing integers is less than or equal to the reciprocal of the smallest integer.

7. **Conclusion:**
   Therefore, we have:
   \[
   \frac{1}{[a, b]} + \frac{1}{[b, c]} + \frac{1}{[c, d]} + \frac{2}{[d, e]} \leq 1
   \]

8. **Equality Condition:**
   Equality holds if:
   \[
   \frac{1}{[a, b]} = \frac{1}{2b}, \quad \frac{1}{[b, c]} = \frac{1}{2c}, \quad \frac{1}{[c, d]} = \frac{1}{2d}, \quad \frac{2}{[d, e]} = \frac{1}{e}
   \]
   This implies:
   \[
   [a, b] = 2b, \quad [b, c] = 2c, \quad [c, d] = 2d, \quad [d, e] = 2e
   \]
   Additionally, \(a = 1\) and \(e = 2d\).

   Considering the constraints, we find the following sets of integers that satisfy the equality condition:
   \[
   (a, b, c, d, e) = (1, 2, 3, 4, 8), (1, 2, 3, 6, 12), (1, 2, 4, 5, 10), (1, 2, 4, 6, 12), (1, 2, 4, 8, 16)
   \]

The final answer is \( \boxed{ (1, 2, 3, 4, 8), (1, 2, 3, 6, 12), (1, 2, 4, 5, 10), (1, 2, 4, 6, 12), (1, 2, 4, 8, 16) } \)