If $a, b, c \in [-1, 1]$ satisfy $a + b + c + abc = 0$, prove that $a^2 + b^2 + c^2 \ge  3(a + b + c)$ . 
When does the equality hold?

1. **Given Condition**: We start with the given condition \(a + b + c + abc = 0\) where \(a, b, c \in [-1, 1]\).

2. **Case Analysis**: We will consider different cases based on the values of \(a, b, c\).

3. **Case 1: One of \(a, b, c\) is zero**:
   - Without loss of generality, assume \(c = 0\).
   - Then the equation simplifies to \(a + b = 0\).
   - Therefore, \(b = -a\).
   - We need to prove \(a^2 + b^2 + c^2 \ge 3(a + b + c)\).
   - Substituting \(b = -a\) and \(c = 0\), we get:
     \[
     a^2 + (-a)^2 + 0^2 = 2a^2
     \]
     \[
     3(a + (-a) + 0) = 3 \cdot 0 = 0
     \]
     \[
     2a^2 \ge 0
     \]
   - This inequality is always true since \(a^2 \ge 0\).

4. **Case 2: \(abc \ge 0\)**:
   - If \(abc \ge 0\), then \(a, b, c\) must all be non-negative or all non-positive.
   - If \(a, b, c \ge 0\), then \(a + b + c \ge 0\).
   - We need to prove \(a^2 + b^2 + c^2 \ge 3(a + b + c)\).
   - Since \(a, b, c \in [-1, 1]\), the maximum value of \(a + b + c\) is 3.
   - However, since \(a + b + c + abc = 0\), if \(a, b, c\) are all positive, \(abc\) must be negative, which contradicts \(abc \ge 0\).
   - Therefore, this case is not possible.

5. **Case 3: \(abc < 0\)**:
   - Assume \(a, b > 0\) and \(c < 0\).
   - Let \(c' = -c > 0\).
   - The equation becomes \(a + b - c' - abc' = 0\).
   - We need to prove \(a^2 + b^2 + c'^2 \ge 3(a + b - c')\).
   - Using the AM-GM inequality:
     \[
     a^2 + b^2 + c'^2 \ge 3\sqrt[3]{a^2 b^2 c'^2}
     \]
   - Since \(a, b, c' \in [0, 1]\), we have:
     \[
     \sqrt[3]{a^2 b^2 c'^2} \le \sqrt[3]{1} = 1
     \]
   - Therefore:
     \[
     a^2 + b^2 + c'^2 \ge 3 \cdot 1 = 3
     \]
   - We need to show \(3 \ge 3(a + b - c')\).
   - Since \(a + b - c' \le 1 + 1 - 0 = 2\), the inequality holds.

6. **Equality Condition**:
   - Equality holds when \(a = b = 1\) and \(c = -1\).
   - Substituting these values:
     \[
     a + b + c + abc = 1 + 1 - 1 + 1 \cdot 1 \cdot (-1) = 0
     \]
     \[
     a^2 + b^2 + c^2 = 1^2 + 1^2 + (-1)^2 = 3
     \]
     \[
     3(a + b + c) = 3(1 + 1 - 1) = 3
     \]
   - Therefore, equality holds when \((a, b, c) = (1, 1, -1)\).

The final answer is \( \boxed{ a^2 + b^2 + c^2 \ge 3(a + b + c) } \) with equality when \((a, b, c) = (1, 1, -1)\).