Find all real numbers $a$ such that the inequality $3x^2 + y^2 \ge  -ax(x + y)$ holds for all real numbers $x$ and $y$.
1. We start with the given inequality:
   \[
   3x^2 + y^2 \geq -ax(x + y)
   \]
   We need to find the values of \(a\) for which this inequality holds for all real numbers \(x\) and \(y\).

2. First, consider the case when \(x = 0\):
   \[
   3(0)^2 + y^2 \geq -a(0)(0 + y) \implies y^2 \geq 0
   \]
   This inequality is always true for any real number \(y\).

3. Now, assume \(x \neq 0\). We divide both sides of the inequality by \(x^2\):
   \[
   \frac{3x^2 + y^2}{x^2} \geq \frac{-ax(x + y)}{x^2}
   \]
   Simplifying, we get:
   \[
   3 + \left(\frac{y}{x}\right)^2 \geq -a \left(1 + \frac{y}{x}\right)
   \]
   Let \(z = \frac{y}{x}\). The inequality becomes:
   \[
   3 + z^2 \geq -a(1 + z)
   \]

4. Rearrange the inequality:
   \[
   3 + z^2 + a(1 + z) \geq 0
   \]
   \[
   z^2 + az + (3 + a) \geq 0
   \]

5. To ensure this inequality holds for all real numbers \(z\), the quadratic expression \(z^2 + az + (3 + a)\) must be non-negative for all \(z\). This is true if and only if the discriminant of the quadratic is non-positive:
   \[
   \Delta = a^2 - 4(3 + a) \leq 0
   \]
   Simplifying the discriminant:
   \[
   a^2 - 12 - 4a \leq 0
   \]
   \[
   a^2 - 4a - 12 \leq 0
   \]

6. Solve the quadratic inequality \(a^2 - 4a - 12 \leq 0\):
   \[
   a^2 - 4a - 12 = 0
   \]
   Using the quadratic formula \(a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):
   \[
   a = \frac{4 \pm \sqrt{16 + 48}}{2} = \frac{4 \pm 8}{2}
   \]
   \[
   a = 6 \quad \text{or} \quad a = -2
   \]

7. The roots of the quadratic equation are \(a = -2\) and \(a = 6\). The quadratic \(a^2 - 4a - 12\) is a parabola opening upwards, so it is non-positive between the roots:
   \[
   -2 \leq a \leq 6
   \]

Conclusion:
The inequality \(3x^2 + y^2 \geq -ax(x + y)\) holds for all real numbers \(x\) and \(y\) if and only if:
\[
-2 \leq a \leq 6
\]

The final answer is \( \boxed{ -2 \leq a \leq 6 } \)