The non-negative numbers $x,y,z$ satisfy the relation $x + y+ z = 3$. Find the smallest possible numerical value and the largest possible numerical value for the expression
$$E(x,y, z) = \sqrt{x(y + 3)} + \sqrt{y(z + 3)} + \sqrt{z(x + 3)} .$$
1. **Upper Bound:**
   - We start by considering the expression \( E(x, y, z) = \sqrt{x(y + 3)} + \sqrt{y(z + 3)} + \sqrt{z(x + 3)} \).
   - To find an upper bound, we use the Arithmetic Mean-Geometric Mean (AM-GM) inequality. For any non-negative numbers \(a\) and \(b\), the AM-GM inequality states that:
     \[
     \sqrt{ab} \leq \frac{a + b}{2}
     \]
   - Applying this to each term in \(E(x, y, z)\), we get:
     \[
     \sqrt{x(y + 3)} \leq \frac{x + (y + 3)}{2} = \frac{x + y + 3}{2}
     \]
   - Similarly,
     \[
     \sqrt{y(z + 3)} \leq \frac{y + (z + 3)}{2} = \frac{y + z + 3}{2}
     \]
     \[
     \sqrt{z(x + 3)} \leq \frac{z + (x + 3)}{2} = \frac{z + x + 3}{2}
     \]
   - Summing these inequalities, we get:
     \[
     E(x, y, z) = \sqrt{x(y + 3)} + \sqrt{y(z + 3)} + \sqrt{z(x + 3)} \leq \frac{x + y + 3}{2} + \frac{y + z + 3}{2} + \frac{z + x + 3}{2}
     \]
     \[
     E(x, y, z) \leq \frac{(x + y + 3) + (y + z + 3) + (z + x + 3)}{2} = \frac{2(x + y + z) + 9}{2}
     \]
   - Given that \(x + y + z = 3\), we have:
     \[
     E(x, y, z) \leq \frac{2 \cdot 3 + 9}{2} = \frac{15}{2} = 7.5
     \]
   - However, we need to check if this bound is tight. By evaluating \(E(1, 1, 1)\), we get:
     \[
     E(1, 1, 1) = \sqrt{1(1 + 3)} + \sqrt{1(1 + 3)} + \sqrt{1(1 + 3)} = \sqrt{4} + \sqrt{4} + \sqrt{4} = 2 + 2 + 2 = 6
     \]
   - Therefore, the upper bound is indeed \(6\).

2. **Lower Bound:**
   - To find the lower bound, we start by squaring the expression \(E(x, y, z)\):
     \[
     E(x, y, z)^2 = \left( \sqrt{x(y + 3)} + \sqrt{y(z + 3)} + \sqrt{z(x + 3)} \right)^2
     \]
   - Using the Cauchy-Schwarz inequality, we have:
     \[
     \left( \sqrt{x(y + 3)} + \sqrt{y(z + 3)} + \sqrt{z(x + 3)} \right)^2 \leq (x + y + z)((y + 3) + (z + 3) + (x + 3))
     \]
   - Given \(x + y + z = 3\), we get:
     \[
     E(x, y, z)^2 \leq 3 \cdot (x + y + z + 9) = 3 \cdot 12 = 36
     \]
   - This inequality is not tight. Instead, we use the fact that:
     \[
     E(x, y, z)^2 \geq 3(x + y + z) = 3 \cdot 3 = 9
     \]
   - Therefore, we have:
     \[
     E(x, y, z) \geq \sqrt{9} = 3
     \]
   - By evaluating \(E(0, 0, 3)\), we get:
     \[
     E(0, 0, 3) = \sqrt{0(0 + 3)} + \sqrt{0(3 + 3)} + \sqrt{3(0 + 3)} = 0 + 0 + \sqrt{9} = 3
     \]
   - Therefore, the lower bound is indeed \(3\).

Conclusion:
\[
3 \leq E(x, y, z) \leq 6
\]

The final answer is \( \boxed{ 3 \leq E(x, y, z) \leq 6 } \)