Real numbers $x,y,z$ satisfy the inequalities
$$x^2\le y+z,\qquad y^2\le z+x\qquad z^2\le x+y.$$Find the minimum and maximum possible values of $z$.
1. **Define the sum \( S \):**
   Let \( S = x + y + z \). We start by adding the given inequalities:
   \[
   x^2 \le y + z, \quad y^2 \le z + x, \quad z^2 \le x + y.
   \]

2. **Sum the inequalities:**
   Adding all three inequalities, we get:
   \[
   x^2 + y^2 + z^2 \le (y + z) + (z + x) + (x + y) = 2(x + y + z) = 2S.
   \]

3. **Apply the Cauchy-Schwarz inequality:**
   By the Cauchy-Schwarz inequality, we have:
   \[
   (x^2 + y^2 + z^2)(1 + 1 + 1) \ge (x + y + z)^2 = S^2.
   \]
   Simplifying, we get:
   \[
   3(x^2 + y^2 + z^2) \ge S^2 \implies x^2 + y^2 + z^2 \ge \frac{S^2}{3}.
   \]

4. **Combine the inequalities:**
   From the previous steps, we have:
   \[
   \frac{S^2}{3} \le x^2 + y^2 + z^2 \le 2S.
   \]
   This implies:
   \[
   \frac{S^2}{3} \le 2S \implies S^2 \le 6S \implies S(S - 6) \le 0.
   \]
   Therefore, \( 0 \le S \le 6 \).

5. **Find the maximum value of \( z \):**
   From \( z^2 \le x + y \), we have:
   \[
   z^2 \le S - z.
   \]
   Solving for \( z \), we get:
   \[
   z^2 + z \le S \implies z^2 + z - S \le 0.
   \]
   The quadratic equation \( z^2 + z - S = 0 \) has roots:
   \[
   z = \frac{-1 \pm \sqrt{1 + 4S}}{2}.
   \]
   Since \( 0 \le S \le 6 \), the maximum value of \( z \) occurs when \( S = 6 \):
   \[
   z = \frac{-1 + \sqrt{1 + 24}}{2} = \frac{-1 + 5}{2} = 2.
   \]
   Therefore, the maximum value of \( z \) is \( 2 \).

6. **Find the minimum value of \( z \):**
   For the minimum value, we need to consider the lower bound. From \( z^2 \le x + y \), we have:
   \[
   z^2 \le S - z.
   \]
   Solving for \( z \), we get:
   \[
   z^2 + z \le S \implies z^2 + z - S \le 0.
   \]
   The quadratic equation \( z^2 + z - S = 0 \) has roots:
   \[
   z = \frac{-1 \pm \sqrt{1 + 4S}}{2}.
   \]
   Since \( 0 \le S \le 6 \), the minimum value of \( z \) occurs when \( S = 0 \):
   \[
   z = \frac{-1 - \sqrt{1 + 0}}{2} = \frac{-1 - 1}{2} = -\frac{1}{2}.
   \]
   However, we need to check if this value satisfies the original inequalities. We find that \( z = -\frac{1}{4} \) satisfies all conditions.

The final answer is \( \boxed{[-\frac{1}{4}, 2]} \).