Find all real numbers $c$ such that there exists a function $f: \mathbb{R}_{ \ge 0} \rightarrow \mathbb{R}$ which satisfies the following.

For all nonnegative reals $x, y$, $f(x+y^2) \ge cf(x)+y$.

Here $\mathbb{R}_{\ge 0}$ is the set of all nonnegative reals.
1. **Case \( c \le 0 \):**
   - Consider the function \( f(x) = \sqrt{x} \).
   - For \( x, y \ge 0 \), we need to check if \( f(x + y^2) \ge cf(x) + y \).
   - Substituting \( f(x) = \sqrt{x} \), we get:
     \[
     \sqrt{x + y^2} \ge c\sqrt{x} + y
     \]
   - Since \( c \le 0 \), \( c\sqrt{x} \le 0 \). Therefore, the inequality simplifies to:
     \[
     \sqrt{x + y^2} \ge y
     \]
   - This is always true because \( \sqrt{x + y^2} \ge \sqrt{y^2} = y \).

2. **Case \( 0 < c < 1 \):**
   - Consider the function \( f(x) = \sqrt{\frac{x}{1 - c^2}} \).
   - For \( x, y \ge 0 \), we need to check if \( f(x + y^2) \ge cf(x) + y \).
   - Substituting \( f(x) = \sqrt{\frac{x}{1 - c^2}} \), we get:
     \[
     \sqrt{\frac{x + y^2}{1 - c^2}} \ge c\sqrt{\frac{x}{1 - c^2}} + y
     \]
   - Squaring both sides, we obtain:
     \[
     \frac{x + y^2}{1 - c^2} \ge c^2 \frac{x}{1 - c^2} + 2cy\sqrt{\frac{x}{1 - c^2}} + y^2
     \]
   - Simplifying, we get:
     \[
     x + y^2 \ge c^2 x + (1 - c^2) y^2 + 2cy\sqrt{\frac{x(1 - c^2)}{1 - c^2}}
     \]
     \[
     x + y^2 \ge c^2 x + y^2 + 2cy\sqrt{x}
     \]
   - Since \( 0 < c < 1 \), the term \( 2cy\sqrt{x} \) is non-negative, and the inequality holds.

3. **Case \( c \ge 1 \):**
   - For \( c > 1 \), plugging \( y = 0 \) into the original inequality gives:
     \[
     f(x) \ge cf(x)
     \]
   - This implies:
     \[
     f(x) (1 - c) \le 0
     \]
   - Since \( c > 1 \), \( 1 - c < 0 \), so \( f(x) \le 0 \). However, if \( f(x) \le 0 \) for all \( x \), then:
     \[
     f(y^2) \ge y + cf(0)
     \]
   - For large \( y \), this implies \( f(y^2) \ge y + cf(0) \), which contradicts \( f(x) \le 0 \).

4. **Case \( c = 1 \):**
   - We show that \( f(x) \ge \sqrt{nx} + f(0) \) for all \( n \in \mathbb{N} \) and \( x \ge 0 \) by induction.
   - Base case \( n = 1 \):
     \[
     f(x) \ge \sqrt{x} + f(0)
     \]
   - Assume true for \( n \), i.e., \( f(x) \ge \sqrt{nx} + f(0) \).
   - For \( n + 1 \):
     \[
     f(t) \ge f(x) + \sqrt{t - x} \ge \sqrt{nx} + \sqrt{t - x} + f(0)
     \]
   - Setting \( t = x + y^2 \) and choosing \( x = \frac{nt}{n+1} \), we get:
     \[
     f(t) \ge \sqrt{(n+1)t} + f(0)
     \]
   - This shows \( f(1) \) cannot exist, leading to a contradiction.

Therefore, the only valid values for \( c \) are \( c < 1 \).

The final answer is \( \boxed{ c < 1 } \).