Find all functions $f:\mathbb{R}\to \mathbb{R}$ such that for all real numbers $a,b,$ and $c$:
 
(i) If $a+b+c\ge 0$ then $f(a^3)+f(b^3)+f(c^3)\ge 3f(abc).$
 
(ii) If $a+b+c\le 0$ then $f(a^3)+f(b^3)+f(c^3)\le 3f(abc).$
 
[i]Proposed by Ashwin Sah[/i]

1. **Initial Assumptions and Simplification:**
   - We are given two conditions for the function \( f: \mathbb{R} \to \mathbb{R} \):
     - If \( a + b + c \ge 0 \), then \( f(a^3) + f(b^3) + f(c^3) \ge 3f(abc) \).
     - If \( a + b + c \le 0 \), then \( f(a^3) + f(b^3) + f(c^3) \le 3f(abc) \).
   - We hypothesize that \( f(x) = mx + n \) for some constants \( m \ge 0 \) and \( n \).

2. **Verification of Hypothesis:**
   - Assume \( f(x) = mx + n \). Then:
     \[
     f(a^3) + f(b^3) + f(c^3) = m(a^3 + b^3 + c^3) + 3n
     \]
     \[
     3f(abc) = 3m(abc) + 3n
     \]
   - Subtracting these, we get:
     \[
     f(a^3) + f(b^3) + f(c^3) - 3f(abc) = m(a^3 + b^3 + c^3 - 3abc)
     \]
   - Using the identity \( a^3 + b^3 + c^3 - 3abc = \frac{1}{2}(a + b + c)((a - b)^2 + (b - c)^2 + (c - a)^2) \), we have:
     \[
     f(a^3) + f(b^3) + f(c^3) - 3f(abc) = \frac{1}{2} m (a + b + c)((a - b)^2 + (b - c)^2 + (c - a)^2)
     \]
   - Since \( m \ge 0 \), this expression is nonnegative when \( a + b + c \ge 0 \) and nonpositive when \( a + b + c \le 0 \), satisfying the given conditions.

3. **Function Properties:**
   - Let \( P(a, b, c) \) denote the assertion of the functional condition.
   - Note that if \( f \) is a solution, then \( f + k \) is also a solution. Thus, we can assume \( f(0) = 0 \) without loss of generality.

4. **Monotonicity:**
   - **Claim 1:** \( f \) is increasing.
     - Proof: Consider \( P(a, -b, 0) \) where \( a \ge b \). Then, we have \( f(a) \ge f(b) \).

5. **Odd Function:**
   - **Claim 2:** \( f \) is odd.
     - Proof: Consider \( P(a^{1/3}, -a^{1/3}, 0) \). This gives \( f(-a) = -f(a) \).

6. **Homogeneity:**
   - **Claim 3:** \( f(\phi^{3i} \ell) = \phi^{3i} f(\ell) \) for all \( i \in \mathbb{Z} \) and \( \ell \in \mathbb{R}^+ \).
     - Proof: Let \( \phi = \frac{\sqrt{5} + 1}{2} \) be the golden ratio. Using the recurrence relation derived from the functional equation, we solve for \( f \) in terms of \( \phi \).

7. **Scaling Property:**
   - **Claim 4:** \( f(2^i \ell) = 2^i f(\ell) \) for all \( i \in \mathbb{Z} \) and \( \ell \in \mathbb{R}^+ \).
     - Proof: Using a similar approach as in Claim 3, we derive a recurrence relation for powers of 2 and solve it.

8. **Conclusion:**
   - Since \( f \) is odd and satisfies the scaling properties, we conclude that \( f(n) = nf(1) \) for all \( n \in \mathbb{R} \).
   - Therefore, \( f(x) = mx \) for some \( m \ge 0 \).

The final answer is \( \boxed{ f(x) = mx } \) for \( m \ge 0 \).