$\definecolor{A}{RGB}{190,0,60}\color{A}\fbox{A1.}$ Find all $f:\mathbb{R}\rightarrow \mathbb{R}$ such that $$\definecolor{A}{RGB}{80,0,200}\color{A} x^4+y^4+z^4\ge f(xy)+f(yz)+f(zx)\ge xyz(x+y+z)$$holds for all $a,b,c\in\mathbb{R}$.

[i]Proposed by [/i][b][color=#FFFF00]usjl[/color][/b].
[color=#B6D7A8]#1733[/color]
1. **Initial Substitution**:
   - Substitute \( x = y = z \) into the given inequality:
     \[
     x^4 + y^4 + z^4 \ge f(xy) + f(yz) + f(zx) \ge xyz(x + y + z)
     \]
     This simplifies to:
     \[
     3x^4 \ge 3f(x^2) \ge 3x^3x
     \]
     Dividing by 3, we get:
     \[
     x^4 \ge f(x^2) \ge x^4
     \]
     Therefore, \( f(x^2) = x^4 \) for \( x \ge 0 \). Hence, \( f(x) = x^2 \) for \( x \ge 0 \).

2. **Negative Values**:
   - Substitute \( y = z = -x \):
     \[
     x^4 + (-x)^4 + (-x)^4 \ge f(x(-x)) + f((-x)(-x)) + f((-x)x) \ge x(-x)(-x + -x + x)
     \]
     Simplifying, we get:
     \[
     3x^4 \ge f(-x^2) + f(x^2) + f(-x^2) \ge -x^3x
     \]
     Since \( f(x^2) = x^4 \), we have:
     \[
     3x^4 \ge 2f(-x^2) + x^4 \ge -x^4
     \]
     Simplifying further:
     \[
     2x^4 \ge 2f(-x^2) \ge -x^4
     \]
     Dividing by 2:
     \[
     x^4 \ge f(-x^2) \ge -\frac{x^4}{2}
     \]
     Therefore, \( f(-x^2) \le x^4 \).

3. **General Case**:
   - Substitute \( xy = a, yz = b, zx = c \):
     \[
     f(a) + f(b) + f(c) \ge ab + ac + bc
     \]
     whenever \( abc > 0 \). Choose \( a, b < 0 \) and \( c > 0 \):
     \[
     f(a) + f(b) \ge ab + (a + b)c - c^2
     \]
     Taking the limit as \( c \to 0 \):
     \[
     f(a) + f(b) \ge ab
     \]
     whenever \( a, b < 0 \). For \( a = b \):
     \[
     f(a) \ge \frac{a^2}{2}
     \]

4. **Conclusion**:
   - Any function \( f \) with \( f(x) = x^2 \) for \( x \ge 0 \) and \( f(x) \in \left[\frac{x^2}{2}, x^2\right] \) for \( x < 0 \) satisfies the condition.

The final answer is \( \boxed{ f(x) = x^2 } \) for \( x \ge 0 \) and \( f(x) \in \left[\frac{x^2}{2}, x^2\right] \) for \( x < 0 \).