Four. (20 points) Let the polynomial
$$
f(x)=x^{3}+a x^{2}+b x+c(a, b, c \in \mathbf{R}) \text {. }
$$

If for any non-negative real numbers $x, y$, we have
$$
f(x+y) \geqslant f(x)+f(y) \text {, }
$$

find the conditions that $a, b, c$ must satisfy.
For any $x, y \geqslant 0$, we have
$$
\begin{array}{l}
f(x+y) \geqslant f(x)+f(y) \\
\Rightarrow 3 x^{2} y+3 x y^{2}-c \geqslant-2 a x y .
\end{array}
$$

Let $x=y=0$. Substituting into the above inequality gives $c \leqslant 0$.
Assume $x>0, y>0$. Then
$$
\begin{array}{l}
3 x^{2} y+3 x y^{2}+(-c) \\
\geqslant 3 \sqrt[3]{3 x^{2} y \cdot 3 x y^{2}(-c)}=-3 x y \sqrt[3]{9 c},
\end{array}
$$

with equality holding if and only if $x_{0}=y_{0}=-\sqrt[3]{\frac{c}{3}}$.
Thus, $-3 x_{0} y_{0} \sqrt[3]{9 c} \geqslant-2 a x_{0} y_{0} \Rightarrow a \geqslant \frac{3}{2} \sqrt[3]{9 c}$.
When $a \geqslant \frac{3}{2} \sqrt[3]{9 c}, c \leqslant 0$, for any $x, y \geqslant 0$,
$$
3 x^{2} y+3 x y^{2}-c \geqslant-2 a x y,
$$

i.e., $f(x+y) \geqslant f(x)+f(y)$.
In summary, the conditions for $a, b, c$ are
$$
a \geqslant \frac{3}{2} \sqrt[3]{9 c}, c \leqslant 0, b \in \mathbf{R} .
$$