11. (20 points) Find all cubic polynomials $P(x)$ such that for all $x, y \geqslant 0$, we have
$$
P(x+y) \geqslant P(x)+P(y) .
$$
11. Let $P(x)=a x^{3}+b x^{2}+c x+d(a \neq 0)$. Then the original inequality is equivalent to
$$
3 a x y(x+y)+2 b x y \geqslant d(\text { for any } x, y \geqslant 0) \text {. }
$$
Let $x, y$ be sufficiently large, we get $a>0$.
Let $x=y=0$, we get $d \leqslant 0$.
Under these conditions, equation (1) can be rewritten as
$$
\begin{aligned}
3 a x^{2} y & +3 a x y^{2}+(-d) \\
& \geqslant(-2 b) x y(\text { for any } x, y \geqslant 0) . \\
& \text { When } 3 \sqrt[3]{9 a^{2}(-d)} \geqslant-2 b, \text { (2) } 8 b^{3} \geqslant 243 a^{2} d
\end{aligned}
$$
By the basic inequality, equation (2) holds.

Conversely, when $3 \sqrt[3]{9 a^{2}(-d)} > 0, d \leqslant 0, 8 b^{3} \geqslant 243 a^{2} d$.