The answer is $p(x)=ax^{3}+bx^{2}+cx+d$ $a>0$ $d\leq \min\{0,\frac{8b^{3}}{243}\}$ and $a=0,b\geq 0$ Becuase $p(x+y)\geq p(x)+p(y)$ is equivlent to $xy(3a(x+y)+2b)\geq d$ Let$x-\to \infty,y-\to \infty$ we get $a\geq 0$ $x=0$ we get that $d\leq 0$ if $b<0$ Let $x=y=-\frac{2b}{9a}$ then we get that $d\leq \frac{8b^{3}}{243}\}$ and if $a\geq 0$ $d\leq \min\{0,\frac{8b^{3}}{243}\}$ we have ${xy(3a(x+y)+2b)\geq xy(6a\sqrt{xy}+2b}\geq 0\geq d$ when $b\geq 0$ ${xy(3a(x+y)+2b)\geq xy(6a\sqrt{xy}+2b}\geq \frac{8b^{3}}{243}\geq d$

Any other solution? Please help me?

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Hint: If you denote $x+y=S, xy=P$, the problem it is equivalent with $P(3aS+2b)\ge d,\text{ }\forall S,P\ge 0$