Three, find all real numbers $k$ such that the inequality
$$
a^{3}+b^{3}+c^{3}+d^{3}+1 \geqslant k(a+b+c+d)
$$

holds for all $a, b, c, d \in[-1,+\infty)$.
(Xu Wanyi, problem contributor)
When $a=b=c=d=-1$, we have
$-3 \geqslant k \times(-4) \Rightarrow k \geqslant \frac{3}{4}$;
When $a=b=c=d=\frac{1}{2}$, we have
$4 \times \frac{1}{8}+1 \geqslant k \times\left(4 \times \frac{1}{2}\right) \Rightarrow k \leqslant \frac{3}{4}$.
Therefore, $k=\frac{3}{4}$.
Next, we prove the inequality
$$
a^{3}+b^{3}+c^{3}+d^{3}+1 \geqslant \frac{3}{4}(a+b+c+d)
$$

for any $a, b, c, d \in[-1,+\infty)$.
First, we prove $4 x^{3}+1 \geqslant 3 x, x \in[-1,+\infty)$.
In fact, from $(x+1)(2 x-1)^{2} \geqslant 0$, we get
$$
4 x^{3}+1 \geqslant 3 x, x \in[-1,+\infty) \text {. }
$$

Therefore, $4 a^{3}+1 \geqslant 3 a, 4 b^{3}+1 \geqslant 3 b$,
$$
4 c^{3}+1 \geqslant 3 c, 4 d^{3}+1 \geqslant 3 d \text {. }
$$

Adding the above 4 inequalities, we obtain inequality (1).
Thus, the desired real number $k=\frac{3}{4}$.