3 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)
Prove that when $a=b=c=d=-1$, we have $-3 \geqslant k \cdot(-4)$, so $k \geqslant \frac{3}{4}$.
When $a=b=c=d=\frac{1}{2}$, we have $4 \cdot \frac{1}{8}+1 \geqslant k \cdot\left(4 \cdot \frac{1}{2}\right)$, so $k \leqslant \frac{3}{4}$. Therefore, $k=\frac{3}{4}$.
Now, 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 that $4 x^{3}+1 \geqslant 3 x, x \in[-1,+\infty)$.
In fact, by $(x+1)(2 x-1)^{2} \geqslant 0$, we get $4 x^{3}+1 \geqslant 3 x, x \in$ $[-1,+\infty)$.
Therefore,
$$
\begin{array}{l}
4 a^{3}+1 \geqslant 3 a \\
4 b^{3}+1 \geqslant 3 b \\
4 c^{3}+1 \geqslant 3 c \\
4 d^{3}+1 \geqslant 3 d
\end{array}
$$

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