4.5.3 $\star \star$ 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 $a, b, c, d \in[-1,+\infty$ ).
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, $k=\frac{3}{4}$.
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 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,
$$
4 a^{3}+1 \geqslant 3 a, 4 b^{3}+1 \geqslant 3 b, 4 c^{3}+1 \geqslant 3 c, 4 d^{3}+1 \geqslant 3 d,
$$

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