Example 4 Find all real numbers $k$ such that $a^{3}+b^{3}+c^{3}+d^{3}+1 \geqslant k(a+b+c+d)$ holds for all $a, b, c, d \in[-1,+\infty$ ).

(2004 China Western Mathematical Olympiad)
When $a=b=c=d=-1$, we have $-3 \geqslant k(-4)$, so $k \geqslant \frac{3}{4}$. When $a=b=c=d=\frac{1}{2}$, we have $4 \times \frac{1}{8}+1 \geqslant k\left(4 \times \frac{1}{2}\right)$, so $k \leqslant \frac{3}{4}$. Therefore, $k=\frac{3}{4}$. Now we prove that
$$a^{3}+b^{3}+c^{3}+d^{3}+1 \geqslant \frac{3}{4}(a+b+c+d)$$
holds 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 $k=\frac{3}{4}$.