Example 4 Find all real numbers $k$ such that $a^{3}+b^{3}+c^{3}+d^{3}+1 \geqslant k(a+b+c+d)$, 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 $k \geqslant 3$. 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, $k=\frac{3}{4}$. Below is the proof 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, from $(x+1)(2 x-1) \geqslant 0$, 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 these four inequalities, we obtain the desired inequality (1).
Thus, the real number $k=\frac{3}{4}$.