If $a = b = c = d = -1$, then $-3 \ge k \cdot (-4)$. Hence $k \ge \frac{3}{4}$.

If $a = b = c = d = \frac{1}{2}$, then $4 \cdot \frac{1}{8} + 1 \ge k \cdot (4 \cdot \frac{1}{2})$. Thus $k \le \frac{3}{4}$, and so $k = \frac{3}{4}$.

At first, we prove that $4x^3 + 1 \ge 3x$, $x \in [-1, +\infty)$.
In fact, from $(x+1)(2x-1)^2 \ge 0$, we have $4x^3 + 1 \ge 3x$, $x \in [-1, +\infty)$.
Therefore
$$4a^3 + 1 \ge 3a,$$
$$4b^3 + 1 \ge 3b,$$
$$4c^3 + 1 \ge 3c,$$
$$4d^3 + 1 \ge 3d.$$
By adding the above 4 inequalities together, we get the inequality (1).