Observe that
$$\begin{aligned} & 4(a^{3} + b^{3} + c^{3} + d^{3}) - (a^{4} + b^{4} + c^{4} + d^{4}) \\ &= -((a - 1)^{4} + (b - 1)^{4} + (c - 1)^{4} + (d - 1)^{4}) \\ & \quad +6(a^{2} + b^{2} + c^{2} + d^{2}) - 4(a + b + c + d) + 4 \\ &= -((a - 1)^{4} + (b - 1)^{4} + (c - 1)^{4} + (d - 1)^{4}) + 52. \end{aligned}$$
Let $x = a - 1, y = b - 1, z = c - 1, t = d - 1$, then it suffices to prove:
under the condition
$$x^2 + y^2 + z^2 + t^2 = 4 \qquad (1)$$
the following inequality holds
$$16 \geq x^4 + y^4 + z^4 + t^4 \geq 4.$$
By the power mean inequality we get
$$x^{4} + y^{4} + z^{4} + t^{4} \geq \frac{(x^{2} + y^{2} + z^{2} + t^{2})^{2}}{4} = 4 \quad (\text{by (1)}).$$
Next,
$$(x^2 + y^2 + z^2 + t^2)^2 = (x^4 + y^4 + z^4 + t^4) + q,$$
where $q$ is a nonnegative real number, so
$$x^{4} + y^{4} + z^{4} + t^{4} \leq (x^{2} + y^{2} + z^{2} + t^{2})^{2} = 16.$$