Solution:
From the Cauchy-Schwarz Inequality, we obtain
$$(a+b+c+d)^{2} \leqslant (a^{3}+b+c+d)\left(\frac{1}{a}+b+c+d\right)$$
Using this, together with the other three analogous inequalities, we get
$$\begin{aligned} \frac{1}{a^{3}+b+c+d}+\frac{1}{a+b^{3}+c+d}+\frac{1}{a+b+c^{3}+d} & +\frac{1}{a+b+c+d^{3}} \\ & \leqslant \frac{3(a+b+c+d)+\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\right)}{(a+b+c+d)^{2}} \end{aligned}$$
So it suffices to prove that
$$(a+b+c+d)^{3} \geqslant 12(a+b+c+d)+4\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\right)$$
or equivalently, that
$$\begin{aligned} & \left(a^{3}+b^{3}+c^{3}+d^{3}\right)+3 \sum a^{2} b+6(a b c+a b d+a c d+b c d) \\ & \quad \geqslant 12(a+b+c+d)+4(a b c+a b d+a c d+b c d) \end{aligned}$$
(Here, the sum is over all possible $x^{2} y$ with $x, y \in \{a, b, c, d\}$ and $x \neq y$.) From the AM-GM Inequality we have
$a^{3}+a^{2} b+a^{2} b+a^{2} c+a^{2} c+a^{2} d+a^{2} d+b^{2} a+c^{2} a+d^{2} a+b c d+b c d \geqslant 12 \sqrt[12]{a^{18} b^{6} c^{6} d^{6}}=12 a$.
Similarly, we get three more inequalities. Adding them together gives the inequality we wanted. Equality holds if and only if $a=b=c=d=1$.