Proof: Since the inequality is a homogeneous inequality, we can assume $a+b+c+d=1$. Notice that
$$\begin{array}{l} (a+b)(a+c)(a+d) \\ \leqslant\left[\frac{(a+b)+(a+c)+(a+d)}{3}\right]^{3} \\ =\left(a+\frac{b+c+d}{3}\right)^{3}=\left(a+\frac{1-a}{3}\right)^{3} \\ =\frac{1}{27}(2 a+1)^{3} . \end{array}$$

Therefore, $\frac{a^{3}}{(a+b)(a+c)(a+d)}$
$$\geqslant \searrow \frac{27 a^{3}}{(2 a+1)^{3}}$$

Thus, to prove the original inequality, it suffices to prove
$$\sum \frac{27 a^{3}}{(2 a+1)^{3}} \geqslant \frac{1}{2}$$

Let $f(x)=\frac{27 x^{3}}{(2 x+1)^{3}}(0<x<1)$.
It is easy to see that the tangent line equation of $f(x)$ at $x=\frac{1}{4}$ is
$$g(x)=1 \times\left(x-\frac{1}{4}\right)+\frac{1}{8}=x-\frac{1}{8} .$$

First, we prove: $\frac{27 x^{3}}{(2 x+1)^{3}} \geqslant x-\frac{1}{8}(0<x<1)$.
Notice that
$$\begin{array}{l} \text { Equation (1) } \Leftrightarrow 216 x^{3} \geqslant(8 x-1)(2 x+1)^{3} \\ \Leftrightarrow 64 x^{4}-128 x^{3}+36 x^{2}+2 x-1 \leqslant 0 \\ \Leftrightarrow(4 x-1)^{2}\left(4 x^{2}-6 x-1\right) \leqslant 0 \\ \Leftrightarrow(4 x-1)^{2}[2 x(2 x-3)-1] \leqslant 0 . \end{array}$$

Since $2 x-3<0$, the above inequality clearly holds. Therefore, Equation (1) is established.
$$\begin{array}{l} \text { Hence } \sum \frac{27 a^{3}}{(2 a+1)^{3}} \geqslant \sum a-\frac{1}{8} \sum 1 \\ =1-\frac{1}{8} \times 4=\frac{1}{2} . \end{array}$$

Therefore, the original inequality is proved.