First, prove the case where $a, b, c, d$ are all non-zero.
From the given theorem: $2(a b+a c+a d+b c+b d+c d) \leqslant 3\left(a^{2}+b^{2}+c^{2}+d^{2}\right)$.
$$\begin{aligned} & \text { The left side of (1) }=\frac{a^{4}}{a b+a c+a d}+\frac{b^{4}}{b c+b d+a b} \\ + & \frac{c^{4}}{c d+a c+b c}+\frac{d^{4}}{a d+b d+c d} \\ \geqslant & \frac{\left(a^{2}+b^{2}+c^{2}+d^{2}\right)^{2}}{2(a b+a c+a d+b c+b d+c d)} \text { (Cauchy inequality } \end{aligned}$$
$$\begin{array}{l} \geqslant \frac{\left(a^{2}+b^{2}+c^{2}+d^{2}\right)^{2}}{3\left(a^{2}+b^{2}+c^{2}+d^{2}\right)}=\frac{a^{2}+b^{2}+c^{2}+d^{2}}{3} \\ = \\ \frac{\frac{1}{2}\left[\left(a^{2}+b^{2}\right)+\left(b^{2}+c^{2}\right)+\left(c^{2}+d^{2}\right)+\left(d^{2}+a^{2}\right)\right]}{3} \\ \geqslant \frac{1}{3} \end{array}$$

If one or two of $a, b, c, d$ are zero, then by removing the zero variables and following the same method, the inequality can still be proven.

Editor's note: The proof uses a variant of the Cauchy inequality:
$$\text { in the form: } \sum_{i=1}^{n} \frac{a_{i}^{2}}{b_{i}} \geqslant \frac{\left(\sum_{i=1}^{n} a_{i}\right)^{2}}{\sum_{i=1}^{n} b_{i}}$$