Prove that by applying $G_{2} \leqslant A_{2}$ twice, we get
$$\begin{aligned} & \frac{a b c+b c d+c d a+a d b}{4} \\ = & \frac{1}{2}\left(a b \cdot \frac{c+d}{2}+c d \cdot \frac{a+b}{2}\right) \\ \leqslant & \frac{1}{2}\left[\left(\frac{a+b}{2}\right)^{2} \cdot \frac{c+d}{2}+\left(\frac{c+d}{2}\right)^{2} \cdot \frac{a+b}{2}\right] \\ = & \frac{a+b}{2} \cdot \frac{c+d}{2} \cdot \frac{a+b+c+d}{4} \\ \leqslant & \left(\frac{\frac{a+b}{2}+\frac{c+d}{2}}{2}\right)^{2} \frac{a+b+c+d}{4}=\left(\frac{a+b+c+d}{4}\right)^{3} . \end{aligned}$$

Then, by $A_{4} \leqslant Q_{4}$, we have
$$\frac{a+b+c+d}{4} \leqslant \sqrt{\frac{a^{2}+b^{2}+c^{2}+d^{2}}{4}}$$

Thus, the original inequality holds.