Prove using the cosine theorem, we know
$$\begin{aligned} B D^{2}= & A D^{2}+A B^{2}-2 A D \cdot A B \cos \angle D A B= \\ & C D^{2}+B C^{2}-2 C D \cdot B C \cos \angle D C B \end{aligned}$$

From the given conditions, we know $60^{\circ} \leqslant \angle D A B, \angle D C B \leqslant 120^{\circ}$, hence $-\frac{1}{2} \leqslant \cos \angle D A B \leqslant \frac{1}{2}, -\frac{1}{2} \leqslant \cos \angle D C B \leqslant \frac{1}{2}$, thus
$$\begin{array}{l} 3 B D^{2}-\left(A B^{2}+A D^{2}+A B \cdot A D\right)= \\ 2\left(A B^{2}+A D^{2}\right)-A B \cdot A D(1+6 \cos \angle D A B) \geqslant \\ 2\left(A B^{2}+A D^{2}\right)-4 A B \cdot A D=2(A B-A D)^{2} \geqslant 0 \end{array}$$

Therefore,
$$\begin{array}{l} \frac{1}{3}\left(A B^{2}+A D^{2}+A B \cdot A D\right) \leqslant \\ B D^{2}=C D^{2}+B C^{2}-2 C D \cdot B C \cos \angle D C B \leqslant \\ C D^{2}+B C^{2}+C D \cdot B C \end{array}$$

Since $A B C D$ is a tangential quadrilateral, we know $A D+B C=A B+C D$, so $|A B-A D| = |C D-B C|$. Combining the above, we have $\left.\frac{1}{3}+A B^{3}-A D^{3}|\leqslant| B C^{3}-C D^{3} \right\rvert$.

The equality holds if $\cos A=\frac{1}{2} ; A B=A D ; \cos C=-\frac{1}{2}$ or $|A B-A D| = |C D-B C|=0$.

Therefore, the equality holds if $A B=A D$ and $C D=B C$.
Similarly, we can prove the other inequality holds, with the same conditions for equality.