It suffices to show that $$|BC^3-CD^3| \leq 3|AB^3-AD^3|,$$ because the other side of the inequality follows by symmetry. By Pitot's Theorem, we have $AB-AD=BC-CD$. WLOG, let $AB \geq AD$. Then we have that $BC \geq CD$, so we must show $BC^3-CD^3 \leq 3(AB^3-AD^3)$. If $AB=AD$, we are done. Therefore, we will assume for the rest of the proof that $AB>AD$, $BC>CD$.
Then we would like to show that $$BC^2+BC\cdot CD+CD^2 \leq 3(AB^2+AB\cdot AD+AD^2).$$
By the Law of Cosines, we have that $$BD^2=BC^2+k_1(BC\cdot CD)+CD^2=AB^2+k_2(AB\cdot AD)+AD^2,$$ where $-1 \leq k_1, k_2 \leq 1$ (this is because $60^{\circ} \leq \angle BAD,\angle BCD \leq 120^{\circ}$). Therefore, $$3(BC^2-BC\cdot CD+CD^2) \leq 3(BC^2+k_1(BC\cdot CD)+CD^2)=3(AB^2+k_2(AB\cdot AD)+AD^2)\leq 3(AB^2+AB\cdot AD+AD^2).$$ Then we wish to show that $$3(BC^2-BC\cdot CD+CD^2)-(BC^2+BC\cdot CD+CD^2) \geq 0 \rightarrow 2(BC-CD)^2 \geq 0,$$ and we are done. $\blacksquare$