By symmetry, we only need to prove the first inequality.
Because quadrilateral $ABCD$ has an incircle, we have $AB + CD = BC + AD$, or $AB - AD = BC - CD$. It suffices to prove that
$$\frac{1}{3}(AB^2 + AB \cdot AD + AD^2) \leq BC^2 + BC \cdot CD + CD^2.$$
By the given condition, $60^{\circ} \leq \angle A, \angle C \leq 120^{\circ}$, and so $\frac{1}{2} \geq \cos A, \cos C \geq -\frac{1}{2}$. Applying the Law of Cosines to triangle $ABD$ yields
$$\begin{aligned} BD^2 &= AB^2 - 2AB \cdot AD \cos A + AD^2 \\ &\geq AB^2 - AB \cdot AD + AD^2 \\ &\geq \frac{1}{3}(AB^2 + AB \cdot AD + AD^2). \end{aligned}$$
The last inequality is equivalent to the inequality $3AB^2 - 3AB \cdot AD + 3AD^2 \geq AB^2 + AB \cdot AD + AD^2$, or $AB^2 - 2AB \cdot AD + AD^2 \geq 0$, which is evident. The last equality holds if and only if $AB = AD$.

On the other hand, applying the Law of Cosines to triangle $BCD$ yields
$$BD^2 = BC^2 - 2BC \cdot CD \cos C + CD^2 \leq BC^2 + BC \cdot CD + CD^2.$$
Combining the last two inequalities gives the desired result.

For the given inequalities to have equality, we must have $AB = AD$. This condition is also sufficient, because all the entries in the equalities are 0. Thus, equality holds if and only if $ABCD$ is a kite with $AB = AD$ and $BC = CD$.