My favorite interpretation comes from the following claim / 'obvious fact'
Claim: Amongst all smooth curves with a fixed perimeter $P$, the area $A$ satisfies the inequality $ 4 \pi A \leq P^2$. Equality holds when the curve is a circle.
Lemma: If the 4 given sides satisfy the cyclic inequalities $a < b+c+d$, then a quadrilateral with sides $a, b, c, d$ exists. Moreover, a cyclic quadrilateral with sides $a, b, c, d$ exists. A sketch of this is given at the end.
This Lemma merely serves to let us know when the question makes geometric sense. Now, let $ABCD$ be a cyclic quad with the desired side lengths, and consider it's circumcircle. Let it have area $X$. Fix the circular arcs along with the intermediate areas between the circular arcs and the quadrilateral. Let these arcs have perimeter $P$, and the intermediate area have total area $I$.
Take any deformation $A^*B^*C^*D^*$, where the arcs and the areas are moved under translation and rotation. This gives a figure with perimeter $P$, and area $I+X^*$, where $X^*$ is the area of the deformed $A^*B^*C^*D^*$. By the initial claim, $I+X^* \leq I+X$. Hence we are done.$ _ \square$
Sketch Proof of Lemma: The first part follows from the triangle inequality. The second part follows by taking a continuous deformation of the quadrilateral, showing that opposite angles must equal $180^\circ$ at some point in time.
One problem is just the dual of the other one, so it is enough to prove one of them to get them both for (almost) free. Assume that we have a fixed perimeter $(2p)$ and we want to maximize the area. Our problem can be written as
$$ \max_{\substack{a+b=p\\ a,b\geq 0}} ab = \large{?} $$
and we may exploit the AM-GM inequality, stating that if $a,b\geq 0$,
$$ \sqrt{ab}\leq\frac{a+b}{2} $$
with a strict inequality if $a\neq b$. It follows that the maximum area ($p^2/4$) is achieved by $a=b$,
i.e. by a square.
Best Answer
We have $$S_{ABCD}\le S_{ABC}+S_{CDA}\le \frac12(AB\cdot BC+CD\cdot DA)\tag{1}$$ Similarly $$S_{ABCD}\le S_{BCD}+S_{BAD}\le \frac12(BC\cdot CD + BA\cdot AD)\tag{2}.$$ Adding (1) and (2), we get $$4S_{ABCD}\le (AB + CD)(BC+AD)\le \frac{(AB+BC+CD+DA)^2}4.\tag{3}$$ Finally, equalities hold in (1) and (2) only if all four angles of $ABCD$ are right angles, which makes $ABCD$ a rectangular; while equality holds in (3) only if $AB+CD=BC+AD$, which makes a rectangular a square.