Another proof that the area is maximized when the quadrilateral is cyclic is the following:

First we consider a cyclic quadrilateral with the given sides lengths (the existence of such a quadrilateral requires a small argument). We now “glue” an arc of the circumcircle on each side.

If you consider another quadrilateral with the same side lengths and the previous arcs glued on each side, the figure has a smaller total area than the circle with the same perimeter (that is, the previous circumcircle) by the classical isoperimetric inequality, hence the quadrilateral has smaller area than the original cyclic quadrilateral.

The same proof works for any number of sides.

Another proof that the area is maximized when the quadrilateral is cyclic is the following:

First we consider a cyclic quadrilateral with the given sides lengths (the existence of such a quadrilateral requires an argument). We now “glue” an arc of the circumcircle on each side.

If you consider another quadrilateral with the same side lengths and the previous arcs glued on each side, the figure has a smaller total area than the circle with the same perimeter (that is, the original circumcircle) by the classical isoperimetric inequality, hence the quadrilateral has smaller area than the original cyclic quadrilateral.

The same proof works for any number of sides.

Another proof that the area is maximized when the quadrilateral is cyclic is the following:

First we find a cyclic quadrilateral with the given sides lengths (the existence of such a quadrilateral requires a small argument). We now “glue” an arc of the circumcircle on each side.

If you now consider another quadrilateral with the same side lengths, and the previous arcs glued on each side, the figure has a smaller total area than the circle with the same perimeter (that is, the previous circumcircle), hence the quadrilateral has smaller area than the original cyclic quadrilateral.

The same proof works for any number of sides.

[I’ll add a figure later.]

Another proof that the area is maximized when the quadrilateral is cyclic is the following:

First we consider a cyclic quadrilateral with the given sides lengths (the existence of such a quadrilateral requires a small argument). We now “glue” an arc of the circumcircle on each side.

If you consider another quadrilateral with the same side lengths and the previous arcs glued on each side, the figure has a smaller total area than the circle with the same perimeter (that is, the previous circumcircle) by the classical isoperimetric inequality, hence the quadrilateral has smaller area than the original cyclic quadrilateral.

The same proof works for any number of sides.