Show that among all quadrilaterals of a given perimeter the square has the largest area.
By Ptolemy's theorem we have that if $a,b,c,d$ are the side lengths of the quadrilateral then $ac+bd \geq d_1d_2$, which implies that $\text{Area}_{\text{quadrilateral}} \leq \dfrac{1}{2}d_1d_2$ where $d_1,d_2$ are the lengths of the diagonals. I then want to show for a given perimeter the maximal area is obtained for equality of the last inequality. We can't just say that the maximal area is for that of a square based on the last inequality since the maximal may not be achieved for a given perimeter. How do I continue?