Show that among all quadrilaterals of a given perimeter the square has the largest area 
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?
 A: 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.
A: This can easily seen by Bretschneider's formula for convex quadrilaterals;
$$
AREA=\sqrt{(s-a)(s-b)(s-c)(s-d)-abcd\cdot \cos^2\left( \frac{\alpha+\beta}{2} \right)}
$$
where $a$, $b$, $c$ and $d$ are sides, $s$ is semiperimeter, $\alpha$ and $\beta$ are opposite angles. $(s-a)(s-b)(s-c)(s-d)$ takes its maximum value on $a=b=c=d$, and $abcd\cdot \cos^2\left( \frac{\alpha+\beta}{2} \right)$ takes its minimum value $0$ on $\alpha=\beta=\frac{\pi}{2}$.
A: Let ABCD our quadrilateral, $AB=a$, $BC=b$, $CD=c$ and $DA=d$.
We can assume that segment BD intersects a line $AC$.
Hence, $S_{ABCD}\leq\frac{1}{2}AC\cdot BD\leq \frac{1}{2}(ac+bd)$.
Let $D'$ is a symmetric to $D$ relative to mid-perpendicular of $AC$.
Hence, $AD'=DC=c$, $CD'=AD=d$ and $S_{ABCD}=S_{ABCD'}\leq\frac{1}{2}(ad+bc)$.
Id est, $S_{ABCD}\leq\frac{1}{4}(ac+bd+ad+bc)=\frac{1}{4}(a+b)(c+d)\leq\frac{(a+b+c+d)^2}{16}$
