Maximum area of specific quadrilateral $ABCD$ is a convex quadrilateral. Lines $AC$ and $BD$ are perpendicular and $AB+CD=8$. Find the maximum area of $ABCD$.
Let $O$ be intersection of the diagonals. Notice that the area of the quadrilateral is $\frac{1}{2} (AO+OC)(BO+OD) \leq \frac{1}{4} (\sum AO^2 + 2AO \cdot OC + BO \cdot OD )$ (By AM-GM).
On the other hand $\sqrt{AO^2+BO^2}+\sqrt{CO^2+DO^2}=8$. The last equation yields: $\sum AO^2 + 2\sqrt{(AO^2+BO^2)(CO^2+DO^2)}=64$.
Can this equations be helpful? Any hint or solution will be appreciated. Thank you in advance.

Let $AO=a$, $BO=b$, $CO=c$, $DO=d$, then problem is the following: $$a,b,c,d> 0,\sqrt{a^2+b^2}+\sqrt{c^2+d^2}=8, S=\frac{1}{2}(a+c)(b+d)={\rm max}$$
$$(a+b)^2+(a-b)^2=2a^2+2b^2 \Rightarrow (a+b)^2\leq 2(a^2+b^2)\Rightarrow\\ a+b\leq \sqrt{2}\sqrt{a^2+b^2}$$
$$c+d\leq\sqrt{2}\sqrt{c^2+d^2}$$
$$a+b+c+d\leq\sqrt{2}(\sqrt{a^2+b^2}+\sqrt{c^2+d^2})=8\sqrt{2}$$
$$a+c=x,b+d=y\Rightarrow x+y\leq 8\sqrt{2}$$
$$(x+y)^2-(x-y)^2=4xy\Rightarrow 4xy\leq (x+y)^2\leq 128\Rightarrow xy\leq 32$$
$$S=\frac{1}{2}xy\leq 16$$
$$S_{\rm max}=16$$
$S=S_{\rm max}=16$ at $x-y=0$, $a-b=0$, $c-d=0$ ($x-y=0$ is automatically true at $a-b=0$ and $c-d=0$).
