Area of quadrilateral: $S \leq \frac{(a+b)(c+d)}4 $ I got stuck on this problem:

Given a convex quadrilateral of area $S$ and sides $a$, $b$, $c$ and $d$, prove that: $$S \leq \frac{(a+b)(c+d)}4$$

What I've done so far was to proof that $$S \leq \frac{(a+c)(b+d)}4$$ using the relation that for a given triangle $ABC$,   $A_{ABC} \leq \frac12 {AB}\cdot{AC}$ or $A_{ABC} \leq \frac12 {AB}\cdot{BC}$ etc.
This is derived from $A_{ABC} = \frac12 AB \cdot AC \cdot \sin (BAC)$ and $0<\sin (BAC)\leq1$.
Anyway, this isn't enough to solve the problem. Some suggestions would be appreciate.
 A: Let $s:=\frac{a+b+c+d}{2}$.  By Bretschneider's Formula, 
$$S=\sqrt{(s-a)(s-b)(s-c)(s-d)-abcd\cos^2\left(\frac{\alpha+\gamma}{2}\right)}\,,$$
where $\alpha$ and $\gamma$ are two opposite angles of the given convex quadrilateral.  Hence,
$$S\leq \sqrt{(s-a)(s-b)(s-c)(s-d)}\;\leq \left(\frac{(s-a)+(s-b)}{2}\right)\left(\frac{(s-c)+(s-d)}{2}\right)\,,$$
where we have used the AM-GM Inequality.  Thus,
$$S\leq \frac{(a+b)(c+d)}{4}\,$$
as required.  The equality holds iff the quadrilateral is cyclic with two pairs of equal adjacent sides: $a=b$ and $c=d$ (or equivalently, the quadrilateral is a kite with two opposite right angles).
You can also modify this proof to show that $S\leq \frac{(a+c)(b+d)}{4}$.  The equality holds iff the quadrilateral is a rectangle.
A: The quadrilateral must be cyclic to maximize area (with given sides), and the area of this cyclic quadrilateral does not depend on the order of the sides.  
Therefore $S$ is at most equal to the bound given in the question, $\frac{(a+c)(b+d)}{4}$ for any permutation of $a,b,c,d$ and the permutation that exchanges $b$ and $c$ gives the claimed additional upper bound of $\frac{(a+b)(c+d)}{4}$.  
From the same argument, $S \leq (a+d)(b+c)/4$ is another bound on the area.
