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.

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.
