How do i prove the following problem:
If a quadrilateral has sides of length $a$, $b$, $c$, and $d$, prove that its area $S$ satisfies the following inequality $$4S\leq (a+c)(b+d)$$ with equality holding only for rectangles.
Hint: Twice the area of a triangle is $a b \sin \alpha$, where $\alpha$ is the angle between the sides of lengths $a$, $b$. But how do i use it?
Thanks in advance.
Hint
The triangle with sides $a$ and $b$ has the area $$A=\frac{1}{2}ab\sin \alpha\leq\frac{1}{2}ab$$ with the expressions equal if and only if $\alpha=\frac{\pi}{2}$. Do the same for each pair of sides and label these areas $A_1$, $A_2$, $A_3$, $A_4$ and note that $$ 2S=A_1+A_2+A_3+A_4$$ and $$ab+ad+cb+cd = (a+c)(b+d)$$