Redefine the variables, so that we are trying to prove the inequality $ab>cd$ It's a well known fact $ab=cd=K$ and that $c=d=\sqrt{K}$. So $b=\frac{K}{a}$ Use $AM-GM$ on $a+\frac{K}{a}$ to get the desired inequality of $a+\frac{K}{a}\geq2\sqrt{K}$. Equality is attained when $a=b$ which occurs when $ABCD$ is a parallelogram. Since a parallelogram is not a trapezoid, the inequality never attains an equality condition.