Let $A_i$ be the area of $Q_i$. Then we show that $S_i\geq 4A_i$. By fixing 3 vertices of the quad, and moving the fourth point while keeping the same area, it is easy to show that the sum of the squares of the sides are minimized when the sides joining at the fourth vertex are of equal length. Thus $S_i$ can be minimized by making the 4 sides equal length, or the quad is a rhombus. Thus it can be further minimized by making the quad a square. Thus $S_i$ is minimum when $Q_i$ is a square, and that is when $S_i=4A_i$. Thus $\sum{S_i}\geq 4\sum{A_i}=4$.