Let S be a square of unit area. Consider any quadrilateral which has one vertex on each side of S. If a,b,c,d are the length of the sides of the quadrilateral, then ${\mathrm{a}}^2+{\mathrm{b}}^2+{\mathrm{c}}^2+{\mathrm{d}}^2$ lies in

$$\begin{gathered} {\mathrm{a}}^2+{\mathrm{b}}^2+{\mathrm{c}}^2+{\mathrm{d}}^2 \\ ={(1-\mathrm{t})}^2+{\mathrm{x}}^2+{(1-\mathrm{x})}^2+{\mathrm{y}}^2+{(1-\mathrm{y})}^2+{\mathrm{z}}^2+{(1-\mathrm{z})}^2+{\mathrm{t}}^2 \end{gathered}$$

consider $\mathrm{S}={\mathrm{x}}^2+(1-{\mathrm{x}}^2)$

$$\begin{gathered} {\mathrm{S}}^1=2\mathrm{x}-2(1-\mathrm{x}) (-1) \\ {\mathrm{S}}^1=2\mathrm{x}-2(1-\mathrm{x}) \end{gathered}$$

S is min or max $\Rightarrow {\mathrm{S}}^1=0$

$$\begin{gathered} 2\mathrm{x}=2-2\mathrm{x} \\ 2\mathrm{x}=1 \Rightarrow \mathrm{x}=\frac{1}{2} \\ {\mathrm{a}}^2+{\mathrm{b}}^2+{\mathrm{c}}^2+{\mathrm{d}}^2 \\ =8{\left(\frac{1}{2}\right)}^2=2=\min \max {\mathrm{a}}^2+{\mathrm{b}}^2+{\mathrm{c}}^2+{\mathrm{d}}^2 \\ =1+1+1+1=4 \\ \therefore 2 \mathrm{S} {\mathrm{a}}^2+{\mathrm{b}}^2+{\mathrm{c}}^2+{\mathrm{d}}^2\le 4 \end{gathered}$$