The $1997$ IIT JEE problem 
Let $S$  be a square of unit area. Consider any quadrilateral whose $4$ vertices lie on each side square $S$. Let the length of the sides of this quadrilateral be $a,b,c,d$. Then prove that $$2 \leq a^2+b^2+c^2+d^2 \leq 4$$ 

This problem appeared in IIT JEE $1997$ (re-exam).I really do not know know how to approach this problem properly.I have managed to prove $a^2 + b^2 + c^2 +d^2  \leq 4$
but not the initial inequality.

Consider this:

You want to find the minimum of:
$$2(a^2+b^2+c^2+d^2-a-b-c-d)+4$$
Or
$$-2.\sum(x(1-x)) +4 $$
Here x, represents a,b,c or d
Notice that it's a quadratic, so maximum of x(1-x) happens at x = 0.5 as it happens between x = 1 and x =0
So max of quadratic is 1/4
Hence,
$$-2.\sum(x(1-x)) +4  = -2.4.1/4 +4 = 2$$
