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.
 A: 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,
$$-2.\sum(x(1-x)) +4  = -2.4.1/4 +4 = 2$$
A: $a^2 = p^2 + s^2$
$ b^2 = (1 – p)^2 + q^2$
$c^2 = (1 – q)^2 + (1 – r)^2$
$ d^2 = r^2 + (1 – s)^2$
$$\implies a^2 + b^2 + c^2 + d^2 = p^2 + (1 - p)^2 + q^2 + (1 – q)^2 + r^2 + (1 – r)^2 + s^2 + (1 – s)^2$$
Where $p$, $q$, $r$, $s$ all vary in the interval $[0, 1]$.
Now consider the function
$y^2 = x^2 + (1 – x)^2, 0 ≤ x ≤ 1,$
$$2y \frac {dy} {dx} = 2x – 2(1 – x) = 0$$
$$\implies x = \frac 12$$
which$$\frac {d^2y} {dx^2} = 4$$ i.e. positive
Hence y is minimum at $x =\frac 1 2$ and its minimum
Value is $\frac 1 4 + \frac 1 4 = \frac 1 2$
Clearly value is maximum at the end points which is $1$.
Minimum value of $a^2 + b^2 + c^2 + d^2 = \frac {1}{2} + \frac {1}{2} + \frac {1}{2} + \frac {1}{2} = 2$
And maximum value is $1 + 1 + 1 + 1 = 4$
QED.
A: The smallest $a^2+b^2+c^2+d^2$ is when the quadrilateral has its 4 vertices on the midpoint of each side of square S.  $a=b=c=d=0.5$, and the sum of the squares is $2$.  The largest is simply achieved when the vertices of the quadrilateral lies on the vertices of square S.  ie, both squares are the same.  The sum of the squares of sides is $4$ then.
Now why exactly is this so?  Let the sides of the square be $w, x, y, z$, and let the subscript 1 and 2 denote a section of the side (e.g. $w_1+w_2=w$).  Then $a^2+b^2+c^2+d^2=w_2^2+x_1^2+x_2^2+y_1^2+y_2^2+z_1^2+z_2^2+w_1^2$. The restriction here lies in that $w+x+y+z$ is fixed, the perimeter of S.  We know that $2\times a^2<(a-1)^2+(a+1)^2$, for any $a>=1$.  Hence, since we are trying to achieve the smallest possible result, we would see each side of the square divided exactly in half, such that $w_1=w_2=...=0.5$.
A: Okay, Here’s my approach, I considered a limiting case for which values of a,b,c,d would be the least ,that’s when they are on the midpoints of side.So when the points are midpoint each side is $1/\sqrt2$.Thus the inequality will be $1/2+1/2+1/ 2 + 1 / 2 =2$.
