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.

$a^2 = p^2 + s^2$
$ 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.
