Any subjective method

Take the lengths - u,1-u ; x,1-x ; y,1-y ; z,1-z ;

between 1-x and y take c
between 1-y and z take d
between 1-z and u take a
and the other b

x² + (1-u)²=b²
(1-x)²+y² = c²
(1-y)²+z²=d²
(1-z)² + u²=a²

expanding and adding we get

4+2(x² - x + y² - y + z² - z + u² - u) = a² + b² + c² + d².

When x=y=z=u=0 or 1, it is max and when they are equal to .5 , it is min..

So substituting, we get the sol..

( I just can't seem to draw a proper pic in paint :()

A²=1 => A=1

a²= (1-x)² + y² n so on..

min value of x,y,z,w .. x=y=z=w=0
so max value of a^2+b^2+c^2+d^2 = 4 .

now take a function m=x^2 +(1-x)^2
=> m' = 2x - 2(1-x) =0
=> x=1/2 is min..

so x=y=z=w=1/2 is the min value.

addind up all of them, a^2+b^2+c^2+d^2 = 4 (1/4 +1/4) =2 = min value of the expression..

hence 2≤a2+b2+c2+d2≤4.. proved :)

Gud....

Actually objective paper solve karte karte dimag nahi chalta....

I solved it in a very shortcut method its not a method either....

thanks 4 the details...