drawn
Let ABCDABCD be a square of unit area.
So AB=BC=CD=DA=1AB=BC=CD=DA=1 unit.
Let PQRSPQRS be a quadrilateral which has one vertex on each side of the square. Here let PQ=b,QR=c,RS=dandSP=aPQ=b,QR=c,RS=dandSP=a
Applying Pythagoras thorem we can write
a^2+b^2+c^2+d^2a2+b2+c2+d2
=x^2+y^2+(1-x)^2+(1-w)^2+w^2+(1-z)^2+z^2+(1-y)^2=x2+y2+(1−x)2+(1−w)2+w2+(1−z)2+z2+(1−y)2
=4+2(x^2+y^2+z^2+w^2-x-y-z-w)=4+2(x2+y2+z2+w2−x−y−z−w)
=2+2(1+x^2+y^2+z^2+w^2-x-y-z-w)=2+2(1+x2+y2+z2+w2−x−y−z−w)
=2+2((x-1/2)^2+(y-1/2)^2+(z-1/2)^2+(w-1/2)^2)=2+2((x−12)2+(y−12)2+(z−12)2+(w−12)2)
Now by the problem we have
0<= x<=1=> 0<=(x-1/2)^2<=1/40≤x≤1⇒0≤(x−12)2≤14
0<= y<=1=> 0<=(y-1/2)^2<=1/40≤y≤1⇒0≤(y−12)2≤14
0<=z<=1=> 0<=(z-1/2)^2<=1/40≤z≤1⇒0≤(z−12)2≤14
0<= w<=1=> 0<=(w-1/2)^2<=1/40≤w≤1⇒0≤(w−12)2≤14
Hence
2<=a^2+b^2+c^2+d^2<=42≤a2+b2+c2+d2≤4