Quadrilateral in Square 
$S$ is a unit square. Four points are taken randomly, one on each side of $S$. A quadrilateral is drawn. Let the  sides of this quadrilateral be $a,b,c,d$. Prove that $2\leq{}a^2+b^2+c^2+d^2\leq{}4$.

My Efforts:
Let 
$\begin{align}m^2+t^2&=a^2&\mathfrak{a}\\n^2+o^2&=b^2&\mathfrak{b}\\p^2+q^2&=c^2&\mathfrak{c}\\r^2+s^2&=d^2&\mathfrak{d}\end{align}$
$\mathfrak{a+b+c+d}\text{ gives}$
$m^2+n^2+o^2+p^2+q^2+r^2+s^2+t^2=a^2+b^2+c^2+d^2$
Since $m+n=o+p=q+r=s+t=1$,
This gives 
$1-2mn+1-2op+1-2qr+1-2st=a^2+b^2+c^2+d^2$
Simplifying,
$4-2{(mn+op+qr+st)}=a^2+b^2+c^2+d^2$
Since the minimum value of $2{(mn+op+qr+st)}$ is $0$, I get
$a^2+b^2+c^2+d^2\leq4$.
Is there any other way to do this? How do I get $2\leq{}a^2+b^2+c^2+d^2$?

Since by AM-GM
$$mn+op+qr+ts\leq\left(\frac{m+n}{2}\right)^2+\left(\frac{o+p}{2}\right)^2+\left(\frac{q+r}{2}\right)^2+\left(\frac{t+s}{2}\right)^2=1,$$ we obtain:
$$a^2+b^2+c^2+d^2=4-2(mn+op+qr+ts)\geq4\cdot1-2=2.$$
