Quadrilateral in Square

Question

$S$ 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 $2\leq{}a^2+b^2+c^2+d^2$?

Answer 1

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: $$a^2+b^2+c^2+d^2=4-2(mn+op+qr+ts)\geq4\cdot1-2=2.$$

Answer 2

For the lower bound, note by CS inequality, $$(a^2+b^2+c^2+d^2)\cdot8=(m^2+n^2+\cdots)(1+1+\cdots)\geqslant(m+n+\cdots)^2=4^2$$