Who loves geometric inequality?

Let there be a quadilateral as shown in the figure by which

$a=\sqrt{p^{2}+1+s^{2}-2s}$

$b=\sqrt{q^{2}+1+p^{2}-2p}$

$c=\sqrt{r^{2}+1+q^{2}-2q}$

$d=\sqrt{s^{2}+1+r^{2}-2r}$

$a^{2}+b^{2}+c^{2}+d^{2}=2[ \ p(p-1) + q(q-1) + r(r-1)+s(s-1)+2 \ ]$

This is minimum when $p(p-1)+q(q-1)+r(r-1)+s(s-1)$ is minimum.

• $p^{2}-p \ \Rightarrow$ is min when $p=\dfrac{1}{2}$

Similarly $q=r=s=\dfrac{1}{2}$

$(a^{2}+b^{2}+c^{2}+d^{2})_{min}=2[\dfrac{1}{2}(\dfrac{-1}{2})+\dfrac{1}{2}(\dfrac{-1}{2})+\dfrac{1}{2}(\dfrac{-1}{2})+\dfrac{1}{2}(\dfrac{-1}{2})+2]$

$(a^{2}+b^{2}+c^{2}+d^{2})_{min}=2(1)$

$\huge{\Rightarrow 2}$