Find minimum value of this expression: $P=\sqrt{a^2+(1-bc)^2}+\sqrt{b^2+(1-ca)^2}+\sqrt{c^2+(1-ab)^2}$ Let $a,b,c\in R$ and satisfying $a^2+b^2+c^2=1$
Find minimum value of this expression: $P=\sqrt{a^2+(1-bc)^2}+\sqrt{b^2+(1-ca)^2}+\sqrt{c^2+(1-ab)^2}$

$3u=a+b+c,3v=ab+bc+ac \implies 9u^2-6v=1$
edit:
$P \ge \sqrt{(a+b+c)^2+(3-(3b+bc+ac))^2}=\sqrt{(9u^2+(3-3v)^2}=\sqrt{1+6v+9-18v+9v^2}=\sqrt{(2-3v)^2+6} $
$3v=ab+bc+ac \le a^2+b^2+c^2 =1 $
$ 2-3v \ge 1 \implies \sqrt{(2-3v)^2+6} \ge \sqrt{7}$
last "=" will hold when $a=b=c=\pm \dfrac{1}{\sqrt{3}}$ 
another "=" will hold when $\dfrac{a}{1-bc}=\dfrac{c}{1-ab}=\dfrac{b}{1-ac}=\dfrac{a+b+c}{3-(ab+bc+ac)}=\pm \dfrac{\sqrt{3}}{2}$
$\sqrt{3}a^2 \pm 2a -\sqrt{3}=0 \implies a=\pm \sqrt{3}$ or $\pm \dfrac{1}{\sqrt{3}} $
so $P_{min}=\sqrt{7}$ will be got when $a=b=c=\pm \dfrac{1}{\sqrt{3}}$ 
