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: $P=\sqrt{a^2+(1-bc)^2}+\sqrt{b^2+(1-ca)^2}+\sqrt{c^2+(1-ab)^2}$

We need to consider only non-negative $a, b, c$ as: $P(a, b, c)=P(-a, -b, -c)$ means if all three are negative we may replace them with positive numbers, if exactly two are negative then we can replace them with a triple having only one negative, and finally if only one is negative (say $a$), $P(a, b, c) \ge P(-a, b, c)$ gives us a better minimum in positive numbers. So WLOG, consider $a, b, c \in [0, 1]$.
By Cauchy-Schwarz, $\left(a^2+(1-bc)^2\right)(3+4) \ge \left(\sqrt3a+2(1-bc)\right)^2$
$\implies \sqrt{a^2+(1-bc)^2} \ge \frac1{\sqrt7} \left(2+\sqrt3a-2bc\right)$
$\implies \sqrt7P \ge 6+\sqrt3(a+b+c) -2(ab+bc+ca)$
Now if $x=a+b+c, \; 2(ab+bc+ca)=x^2-1$ so $\sqrt7P \ge 7+x(\sqrt3-x)$.  As $x$ is non-negative and $x \le \sqrt{(a^2+b^2+c^2)(1+1+1)} = \sqrt3$, the minimum is in fact when $x=\sqrt3$, giving $P\ge \sqrt7$.
We may note that all inequalities used achieve equality simultaneously when $a=b=c=\frac1{\sqrt3}$.
