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

Question

Let $a,b,c\in R$ and satisfying $a^2+b^2+c^2=1$

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

Answer 1

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 $a=b=c=\frac1{\sqrt3}$.

Answer 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 $a=b=c=\pm \dfrac{1}{\sqrt{3}}$