Let $a,b,c>0$ so that $a+b+c=1$... 
Let $a,b$ and $c$  be positive real numbers such that $a+b+c=1$. Prove that $$\frac{a}{b}+\frac{b}{a}+\frac{b}{c}+\frac{c}{b}+\frac{c}{a}+\frac{a}{c}+6\geq 2\sqrt{2}\left ( \sqrt{\frac{1-a}{a}}+\sqrt{\frac{1-b}{b}} + \sqrt{\frac{1-c}{c}}\right )$$

What I think is we should replacing $1-a ,1-b ,1-c$ with $b+c,c+a, a+b$ respectively on the right hand side, I have no idea if this is true, any help will be appraciated.
 A: $$\begin{align}\\&\frac ab+\frac ba+\frac bc+\frac cb+\frac ca+\frac ac+6-2\sqrt 2\left(\sqrt{\frac{1-a}{a}}+\sqrt{\frac{1-b}{b}}+\sqrt{\frac{1-c}{c}}\right)\\&=\frac{b+c}{a}+\frac{a+c}{b}+\frac{b+a}{c}+6-2\sqrt 2\left(\sqrt{\frac{1-a}{a}}+\sqrt{\frac{1-b}{b}}+\sqrt{\frac{1-c}{c}}\right)\\&=\frac{1-a}{a}+\frac{1-b}{b}+\frac{1-c}{c}+6-2\sqrt 2\left(\sqrt{\frac{1-a}{a}}+\sqrt{\frac{1-b}{b}}+\sqrt{\frac{1-c}{c}}\right)\\&=\frac{1-a}{a}-2\sqrt{2}\sqrt{\frac{1-a}{a}}+2+\frac{1-b}{b}-2\sqrt{2}\sqrt{\frac{1-b}{b}}+2+\frac{1-c}{c}-2\sqrt 2\sqrt{\frac{1-c}{c}}+2\\&=\left(\sqrt{\frac{1-a}{a}}-\sqrt 2\right)^2+\left(\sqrt{\frac{1-b}{b}}-\sqrt 2\right)^2+\left(\sqrt{\frac{1-c}{c}}-\sqrt 2\right)^2\ge 0\end{align}$$
A: We have $$\frac{a}{b}+\frac{b}{a}+\frac{b}{c}+\frac{c}{b}+\frac{c}{a}+\frac{a}{c}=\sum \frac{b+c}{a}=\sum \frac{1-a}{a}$$ so $$\frac{a}{b}+\frac{b}{a}+\frac{b}{c}+\frac{c}{b}+\frac{c}{a}+\frac{a}{c}+6- 2\sqrt{2}\left ( \sqrt{\frac{1-a}{a}}+\sqrt{\frac{1-b}{b}} + \sqrt{\frac{1-c}{c}}\right )=\sum \left ( \frac{1-a}{a}-2\sqrt{2}\sqrt{\frac{1-a}{a}}+2 \right )=\sum \left ( \sqrt{\frac{1-a}{a}}-\sqrt{2} \right )^{2} \geq 0$$
Equality occurs when $$\sqrt{\frac{1-a}{a}}=\sqrt{\frac{1-b}{b}} = \sqrt{\frac{1-c}{c}}$$
Therefore $a=b=c=\frac{1}{3}$
AND WE'RE DONE
