For positive real numbers $a,b,c$ such that $a+b+c=1$. Prove that...

For positive real numbers $a,b,c$ such that $a+b+c=1$. Prove that $$\frac{a}{b}+\frac{a}{c}+\frac{b}{a}+\frac{b}{c}+\frac{c}{a}+\frac{c}{b}+6\ge2\sqrt2\Big(\sqrt\frac{1-a}{a}+\sqrt\frac{1-b}{b}+\sqrt\frac{1-c}{c}\Big)$$

Here's what I've done so far:
$\frac{a}{b}+\frac{a}{c}+\frac{b}{a}+\frac{b}{c}+\frac{c}{a}+\frac{c}{b}+6\ge2\sqrt2\Big(\sqrt\frac{1-a}{a}+\sqrt\frac{1-b}{b}+\sqrt\frac{1-c}{c}\Big)$
$=\frac{b+c}{a}+\frac{a+c}{b}+\frac{a+b}{c}+6\ge2\sqrt2\Big(\sqrt\frac{1-a}{a}+\sqrt\frac{1-b}{b}+\sqrt\frac{1-c}{c}\Big)$
$=\frac{a+b+c}{a}+\frac{a+b+c}{b}+\frac{a+b+c}{c}+3\ge2\sqrt2\Big(\sqrt\frac{1-a}{a}+\sqrt\frac{1-b}{b}+\sqrt\frac{1-c}{c}\Big)$
$=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+3\ge2\sqrt2\Big(\sqrt\frac{1-a}{a}+\sqrt\frac{1-b}{b}+\sqrt\frac{1-c}{c}\Big)$
Not sure where to go from here, any help's appreciated. I think the $AM-GM$ inequality should be used here in some way.

Hint: $$(\frac{a}{b}+\frac{c}{b})+(\frac{a}{c}+\frac{b}{c})+(\frac{b}{a}+\frac{c}{a})+6$$ $$ = \frac{1-b}{b}+\frac{1-c}{c}+\frac{1-a}{a}+2+2+2$$

Now check if $$\frac{1-x}{x}+2\geq 2\sqrt2 \sqrt\frac{1-x}{x}$$ is true?