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.
 A: 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?

A: $\frac{a}{b}+\frac{a}{c}+\frac{b}{a}+\frac{b}{c}+\frac{c}{a}+\frac{c}{b}+6\ge2\sqrt2\Big(\sqrt\frac{1-a}{a}+\sqrt\frac{1-b}{b}+\sqrt\frac{1-c}{c}\Big)$
$=\frac{b+c}{a}+\frac{a+c}{b}+\frac{a+b}{c}+6\ge2\sqrt2\Big(\sqrt\frac{1-a}{a}+\sqrt\frac{1-b}{b}+\sqrt\frac{1-c}{c}\Big)=$  (1)
We know that $a+b+c=1$, so
$b+c=1-a$,
$a+c=1-b$,
$a+b=1-c$.
$(1)=\frac{1-a}{a}+2+\frac{1-b}{b}+2+\frac{1-c}{c}+2\ge2\sqrt2\Big(\sqrt\frac{1-a}{a}+\sqrt\frac{1-b}{b}+\sqrt\frac{1-c}{c}\Big)$
We know from the $A.M-G.M$ inequality that
$\frac{1-x}{x}+2\ge2\sqrt{2*\frac{1-x}{x}}$
So we have
$\frac{1-a}{a}+2\ge2\sqrt{2*\frac{1-a}{a}}$
$\frac{1-b}{b}+2\ge2\sqrt{2*\frac{1-b}{b}}$
$\frac{1-c}{c}+2\ge2\sqrt{2*\frac{1-c}{c}}$
Summing them we get
$\frac{1-a}{a}+2+\frac{1-b}{b}+2+\frac{1-c}{c}+2\ge2\sqrt{2*\frac{1-a}{a}}+2\sqrt{2*\frac{1-b}{b}}+2\sqrt{2*\frac{1-c}{c}}$
$=\frac{1-a}{a}+2+\frac{1-b}{b}+2+\frac{1-c}{c}+2\ge2\sqrt2\Big(\sqrt\frac{1-a}{a}+\sqrt\frac{1-b}{b}+\sqrt\frac{1-c}{c}\Big)$
So we've proved the original inequality.
A: The key is to write:
$$\frac ab+\frac ba+\frac bc+\frac cb+\frac ca+\frac ac=\frac{b+c}a+\frac{c+a}b+\frac{a+b}c=\frac{1-a}a+\frac{1-b}b+\frac{1-c}c$$
and
$$\frac{1-a}a+2\ge2\sqrt2\sqrt{\frac{1-a}a}$$
by AM-GM.
Summing similar inequalities yields the result.
