How to prove $\sqrt{(a-1)(b-1)}+\sqrt{(a-1)(c-1)}+\sqrt{(b-1)(c-1)}\geq a+b+c+\sqrt{ab}+\sqrt{ac}+\sqrt{bc}$? I am trying to prove this inequality:
$$\sqrt{(a-1)(b-1)}+\sqrt{(a-1)(c-1)}+\sqrt{(b-1)(c-1)}\geq a+b+c+\sqrt{ab}+\sqrt{ac}+\sqrt{bc}$$
if there is condition $a+b+c=1$. $a, b, c$ are positive rational numbers.
I have tried it in several ways, but I can't find a solution. (I think that it may be solved using AM-GM inequality.)
What have I done: (I have done much more, but it isn't useful in this example...)
$\sqrt{(a-1)(b-1)}+\sqrt{(a-1)(c-1)}+\sqrt{(b-1)(c-1)}=\sqrt{ab+c}+\sqrt{ac+b}+\sqrt{bc+a}$
I have also done some googling, but no success. I think that it may be solved using AM-GM inequality.
I also thought, that $\sqrt{a}+\sqrt{b}+\sqrt{c}\geq a+b+c$ if the condition is true. Then we write:
$$\sqrt{ab+c}+\sqrt{ac+b}+\sqrt{bc+a} \geq a+b+c+\sqrt{ab}+\sqrt{ac}+\sqrt{bc}$$
And somehow reduce to this:
$$\sqrt{a}+\sqrt{b}+\sqrt{c}\geq a+b+c$$
which is true.
Thank you for the answer and effort!
 A: We assume throughout that $a,b,c$ are positive. Using the condition $a+b+c = 1$, we may write $$\sqrt{(a-1)(b-1)} + \sqrt{(b-1)(c-1)} + \sqrt{(a-1)(c-1)} = \sqrt{(b+c)(a+c)} + \sqrt{(a+c)(a+b)} + \sqrt{(b+c)(a+b)}$$ Note that $$(b+c)(a+c) = c^2 + ab + c(a+b) \ge c^2 + ab + 2c\sqrt{ab} = (c+\sqrt{ab})^2$$ using the two-variable AM-GM inequality. Thus, $$\sqrt{(b+c)(a+c)} \ge c + \sqrt{ab}$$ and summing the analogous inequalities cyclically yields $$\sqrt{(b+c)(a+c)} + \sqrt{(a+c)(a+b)} + \sqrt{(b+c)(a+b)} \ge a+b+c+ \sqrt{ab}+\sqrt{bc}+\sqrt{ac}$$
A: Another way for positive variables. 
We need to prove that
$$\sum_{cyc}2\sqrt{(1-a)(1-b)}\geq\sum_{cyc}(2a+2\sqrt{ab})$$ or
$$\sum_{cyc}(1-a+1-b-2a-2\sqrt{ab})\geq\sum_{cyc}(1-a-2\sqrt{(1-a)(1-b)}+1-b)$$ or
$$\sum_{cyc}(\sqrt{a}-\sqrt{b})^2\geq\sum_{cyc}(\sqrt{1-a}-\sqrt{1-b})^2$$ or
$$\sum_{cyc}\frac{(a-b)^2}{(\sqrt{a}+\sqrt{b})^2}\geq\sum_{cyc}\frac{(a-b)^2}{(\sqrt{1-a}+\sqrt{1-b})^2},$$ which is true because
$$\sqrt{1-a}+\sqrt{1-b}=\sqrt{b+c}+\sqrt{a+c}\geq\sqrt{b}+\sqrt{a}$$ and we are done!
