prove $abc \ge 8$ for $\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}=1$ I am reading this book about inequalities and the chapter about AM-GM inequalities includes this problem:
Let $a,b,c$ be positive numbers for which  $\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}=1$, prove that
$$abc \ge 8$$
The book does not provide full solutions but only hints, and the one for this question is that it is similar to this problem:
Let $a,b,c$ be positive numbers with $a+b+c=1$, prove that
$$(\frac{1}{a}-1)(\frac{1}{b}-1)(\frac{1}{c}-1) \ge 8$$
again there's no solution provided but what I came up with is the following (you can correct me if it is wrong):
$$a+b=1-c$$
$$a+c=1-b$$
$$b+c=1-a$$
$$(\frac{1}{a}-1)(\frac{1}{b}-1)(\frac{1}{c}-1)=(\frac{1-a}{a})(\frac{1-b}{b})(\frac{1-c}{c})=(\frac{b+c}{a})(\frac{a+c}{b})(\frac{a+b}{c})=(\frac{b}{a}+\frac{c}{a})(\frac{a}{b}+\frac{c}{b})(\frac{a}{c}+\frac{b}{c})$$
By AM-GM we have for each term:
$$(\frac{b}{a}+\frac{c}{a}) \ge 2\sqrt{\frac{bc}{a^2}}$$
$$(\frac{a}{b}+\frac{c}{b}) \ge 2\sqrt{\frac{ac}{b^2}}$$
$$(\frac{a}{c}+\frac{b}{c}) \ge 2\sqrt{\frac{ab}{c^2}}$$
By multiplying the three inequalities we have
$$(\frac{b}{a}+\frac{c}{a})(\frac{a}{b}+\frac{c}{b})(\frac{a}{c}+\frac{b}{c}) \ge 8 \sqrt{\frac{a^2b^2c^2}{a^2b^2c^2}}=8$$ 
as desired.
Can someone please provide me a proof for the first problem as I cannot find any way around it, Thanks.

Define 
\begin{align*}
x&=\frac{1}{1+a}\\
y&=\frac{1}{1+b}\\
z&=\frac{1}{1+c}
\end{align*}
Then your problem transforms into: given $x+y+z=1$ prove $(\frac{1}{x}-1)(\frac{1}{y}-1)(\frac{1}{z}-1) \geq 8$. I believe you know the rest.
