If $ a,b,c \gt 0 $ and $ \frac{1}{a+1} + \frac{1}{b+1} + \frac{1}{c+1} = 1 $ Prove $ abc \geq 8 $ If $ a,b,c \gt 0 $  and  $ \frac{1}{a+1} + \frac{1}{b+1} + \frac{1}{c+1} = 1 $
Prove $ abc \geq 8 $
I have tried AM-GM and substituting the 1, without results.

Let
$$x = \frac{1}{a+1}, \quad y = \frac{1}{b+1},  \quad z = \frac{1}{c+1},$$
then $x+y+z=1$ and
$$a = \frac{1-x}{x} = \frac{y+z}{x}.$$
The inequality become
$$\frac{(x+y)(y+z)(z+x)}{xyz} \geqslant 8.$$
Which is true because
$$\frac{(x+y)(y+z)(z+x)}{xyz} \geqslant \frac{2\sqrt{xy} \cdot 2\sqrt{yz} \cdot 2\sqrt{zx}}{xyz} = 8.$$
The proof is completed.
