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.

This simplifies to $$a+b+c+2=abc$$
$$⇒\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}+\frac{2}{abc}=1$$
Applying AM-GM inequality we get
$$\frac{1}{4}=\frac{\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}+\frac{2}{abc}}{4}≥(\frac{2}{a^3b^3c^3})^\frac{1}{4}$$
Riasing both sides to the power of 4 we get
$$\frac{1}{256}≥(\frac{2}{a^3b^3c^3})$$
$$⇒abc≥8$$
