If $x,y,z>0$ and $x+y+z=1$ then $\frac{xyz}{(1-x)(1-y)(1-z)}\le\frac{1}{8}$ If $x,y,z>0$ and $x+y+z=1$,
then: $$\frac{xyz}{(1-x)(1-y)(1-z)}\le\frac{1}{8}$$

$$\frac{xyz}{(1-x)(1-y)(1-z)}=\frac{x}{(1-x)}\frac{y}{(1-y)}\frac{z}{(1-z)}$$
Let $\frac{x}{(1-x)}=a$, $\frac{y}{(1-y)}=b$, $\frac{z}{(1-z)}=c$
I am stuck here.

You have: $$x+y+z=1$$
So, write: $$1-x=y+z$$
Using inequality $AP\geq GP$, $$y+z\geq 2\sqrt{yz}$$
So, $$(1-x)\geq2 \sqrt{yz}$$ Similarly,  $$(1-y)\geq2 \sqrt{xz} \hspace{0.5cm} {and} \hspace{0.5cm}  (1-z)\geq2 \sqrt{xy}$$
Multiplying all: $$(1-x)\cdot (1-y) \cdot (1-z) \geq 8 \cdot xyz$$

Divide:$$\frac{(1-x)\cdot (1-y)\cdot (1-z)}{xyz}\geq 8$$ Or,
$$\frac{xyz}{(1-x)(1-y)(1-z)}\leq \frac 1 8$$ In-equality will be reversed on dividing (why?)

