If $a,b,c\in(0,1)$ satisfy $a+b+c=2$, prove that $\frac{abc}{(1-a)(1-b)(1-c)}\ge 8.$

Hint: The inequality is: $$\frac{a}2\cdot\frac{b}2\cdot\frac{c}2 \geqslant (1-a)\cdot(1-b)\cdot(1-c)$$

which follows from Karamata's inequality as $\left(1-a, 1-b, 1-c\right) \succ \left(\dfrac{a}2, \dfrac{b}2, \dfrac{c}2\right)$, and $t\mapsto \log t$ is concave.

You may set $$u=1-a,\; v= 1-b,\; w = 1-c$$

$$\Rightarrow u+v+w= 1 \text{ and } u,v,w\in (0,1)$$

To show is now

$$(1-u)(1-v)(1-w)\geq 8uvw$$

or, after expanding and using $u+v+w=1$

$$uv+vw+wu \geq 9uvw$$

or, because auf AM-HM

$$\frac 1 3 = \frac{u+v+w}{3}\geq \frac 3{\frac 1w + \frac 1u + \frac 1v}$$ which is true.

Your inequality is the same as $$\sqrt{(1 - a)(1 - b)}\sqrt{(1 - b)(1 - c)}\sqrt{(1 - a)(1 - c)} \leq {abc \over 8}$$ By the AM-GM inequality you have $$\sqrt{(1 - a)(1 - b)} \leq {1 - a + 1 - b \over 2} = {c \over 2}$$ $$\sqrt{(1 - b)(1 - c)} \leq {1 - b + 1 - c \over 2} = {a \over 2}$$ $$\sqrt{(1 - a)(1 - c)} \leq {1 - a + 1 - c \over 2} = {b \over 2}$$ Multiplying these together gives the desired inequality.