If $x + y + z = 2$, then show $\frac{x y z}{(1-x)(1-y)(1-z)} \geq 8$, added a second question(Problem 2). Problem number 1:
The problem is that $x, y, z$ are proper fractions, and each one of them is greater than zero.

Given $x + y + z = 2$, prove 
  $$\frac{x y z}{(1-x)(1-y)(1-z)} \geq 8$$

I have tried to solve this using AM $\geq$ GM inequality.
Attempt : 
$$\frac{\dfrac{1-x}{x} + \dfrac{1-y}{y} + \dfrac{1-z}{z}}{3}\geq\left(\frac{(1-x)(1-y)(1-z)}{xyz}\right)^{1/3}$$
What should I do to calculate the value of $\dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{z}$?
Doubt: 


*

*Let $a, b, c, d$ all be positive numbers.

*Let $a > c$ and $b > d$.

*Can we say that $\dfrac {a}{b} > \dfrac{c}{d}$?


Problem number 2:
$x,y,z$ are unequal positive quantities, prove that 
$(1+x^3)(1+y^3)(1+z^3) > (1+xyz)^3$
My attempt:
$ \frac{x^3+y^3+z^3}{3} > xyz$
$ => (1+x^3)+(1+y^3)+(1+z^3) > 3(1+xyz)$
now by cubing the both sides we get,
$ => \frac{(1+x^3)+(1+y^3)+(1+z^3)}{27} > (1+xyz)^3$ ---(eqn. 1)
$ \frac{(1+x^3)+(1+y^3)+(1+z^3)}{3} > {\biggl((1+x^3)(1+y^3)+(1+z^3)\biggl)}^{\frac{1}{3}}$
$=> \frac{(1+x^3)+(1+y^3)+(1+z^3)}{27} > {\biggl((1+x^3)(1+y^3)+(1+z^3)\biggl)}$ ---(eqn. 2)
So, now if we do : $\frac{(eqn. 1)}{(eqn. 2)}$ 
then the result will be : 
${\biggl((1+x^3)(1+y^3)+(1+z^3)\biggl)} > (1+xyz)^3$
and this is the correct result for all $x,y,z$ greater than zero.
But if we do $\frac{(eqn. 2)}{(eqn. 1)}$ then the result will be :
${\biggl((1+x^3)(1+y^3)+(1+z^3)\biggl)} < (1+xyz)^3$
So, my question is, how to prove it correctly without doing the division operation, cause when we divide the two equations then we get two separate results. 

$\dfrac ab-\dfrac cd=\dfrac{ad-bc}{be}$ will be $\ge0$
only if $ad-bc>0\iff \dfrac ab>\dfrac cd$
Alternatively
Let $a=c+p,b=d+q,p,q>0$
$\dfrac ab-\dfrac cd=\dfrac{c+p}{d+q}-\dfrac cd=\dfrac{d(c+p)-c(d+q)}{...}$
$=\dfrac{dp-cq}{...}$
will be $>0$
only if $dp-cq>0\iff \dfrac dc>\dfrac qp$
as $c,d,p,q>0$
