if $x, y, z$ are $3 \leq x \leq y \leq z$ and they are odd numbers, show that $1 < (xyz - 1) / (x-1)(y-1)(z-1) < 3$ There are three question. Maybe there are related.
(1) if x, y, z are integers greater than 2, showing that
$$ \frac{xyz - 1}{(x-1)yz} < \frac{xyz}{(x-1)(y-1)(z-1)}$$
(2) if $x, y, z$ are $3 \le x \le y \le z$ and they are odd numbers, showing that $$1 < \frac{xyz - 1}{(x-1)(y-1)(z-1)} < 3$$
(3) if function $g(x) = x^3 + Ax^2 + Bx + C$ satisfy the following two conditions, get $A, B, C$

*

*Equation $g(x) = 0$ have three different integers in their roots, and all of roots are odd numbers greater than 3

*$\frac{A+B+2C+2}{A+B+C+1}$ is an integer.

(1) was easy.. $xyz - 1 < xyz$ and $(x-1)yz > (x-1)(y-1)(z-1)$ so it is correct.
But I'm stuck with (2), especially $\frac{xyz - 1}{(x-1)(y-1)(z-1)} < 3$
 A: Presumably part 3 wants to use part 2 to conclude that $(C+2)/(A+B+C+1)=2$.
If so, I believe that part 2's condition should be $ 3 \leq x < y < z$, reflecting the "three different integers in their roots". As such, I will make this change to the question, and proceed accordingly.

Hints/guide. If you're stuck, show what you've tried.
(2) $1 < \frac{xyz-1}{(x-1)(y-1)(z-1)}$ is obvious by expanding it out to get $ xy + yz + zx > x + y + z$.
For the other inequality, show that $ f(x,y,z) =3(x-1)(y-1)(z-1) - (xyz-1)$ is increasing in each variable.

 The coefficient of $z$ is $2xy-3x-3y+3 = x(y-3)+y(x-3)+3 > 0$.
 Similarly for the coefficient of $x$, $y$.

Hence we just need to verify that $ f(3,5,7)>0$ .
(3) Let $g(X) = X^3+AX^2+BX + C = (X-x)(X-y)(X-z)$.
We are given that $\frac{A+B+2C+2}{A+B+C+1} = 1 + \frac{-xyz+1}{(1-x)(1-y)(1-z)}$ is an integer.
The previous part tells us that the second term is bounded strictly between 1 and 3, so has to be 2. Thus: $xyz-1 = 2 (x-1)(y-1)(z-1)$.
Show that has a unique solution $(3, 5, 15)$, hence $A, B, C$ can be determined.

 You could modify the $f(x,y,z)$ approach above to consider $h(x,y,z) = 2 (x-1)(y-1)(z-1) - (xyz-1)$ and look at the finitely many cases where it is non-positive.


 As before, the coefficient of $z$ is $xy-2x-2y+2 =(x-2)(y-2)-2 > 0$.
$h(3,7,9) >0$, so any solution is restricted to $ x = 3, y = 5$.
 We verify that this gives $z=15$.


Note: We can actually remove the "odd numbers" requirement.
(2) holds because $f(3,4,5) =13 > 0$.
(3) holds after checking the additional cases to verify that there are no extra solutions.

 The only use of odd numbers was to force the coefficient $xy-2x-2y+2$ to be positive.
 In the case of $ x = 3, y = 4$, the coefficient is actually $1\times 2 - 2 = 0 $, so $h(3,4, z) = -11 \, \forall z$. This clearly yields no solutions.
 The rest follows in a similar manner as before: $h(3,6,10) > 0, h(3,7,8) > 0, h(4,5,6) > 0    $

