If $A < B < C < D$ where $A, B, C, D \in \mathbb{N}$, does the following series of inequalities hold? PROBLEM STATEMENT

If $A < B < C < D$ where $A, B, C, D \in \mathbb{N}$, does the following series of inequalities hold?
  $$\frac{A}{D}<\frac{A}{C}<\frac{A}{B}<\frac{B}{D}<\frac{B}{C}<\frac{C}{D}<1$$
  $$1<\frac{D}{C}<\frac{C}{B}<\frac{D}{B}<\frac{B}{A}<\frac{C}{A}<\frac{D}{A}$$

MY ATTEMPT
Since $A, B, C, D \in \mathbb{N}$, we obtain
$$\frac{A}{D} < \frac{B}{D} < \frac{C}{D} < 1$$
$$\frac{A}{C} < \frac{B}{C} < 1 < \frac{D}{C}$$
$$\frac{A}{B} < 1 < \frac{C}{B} < \frac{D}{B}$$
$$1 < \frac{B}{A} < \frac{C}{A} < \frac{D}{A}$$
Summarizing the first and the second:
$$\frac{A}{C} < \frac{B}{C} < 1 < \frac{D}{C} < \frac{D}{B} < \frac{D}{A}$$
Summarizing the third and the fourth:
$$\frac{A}{D} < \frac{A}{C} < \frac{A}{B} < 1 < \frac{C}{B} < \frac{D}{B}$$
Summarizing the first and the third:
$$\frac{A}{B} < 1 < \frac{C}{B} < \frac{D}{B} < \frac{D}{A}$$
Summarizing the second and the fourth:
$$\frac{A}{D} < \frac{A}{C} < \frac{B}{C} < 1 < \frac{D}{C}$$
Summarizing the first and the fourth:
$$\frac{A}{D} < \frac{B}{D} < \frac{C}{D} < 1 < \frac{B}{A} < \frac{C}{A} < \frac{D}{A}$$
Summarizing the second and the third:
$$\frac{A}{C} < \frac{B}{C} < 1 < \frac{C}{B} < \frac{D}{B}.$$
Following a different approach, we also have
$$\frac{A}{D}<\frac{A}{C}<\frac{A}{B}<1$$
$$\frac{B}{D}<\frac{B}{C}<1<\frac{B}{A}$$
$$\frac{C}{D}<1<\frac{C}{B}<\frac{C}{A}$$
$$1<\frac{D}{C}<\frac{D}{B}<\frac{D}{A}.$$
CONCLUSION
We are therefore sure about the validity of the series of inequalities
$$\frac{A}{D} < \frac{A}{C} < \frac{B}{C} < 1 < \frac{C}{B} < \frac{D}{B} < \frac{D}{A}.$$
(Note that $\min(A,B,C,D)=A$ and $\max(A,B,C,D)=D$.  Therefore, $A/D$ is the minimum possible fraction with numerator and denominator (distinct from the numerator) coming from the set $\{A,B,C,D\}$.  Reciprocally, $D/A$ is the maximum possible fraction.)
However, we cannot conclusively "decide" which member of the (unordered) pairs
$$\left\{\frac{D}{C},\frac{C}{B}\right\}$$
and
$$\left\{\frac{D}{B},\frac{B}{A}\right\}$$
is larger (or equivalently, smaller).
QUESTION

Is my analysis of the problem correct?  Or is the problem "decidable" using some other approach/in another context?

 A: You are correct. Consider the following examples:
$$A=1,B=3,C=4,D=5$$
In which $$\frac{D}{C}<\frac{C}{B},\frac{D}{B}<\frac{B}{A}$$
Also:
$$A=2,B=3,C=4,D=6$$
yields $$\frac{D}{C}>\frac{C}{B}, \frac{D}{B}>\frac{B}{A}$$
A: You are correct; here's a simpler way to analyze the problem.
From $C<D$ we get $1/D<1/C$, so $A/D<A/C$. Similarly, $A/C<A/B$ and $B/D<B/C$. Also $C/D<1$, since $C<D$. Moreover $B/D<C/D$, because $B<C$.
Note also that taking reciprocals of positive numbers reverses inequalities, so the second set is essentially the same as the first one.
The ones that could go wrong are
$$
\frac{A}{B}<\frac{B}{D}
$$
and
$$
\frac{B}{C}<\frac{C}{D}
$$
(together with the similar ones between the reciprocals).
The first one is equivalent to $AD<B^2$, the second one to $BD<C^2$.
So we'd like to find $AD\ge B^2$, which is obtained, for instance, with $A=2$, $D=8$ and $B=4$. Now $BD=32$, so we can choose $C=5$ and $BD>C^2$.
You may try and find examples where those inequalities both hold, or just one of them.
