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?
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.