How to solve 3D system of nonlinear inequalities? I have a system of 3 nonlinear inequalities in 3 variables, and I need to find $x, y, z$, in terms of the parameters, that satisfies the equations. The system is:
\begin{equation}
  \begin{aligned}
    \frac{1}{1+z^m}-ax < 0\\
    x-by<0\\
    y-cz<0
  \end{aligned}
\end{equation}
Where $a, b, c > 0$ are constants and $m$ a positive integer. How can I go about solving for x, y, z in terms of the parameters? Thanks for any help with this!
 A: Assume the additional constraint $z > 0$.

The last two inequalities 
$$
\left\lbrace
\begin{align*}
&x-by < 0\\[4pt]
&y-cz < 0\\[4pt]
\end{align*}
\right.
$$
of the given system trap $y$ by
$$
\frac{x}{b} < y < cz
$$
so we must have $x < bcz$.

Using the first inequality 
$$
\frac{1}{(1+z^m)}-ax < 0
$$
of the given system together with $x < bcz$, we get that $x$ is trapped by
$$
\frac{1}{a(1+z^m)} < x < bcz
$$
so a necessary condition on $z$ is
$$
\frac{1}{a(1+z^m)} < bcz
$$
Reversing the process, we can solve the given system as follows . . .

Let $z$ be any positive real number such that
$$
\frac{1}{a(1+z^m)} < bcz
$$
Note that the above inequality holds for all sufficiently large $z$ since, as $z$ approaches infinity, the $\text{LHS}$ approaches zero and the $\text{RHS}$ approaches infinity.

For the chosen $z$, choose any $x$ such that
$$
\frac{1}{a(1+z^m)} < x < bcz
$$
Then for the chosen $x,z$, we have
$$
\frac{x}{b} < cz
$$
hence if we now choose any $y$ such that
$$
\frac{x}{b} < y < cz
$$
it follows that the system
$$
\left\lbrace
\begin{align*}
&\frac{1}{(1+z^m)}-ax < 0\\[4pt]
&x-by < 0\\[4pt]
&y-cz < 0\\[4pt]
\end{align*}
\right.
$$
is satisfied, and moreover, with the additional constraint $z > 0$, the choices of $x,y,z$ as described yield all solutions.

As an example, the triple $(x,y,z)$ given by
$$
\left\lbrace
\begin{align*}
&x=\frac{1}{a}\\[4pt]
&y=\frac{2}{ab}\\[4pt]
&z=\frac{3}{abc}\\[4pt]
\end{align*}
\right.
\qquad\qquad\;\;
$$
always satisfies the given system.
