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!
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.