Solving System of Inequalities with Examples

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:
$\begin{array}{r}\frac{1}{1+z^m}-<!--\; -\; -->ax<0\\ x-<!--\; -\; -->by<0\\ y-<!--\; -\; -->cz<0\end{array}$
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?

Trying to solve a system of inequalities
I have this kind of system of inequalities (with one equality):
1. $x+y+z=1$
2. $ax+by+cz\le <!--\; \le \; -->2$
3. $a^{2}x+b^{2}y+c^{2}z\le <!--\; \le \; -->6$
4. $a^{3}x+b^{3}y+c^{3}z\le <!--\; \le \; -->14$
and so on.. (I could continue with 5. 6. 7.... the power always increases by one and I know the value on the right).
x,y,z are my unknowns and a,b,c are known.
Could I get a solution (or an approximation) for x,y and z as the number of inequalities grows, I don't see it?

Solving System on Inequalities Numerically
I am stuck on a problem that says given for the range [750,1800] (where each element has a spacing of 5 units: [750,755,760,...,1795,1800]), to select 4 elements from this range who are all at least 150 units apart from each other.

how to solve for x in an inequality of this form:
$ax+b<cx+d<ex+f$ (a, b, c, d, e, f are constants; inequality signs may be "$\le <!--\; \le \; -->$") I was taught that we only needed to solve $ax+b<cx+d$ and $cx+d<ex+f$ and not $ax+b<ex+f$ but I do not understand why.

Solve the following system of equations
$\{\begin{array}{r}\sqrt{y^2-<!--\; -\; -->8x+9}-<!--\; -\; -->\sqrt[3]{xy-<!--\; -\; -->6x+12}=1\\ \sqrt{2(x-<!--\; -\; -->y{)}^2+10x-<!--\; -\; -->6y+12}-<!--\; -\; -->\sqrt{y}=\sqrt{x+2}\end{array}$

detecting hidden equality constraints in optimization
I have a system of linear inequalities $Ax\le <!--\; \le \; -->b$. It is possible that some inequality constraints actually create a linear equality constraint. for e.g. $ax\le <!--\; \le \; -->b$ and $ax\ge <!--\; \ge \; -->b$ implies $ax=b$. A more involved linear combination of inequalities can also lead to equality constraints. I want to detect the final equality constraint given that $Ax\le <!--\; \le \; -->b$. Can it be done?

Given a system of inequalities $Ax\le <!--\; \le \; -->b$ , how can I derive the upper and lower bound on $x$?
I have the following system of inequalities $Ax\le <!--\; \le \; -->b$, where $x_1$ and $x_2$ are unknown, and the $a$'s and $и$'s are constants.
$\left[\begin{array}{cc}{a}_{11}& a_{12}\\ a_{21}& a_{22}\end{array}\right]\left[\begin{array}{c}{x}_1\\ x_2\end{array}\right]\le <!--\; \le \; -->\left[\begin{array}{c}{b}_1\\ b_2\end{array}\right]$

Find one solution for system of inequalities (if exists)
If there exists a solution for following system of inequalities, find at least one, else show that there is no solution exists.
$x_4(x_1+x_7)>x_3(x_2+x_8)$
$x_3(x_5+x_8)>x_7(x_4+x_1)$
$x_7(x_6+x_4)>x_8(x_5+x_3)$
$x_8(x_3+x_2)>x_4(x_6+x_7)$
for all $i$, $0<x_i<1$ and $\underset{i}{\sum <!--\; \sum \; -->}{x}_i=1$

Property of a system of two inequalities
$\{\begin{array}{l}a+b>1\\ a-<!--\; -\; -->b>1\end{array}$
can I sum the second inequality to the first getting $a>1$? Or this property can be used only equations?

For what values of the parameter t does the following system of equations have
1. no solution
2. more than one solution
3. exactly one solution
First of all I wrote that as an augmented coefficient matrix:
$A=\left(\begin{array}{ccccc}1& 1& t& |& -<!--\; -\; -->1\\ 3& (t+1)& (t-<!--\; -\; -->1)& |& -<!--\; -\; -->1\\ t& 2& 1& |& 0\end{array}\right)$
I suppose there'd be no solution if one of the lines showed inequality and exactly one if the rank of A is equal to the number of unknowns, but I don't know how to apply this here. Furthermore: what's the requirement for this system to have more than one solution?

These four inequalities define a hollow (zero volume) cube. Is it possible to describe a hollow cube using a smaller system of (real scalar generalized-polynomial) inequalities?
$\begin{array}{rcl}{x}^2& \le <!--\; \le \; -->& 1\\ y^2& \le <!--\; \le \; -->& 1\\ z^2& \le <!--\; \le \; -->& 1\\ (x^2-<!--\; -\; -->1{)}^2(y^2-<!--\; -\; -->1{)}^2(z^2-<!--\; -\; -->1{)}^2& \le <!--\; \le \; -->& 0\end{array}$

A bowl is described by by the inequalities
$z\ge <!--\; \ge \; -->x^2+y^2-<!--\; -\; -->1,z\le <!--\; \le \; -->\sqrt{x^2+y^2}+1,z\ge <!--\; \ge \; -->0.$
Sketch the bowl and compute its volume.

Exercise: Let $P\in <!--\; \in \; -->{\mathbb{R}}^V$ be defined by the inequalities
$\begin{array}{rl}{x}_u\le <!--\; \le \; -->1& \text{ for every }u\in <!--\; \in \; -->V,(1)\\ x_u+x_v\ge <!--\; \ge \; -->1& \text{ for every edge }uv\in <!--\; \in \; -->E,(2)\end{array}$
node set $V=\{1,2,3,4,5,6,7,8,9\}$
Starting from the system $(1)-<!--\; -\; -->(2)$, give the cutting-plane proof of the inequality
$x_1+x_2+x_6\ge <!--\; \ge \; -->2$
What I've tried: I know that I need to show that there exists a nonnegative combination from the inequalities $(1)$ and $(2)$ such that $x_1+x_2+x_3\ge <!--\; \ge \; -->3$ holds. I don't know how unfortunately.
Question: How do I solve this exercise?

Let $P$ be a polyhedron in $[0,1{]}^n$ defined by the constraints$Ax\le <!--\; \le \; -->b$ for $A\in <!--\; \in \; -->{\mathbb{R}}^{m\times <!--\; \times \; -->n}$, $x\in <!--\; \in \; -->{\mathbb{R}}^n$, and $b\in <!--\; \in \; -->{\mathbb{R}}^m$.
In the solutions of an exercise, the following is mentioned:
"Since the first $n$ constraints are linearly independent, they correspond to a basic solution of the system which, a priori, may be feasible or infeasible. This solution is obtained by replacing inequalities with equalities and computing the unique solution of this linear system."
So I am quite confused about this:
1) Why does "linear independent constraints" imply that "basic solution"?
2) Is a basic solution not always feasible?

I don't understand why it's $|a-<!--\; -\; -->b|<c<a+b$ with the absolute value on the left.
So, we have this system of inequalities:
- $a+b>c$
- $a+c>b$
- $b+c>a$

Solve the system of inequalities and indicate all the integers which are in the solution set:
$\begin{array}{rl}3-2a& <13\\ 5a& <17\end{array}$

Suppose I have two systems of $n$ homogeneous inequalities of $k$ variables:
$Ax\ge <!--\; \ge \; -->0$ and $Bz\ge <!--\; \ge \; -->0,$
where both $A$ and $B$ are $n\times <!--\; \times \; -->k$ matrices such that for any $i=1,\dots <!--\; \dots \; -->,n$, and $j=1,\dots <!--\; \dots \; -->,k$ we have $|a_{ij}-<!--\; -\; -->b_{ij}|\le <!--\; \le \; -->\mathrm{ϵ<!--\; ϵ\; -->}$, where $\mathrm{ϵ<!--\; ϵ\; -->}>0$ is very small.
How close are solutions of systems of homogeneous linear inequalities with close coefficients?

Suppose that you are trying to solve the optimization problem:
$\begin{array}{lll}\text{maximize}& v\cdot <!--\; \cdot \; -->x& \\ \text{subject to}& Ax\ge <!--\; \ge \; -->b& A\in <!--\; \in \; -->{\mathbb{R}}^{m\times <!--\; \times \; -->n}\end{array}$
(i.e. trying to solve an optimization problem in $n$ variables with $m$ linear inequality constraints).
This problem can be reduced to running a solution finding algorithm on a different system of linear equations in $k$ variables. What is the smallest value of $k$ for which this can be done.

Basic proof showing a system of equations is inconsistent
I have 8 variables, all between 0 and 1. I have a system of 4 inequalities as follows
$\mid <!--\; \mid \; -->a_1-<!--\; -\; -->x_1\mid <!--\; \mid \; -->-<!--\; -\; -->\mid <!--\; \mid \; -->b_1-<!--\; -\; -->x_1\mid <!--\; \mid \; -->+b_2-<!--\; -\; -->a_2<0$
$\mid <!--\; \mid \; -->b_1-<!--\; -\; -->y_1\mid <!--\; \mid \; -->-<!--\; -\; -->\mid <!--\; \mid \; -->a_1-<!--\; -\; -->y_1\mid <!--\; \mid \; -->+a_2-<!--\; -\; -->b_2<0$
$\mid <!--\; \mid \; -->a_1-<!--\; -\; -->y_1\mid <!--\; \mid \; -->-<!--\; -\; -->\mid <!--\; \mid \; -->a_1-<!--\; -\; -->x_1\mid <!--\; \mid \; -->+x_2-<!--\; -\; -->y_2<0$
$\mid <!--\; \mid \; -->b_1-<!--\; -\; -->x_1\mid <!--\; \mid \; -->-<!--\; -\; -->\mid <!--\; \mid \; -->b_1-<!--\; -\; -->y_1\mid <!--\; \mid \; -->+y_2-<!--\; -\; -->x_2<0$
Is there a way to prve it without expanding the absolute values?

Let $(n_k)$ be a sequence of natural numbers starting at $n_1=2$ and growing as follows
$n_k\le <!--\; \le \; -->n_{k+1}\le <!--\; \le \; -->n_k^2+2$
As far as I know this sort of dynamical system may have no closed-form solution, but even if it has one, the $=$ replaced in it by $\le <!--\; \le \; -->$ would include more sequences than the above recurrence allows.