Solving System of Inequalities with Examples

Can systems of polynomial inequalities be reduced to systems of quadratic inequalities?
Firstly, any system of polynomial inequalities can be reduced to a system of quadratic inequalities by increasing the number of variables and equations (for example, by setting a square of a variable equal to a new variable).

Instead of solving the equality Ax=b, I want to solve the inequality Ax>=b. In general, how can I solve this problem with n unknowns, or when A has n columns?
When A has 1 column (or, when each equation in this system has only 1 unknown x) I can solve it easily by isolating x in each inequality and combining the inequalities to yield 1 inequality that bounds x. When choosing a satisfactory x, I simply refer to the yielded inequality.
When A has 2 columns (or, when each equations has 2 unknowns x and y), I graph the constraints and find the shaded regions, and use the appropriate equations for different x intervals (i.e., for x = 0 to x = 10 use x + y <= 50 and for x = 11 to x = 20 use 10 x +10 y <= 30). When choosing a satisfactory pair (x , y), I plug in x to the equation appropriate for the x interval, and I can choose any y value within the bounds for y.

How to solve linear system of equations which have inequality constraints?
$y=mi{n}_x\frac{1}{2}||Dx-<!--\; -\; -->d|{|}_2^2$
$s.t.$
$Ax\le <!--\; \le \; -->c$

Counting integer solutions for a system of inequalities
I wish to enumerate the number of solutions of the system of equations and inequalities for 3 non-negative integer unknowns $x,y,z\ge <!--\; \ge \; -->0$: ($a$,$b$ specified)
$\begin{array}{rl}x+y+z& =a\\ x+y& >b\\ y+z& >b\end{array}$
Is there an elegant way of finding the number of solutions or must I use an exhaustive numerical algorithm?

If we consider N as a subset of the usual metric space of real numbers R, we can think in a fundamental system of neighbourhoods (FSN) of N. I need to prove that does not exist a countable FSN, and as a suggestion Dieudonne (The autor of the book Modern analysis) says: Use contradiction. If ($(a_{nm})$) is a doble sequence of positive numbers, the sequence $(b_n)=\frac{a_{nn}}{2}$ is such that for no integer $m$ is valid the inequality $b_n\ge <!--\; \ge \; -->a_{mn}$ for all the integers $n$.
How can I use this fact to prove it?

Let's say that we have a system of linear inequalities:
$\left[\begin{array}{cccc}{c}_{1,1}& c_{1,2}& \dots <!--\; \dots \; -->& c_{1,n}\\ c_{2,1}& c_{2,2}& \dots <!--\; \dots \; -->& c_{2,n}\\ ⋮<!--\; ⋮\; -->& ⋮<!--\; ⋮\; -->& \ddots <!--\; \ddots \; -->& ⋮<!--\; ⋮\; -->\\ c_{m,1}& c_{m,2}& \dots <!--\; \dots \; -->& c_{m,n}\end{array}\right]\times <!--\; \times \; -->\left[\begin{array}{c}{x}_1\\ x_2\\ ⋮<!--\; ⋮\; -->\\ x_n\end{array}\right]\ge <!--\; \ge \; -->\left[\begin{array}{c}{b}_1\\ b_2\\ ⋮<!--\; ⋮\; -->\\ b_m\end{array}\right]$
It can be represented in a matrix form:
$\mathbf{C}\mathbf{x}\ge <!--\; \ge \; -->\mathbf{b}$
Does it hold that:
$\mathbf{x}\ge <!--\; \ge \; -->{\mathbf{C}}^{-<!--\; -\; -->1}\mathbf{b}$
and that $\mathbf{C}$ is invertible if and only if the whole system is solvable?
P.S. All the numbers $x_i,b_i,c_{i,j}$ are real. Would restricting them to be integers change the answer?
EDIT 1: for matrices $\mathbf{x}$ and $\mathbf{y}$ it holds that $\mathbf{x}\ge <!--\; \ge \; -->\mathbf{y}$ if and only if every element of $\mathbf{x}$ is $\ge <!--\; \ge \; -->$ to corresponding element in $\mathbf{y}$.
EDIT 2: the $x_i$ are bounded to $[-<!--\; -\; -->2,2]$.

System of inequalities - proving $n=k$
Let's say we have the following inequalities:
$n<x+1\le <!--\; \le \; -->k+1$
and
$k<x+1\le <!--\; \le \; -->n+1$
How to prove that $n=k$?

System of inequalities with 3 variables
$\begin{array}{c}a-<!--\; -\; -->b+c>0\\ a+b+c<4\\ 9a-<!--\; -\; -->3b+c<-<!--\; -\; -->5\end{array}$

How can we express union of two or more constraints, for example $x\ge <!--\; \ge \; -->0\vee <!--\; \vee \; -->y\ge <!--\; \ge \; -->0$, in an inequality system as below:
$$\begin{gathered} f_1(x,y)\ge <!--\; \ge \; -->0 \\ {f}_2(x,y)\ge <!--\; \ge \; -->0 \\ {f}_3(x,y)\ge <!--\; \ge \; -->0 \\ ⋮<!--\; ⋮\; --> \end{gathered}$$
where $f_i\in <!--\; \in \; -->{\mathbb{C}}^1$. This feasible set is not necessarily a convex set, hence there is not any linear inequality system for this problem. Is there any non-linear one?

Simplifying a system of linear inequalities
Given two inequalities
$y\ge <!--\; \ge \; -->m_1(x-<!--\; -\; -->x_1)+y_1$
and
$y\ge <!--\; \ge \; -->m_2(x-<!--\; -\; -->x_2)+y_2$
Is there anyway to solve for the space that they both exist in?

Let $a,b,c$ be whole numbers so that
$1\le <!--\; \le \; -->a\le <!--\; \le \; -->3$
$2\le <!--\; \le \; -->b\le <!--\; \le \; -->4$
$3\le <!--\; \le \; -->c\le <!--\; \le \; -->5$
Find number of solutions of the equation, $a+b+c=10$. Note: Please use FPC/PnC to answer this instead of the binomial theorem

$|z-<!--\; -\; -->a_k|\le <!--\; \le \; -->R_k$
where $z=x+iy$ (complex number) and $a_k$ and $R_k$ are real numbers for $k=1,\dots <!--\; \dots \; -->,n$. Basically the inequality above shows circle with center $a_k$ and radius $R_k$. The question here is, if I write n inequalities as a system of inequalities and then solve this system, the solution will be the intersection of $n$ inequalities. But I want to find the union of $n$ inequalities. Is there any way to do that?

Consider the known parameters $a_1,...,a_4;d_1,d_2,d_3$ such that
$0<a_i<1$ $\mathrm{\forall <!--\; \forall \; -->}i\in <!--\; \in \; -->\{1,...,4\}$ and $\underset{i=1}{\overset{4}{\sum <!--\; \sum \; -->}}{a}_i=1$
$0<d_i<1$ and $\underset{i=1}{\overset{3}{\sum <!--\; \sum \; -->}}{d}_i=1$
The system, with unknowns $y_j^i$ $\mathrm{\forall <!--\; \forall \; -->}i\in <!--\; \in \; -->\{1,...,4\}$, $\mathrm{\forall <!--\; \forall \; -->}j\in <!--\; \in \; -->\{1,2,3\}$, is
$\{\begin{array}{l}{d}_1=a_1{y}_1^1+a_2{y}_1^2+...+a_4{y}_1^4\\ d_2=a_1{y}_2^1+a_2{y}_2^2+...+a_4{y}_2^4\\ d_3=a_1{y}_3^1+a_2{y}_3^2+...+a_4{y}_3^4\\ y_1^i+y_2^i+y_3^i=1\mathrm{\forall <!--\; \forall \; -->}i\in <!--\; \in \; -->\{1,...,4\}\\ 0<y_j^i<1\mathrm{\forall <!--\; \forall \; -->}i\in <!--\; \in \; -->\{1,...,4\}\text{, }\mathrm{\forall <!--\; \forall \; -->}j\in <!--\; \in \; -->\{1,2,3\}\end{array}$
Can some inequalities help to pin down an unique solution in a linear system of equations with infinite solutions?

Underdetermined system with inequality constraints
$Ax=b,$
where $A\in <!--\; \in \; -->{\mathbf{R}}^{m\times <!--\; \times \; -->n}$ with $m<n$, subject to
$0⪯<!--\; ⪯\; -->x⪯<!--\; ⪯\; -->c.$
1. I would like to know if there is any way to express the feasible set for this problem analytically.
2. Is there any way to obtain any of feasible solutions in closed form?

Given linear system:
$$\begin{gathered} 4+{\mathrm{\delta <!--\; \delta \; -->}}_1-<!--\; -\; -->{\mathrm{\delta <!--\; \delta \; -->}}_2\ge <!--\; \ge \; -->0 \\ 2-<!--\; -\; -->{\mathrm{\delta <!--\; \delta \; -->}}_1+{\mathrm{\delta <!--\; \delta \; -->}}_2\ge <!--\; \ge \; -->0 \\ 1+{\mathrm{\delta <!--\; \delta \; -->}}_1-<!--\; -\; -->{\mathrm{\delta <!--\; \delta \; -->}}_2+{\mathrm{\delta <!--\; \delta \; -->}}_3\ge <!--\; \ge \; -->0 \end{gathered}$$
How from there it follows that ${\mathrm{\delta <!--\; \delta \; -->}}_2={\mathrm{\delta <!--\; \delta \; -->}}_3=0$ and ${\mathrm{\delta <!--\; \delta \; -->}}_1\ne <!--\; \ne \; -->0$?

Subsystem of infeasible system of linear inequalities
Suppose that we have a system of linear inequalities described by
$Ax\le <!--\; \le \; -->b$
where $A$ is $m\times <!--\; \times \; -->n$ matrix.
I want to show that if the system is infeasible then it has a subsystem of at most rank(A)+1 inequalities such that the subsystem is also infeasible.
My is initial thought was to use Farkas Lemma on the original system and modify the certificate y to obtain y′ as a certificate for a subsystem of at most rank(A)+1 inequalities, but I am stuck from there, maybe my initial thought it wrong?

Power series and ratio test: confused about the interval of convergence
I have a question for you. I was asked to find the Maclaurin series of $\ln <!--\; \; -->(\sin <!--\; \; -->x/x)$ and to evaluate its convergence. After finding the power series, I've applied the ratio test and I've found that the series converges for $|-<!--\; -\; -->x^2/6+x^4/120|<1$. When I solve the system of inequalities, I find that it is actually impossible, because the first inequality is verified for every real value of x, while the second one has no solution. How can it be? Where do I go wrong?

Prove that if the linear system $Ax=b$ is perturbed in both $A$ and $b$
$(A+\mathrm{\delta <!--\; \delta \; -->}A)(x+\mathrm{\delta <!--\; \delta \; -->}x)=b+\mathrm{\delta <!--\; \delta \; -->}b$
then
$\frac{\Vert <!--\; \Vert \; -->\mathrm{\delta <!--\; \delta \; -->}x\Vert <!--\; \Vert \; -->}{\Vert <!--\; \Vert \; -->x\Vert <!--\; \Vert \; -->}\le <!--\; \le \; -->\frac{\mathrm{\kappa <!--\; \kappa \; -->}(A)}{1-<!--\; -\; -->\mathrm{\kappa <!--\; \kappa \; -->}(A)\frac{\Vert <!--\; \Vert \; -->\mathrm{\delta <!--\; \delta \; -->}A\Vert <!--\; \Vert \; -->}{\Vert <!--\; \Vert \; -->A\Vert <!--\; \Vert \; -->}}(\frac{\Vert <!--\; \Vert \; -->\mathrm{\delta <!--\; \delta \; -->}A\Vert <!--\; \Vert \; -->}{\Vert <!--\; \Vert \; -->A\Vert <!--\; \Vert \; -->}+\frac{\Vert <!--\; \Vert \; -->\mathrm{\delta <!--\; \delta \; -->}b\Vert <!--\; \Vert \; -->}{\Vert <!--\; \Vert \; -->b\Vert <!--\; \Vert \; -->})$

Converting inequalities into equalities by adding more variables
So my question is, it is valid to do so? What I'm trying to do is to find the range of x where the following is true:
$\mathbb{A}x\le <!--\; \le \; -->0,x\ge <!--\; \ge \; -->0$
where $\mathbb{A}$ is a specific $m\times <!--\; \times \; -->n$ coefficient matrix and $x$ is an $n\times <!--\; \times \; -->1$ vector. But is this the same as solving the following:
$\stackrel{~<!--\; ~\; -->}{\mathbb{A}}\stackrel{~<!--\; ~\; -->}{x}=0,\stackrel{~<!--\; ~\; -->}{x}\ge <!--\; \ge \; -->0$
where $\stackrel{~<!--\; ~\; -->}{\mathbb{A}}$ is an $m\times <!--\; \times \; -->(n+m)$ matrix and $\stackrel{~<!--\; ~\; -->}{x}$ is an $(n+m)\times <!--\; \times \; -->1$ vector. By solution, I mean writing the elements of the original $x$ in terms of the elements that are in $\stackrel{~<!--\; ~\; -->}{x}$ but not $x$.

Minimal set of inequalities
I have a set of m linear inequalities in Rn, of the form
$Ax\le <!--\; \le \; -->b$
These are automatically generated from the specification of my problem. Many of them could be removed because they are implied by the others.
I would like to find the minimal set of inequalities,
$A^{\prime }x\le <!--\; \le \; -->b^{\prime }$
such that a solution to the first problem is also a solution to the second, and vice versa.