Solving System of Inequalities with Examples

Let $A=C^{\ast <!--\; \ast \; -->}(1,T_1,T_2,...|T_i^{\ast <!--\; \ast \; -->}=T_i,||T_i||⩽<!--\; ⩽\; -->1)$ - universal enveloped $C^{\ast <!--\; \ast \; -->}$-algebra of countable family of selfadjoint operators. I want to know as more as possible about that algebra, so, any links are welcome.
In that $C^{\ast <!--\; \ast \; -->}$∗-algebra we have a family of Banach subspaces $B_n$ which defines like $B_n=\{${ All words generated by $T_i$ which contain at least one $T_n$ and does not contain $T_{n+1}$}.
Let $e_n\in <!--\; \in \; -->B_n$ some system of vectors. It is obviously linear independent, but moreover, it is Schauder base system. It is exist simple argument: let ${\mathrm{\pi <!--\; \pi \; -->}}_n$ - is representation of $A$ which seems like ${\mathrm{\pi <!--\; \pi \; -->}}_n(T_k)=T_k$ if $k⩽<!--\; ⩽\; -->n$ and ${\mathrm{\pi <!--\; \pi \; -->}}_n(T_k)=0$ otherwise. So, we have inequality
$\mathrm{\forall <!--\; \forall \; -->}(a_n)\subset <!--\; \subset \; -->\mathbb{C}\mathrm{\forall <!--\; \forall \; -->}n\in <!--\; \in \; -->\mathbb{N}\mathrm{\forall <!--\; \forall \; -->}m\le <!--\; \le \; -->n\Vert \underset{k=1}{\overset{m}{\sum <!--\; \sum \; -->}}{a}_k{e}_k\Vert \le <!--\; \le \; -->\Vert {\mathrm{\pi <!--\; \pi \; -->}}_m(\underset{k=1}{\overset{n}{\sum <!--\; \sum \; -->}}{a}_k{e}_k)\Vert \le <!--\; \le \; -->\Vert \underset{k=1}{\overset{n}{\sum <!--\; \sum \; -->}}{a}_k{e}_k\Vert$
and it is sufficient condition to be $e_n$ Schauder basic system.
So, my question: does this base system have some good properties?

Solve a system of inequalities
$\{\begin{array}{ll}{\log }_2^2<!--\; \; -->(-<!--\; -\; -->{\log }_2<!--\; \; -->x)+{\log }_2<!--\; \; -->{\log }_2^2<!--\; \; -->x\le <!--\; \le \; -->3& \\ -<!--\; -\; -->4|x^2-<!--\; -\; -->1|-<!--\; -\; -->3\ge <!--\; \ge \; -->\frac{1}{x^2-<!--\; -\; -->1}& \end{array}$
What I've tried:Make substitution $t=x^2-<!--\; -\; -->1$ and solve second inequality:
$-<!--\; -\; -->4|t|-<!--\; -\; -->3\ge <!--\; \ge \; -->\frac{1}{t}<=>-<!--\; -\; -->4|t|-<!--\; -\; -->3-<!--\; -\; -->\frac{1}{t}\ge <!--\; \ge \; -->0$
Where I got interval for $t$: $-<!--\; -\; -->\frac{1}{4}\le <!--\; \le \; -->t<0$
And then for $x$: $x\in <!--\; \in \; -->(-<!--\; -\; -->1;-<!--\; -\; -->\frac{\sqrt{3}}{2}]\cup <!--\; \cup \; -->[\frac{\sqrt{3}}{2};1)$
But I have no thoughts how to solve first inequality.

Suppose there are $3$ runners $R1$,$R2$ and $R3$. The betting odds paid on each runner are $r1:1$, $r2:1$ and $r3:1$. This means that if, for example, you bet $\mathrm{\$<!--\; \$\; -->}2$ for runner $1$ winning, you will make $2((r1+1)/1)-<!--\; -\; -->2$ if runner 1 ends up winning.
The actual probabilities of each runner winning are unspecified. How much should you bet on each runner so that you always end up with a profit?
Is it simply a matter of solving the following system of inequalities (let $a,b$ and $c$ be the amount you bet for each of the runners) for $a,b$ and $c$:
$$\begin{gathered} a(r1+1)-<!--\; -\; -->a-<!--\; -\; -->b-<!--\; -\; -->c>0 \\ b(r2+1)-<!--\; -\; -->b-<!--\; -\; -->a-<!--\; -\; -->c>0 \\ c(r3+1)-<!--\; -\; -->c-<!--\; -\; -->b-<!--\; -\; -->a>0 \end{gathered}$$

Graph the following system of inequalities. Show (by shading in) the feasible region.
$x+2y\le <!--\; \le \; -->12$
$2x+y\le <!--\; \le \; -->12$
$x\ge <!--\; \ge \; -->0,y\ge <!--\; \ge \; -->0$

Solve $\frac{1}{x^2-<!--\; -\; -->1}⩾<!--\; ⩾\; -->\frac{1}{x+1}$

What is the domain of $1/\sqrt{x^2-<!--\; -\; -->4}$ using the set builder notation?
I'm confused when it comes to getting the domain of a function using set builder notation. Please explain also how to do the system of inequalities when getting the domain of the given function
$x\mapsto <!--\; \mapsto \; -->\frac{1}{\sqrt{x^2-<!--\; -\; -->4}}$

For linear equalities $Ax=b$, the solution can be written as $x=(A^{T}A{)}^{-<!--\; -\; -->1}{A}^{T}b$.
For linear inequalities system $Ax\ge <!--\; \ge \; -->b$, on what condition the solution is $x\ge <!--\; \ge \; -->(A^{T}A{)}^{-<!--\; -\; -->1}{A}^{T}b$?

Unique solution for parametric system of two equations
The system is:
$$\begin{gathered} x-<!--\; -\; -->2y+z=2\mathrm{\alpha <!--\; \alpha \; -->} \\ 3(xy+xz+yz)=3\mathrm{\alpha <!--\; \alpha \; -->}-<!--\; -\; -->4 \end{gathered}$$

This two equations. How we can solve them?
$\{\begin{array}{l}{\displaystyle \frac{1}{\sqrt{x+1}+1}+\frac{1}{\sqrt{y+1}+1}=\frac{2}{3}}\\ {\displaystyle \sqrt{\frac{1}{x^2}+\frac{1}{y\phantom{{}^2}}}+\sqrt{\frac{1}{y^2}+\frac{1}{x}}-<!--\; -\; -->\frac{2}{\sqrt{xy}}=\frac{2}{3}}\end{array}$

Need to find a solution to the following system of linear inequalities:
$\begin{array}{rl}{x}_1-<!--\; -\; -->x_2& \le <!--\; \le \; -->1\\ x_1-<!--\; -\; -->x_4& \le <!--\; \le \; -->-<!--\; -\; -->4\\ x_2-<!--\; -\; -->x_3& \le <!--\; \le \; -->2\\ x_2-<!--\; -\; -->x_5& \le <!--\; \le \; -->7\\ x_2-<!--\; -\; -->x_6& \le <!--\; \le \; -->5\\ x_3-<!--\; -\; -->x_6& \le <!--\; \le \; -->10\\ x_4-<!--\; -\; -->x_2& \le <!--\; \le \; -->2\\ x_5-<!--\; -\; -->x_1& \le <!--\; \le \; -->-<!--\; -\; -->1\\ x_5-<!--\; -\; -->x_4& \le <!--\; \le \; -->3\\ x_6-<!--\; -\; -->x_3& \le <!--\; \le \; -->-<!--\; -\; -->8\end{array}$
Is there any systematic way to find a solution?

Finding solution of a set of linear equations and inequalities.
Set of 28 linear equations with 28 unknows ($x_i$), so looking like this:
$b_i$=$\underset{n=1}{\overset{28}{\sum <!--\; \sum \; -->}}{x}_i\cdot <!--\; \cdot \; -->a_i$, for $i=1...28$ and where $a_i$ is either 0 or 1.
Unfortunately, the rank of the A matrix (28x28), containing the parameters $a_{i,j}$ is 17. However, I do know for every of the 28 unknowns $x_i$ that it should be a whole number and that:
1 $\le <!--\; \le \; -->$ $x_i$ and $x_i$ $\le <!--\; \le \; -->$ 26.
So the way I see it is that I have 17 independent equations and 2*28=56 inequalities. Is it possible to solve this problem and to find values for $x_i$, given these 17 equations and 56 inequalities?

For a linear system of equations constrained by inequalities, is $Ax\le <!--\; \le \; -->b=>y^{T}Ax\le <!--\; \le \; -->y^{T}b$ acceptable? Or does that not generally hold.
($y^T$ being the transpose of $y$).

Solve a system of linear inequalities without graphing, i.e. using pure algebra?
$y\ge <!--\; \ge \; -->2x+1$
$y>(x/2)-<!--\; -\; -->1$

How do you find all solutions of the following system of inequalities assuming positive and real $x$ and $y$?
$xy<1$
$2+x(y-<!--\; -\; -->1)>0$
I have rearranged the first inequality to get $y<\frac{1}{x}$ and solved the second inequality for $y$ to get
$y>1-<!--\; -\; -->\frac{2}{x}$
Thus $\frac{1}{x}\ge <!--\; \ge \; -->1-<!--\; -\; -->\frac{2}{x}\Rightarrow <!--\; \Rightarrow \; -->x\le <!--\; \le \; -->3$ and the first one gives $y<\frac{1}{3}$.
Similarly solving the second equation for $x$ gives $x>\frac{2}{1-<!--\; -\; -->y}$ so assuming that $y\ne <!--\; \ne \; -->1$ we get $\frac{1}{y}\ge <!--\; \ge \; -->\frac{2}{1-<!--\; -\; -->y}$ and then $y\le <!--\; \le \; -->\frac{1}{3}$ and the first one gives $x<3$.
Now assuming $y=1$, we get $x<1$.
So the solutions found are:
$x\le <!--\; \le \; -->3\text{ AND }y<\frac{1}{3}$
$x<3\text{ AND }y\le <!--\; \le \; -->\frac{1}{3}$
$x<1\text{ AND }y=1$
This seemed like a good method but I have missed the solution $x=1,y<1$
How do you solve such a system to include all solutions?

Find the number of solutions to
$a-<!--\; -\; -->b^2\ge <!--\; \ge \; -->\frac{1}{4},b-<!--\; -\; -->c^2\ge <!--\; \ge \; -->\frac{1}{4},c-<!--\; -\; -->d^2\ge <!--\; \ge \; -->\frac{1}{4},d-<!--\; -\; -->a^2\ge <!--\; \ge \; -->\frac{1}{4}$

Consider the system of linear inequalities: $Ax⪯<!--\; ⪯\; -->b$.
The alternative system of inequalities is: $\mathrm{\lambda <!--\; \lambda \; -->}⪰<!--\; ⪰\; -->0,A^T\mathrm{\lambda <!--\; \lambda \; -->}=0,b^T\mathrm{\lambda <!--\; \lambda \; -->}\le <!--\; \le \; -->0$
These are strong alternatives since the optimum in the related system (A) below is achieved unless it is unbounded below.
(A) minimize s, subject to: $f_i(x)-<!--\; -\; -->s\le <!--\; \le \; -->0$ for $i=1,\dots <!--\; \dots \; -->,m;Ax=b$.

I have to graphically represent the following subset in the complex plane being z a complex number: $A=1<|z|<2$
What I did previously was solve it like it was a regular inequality system in the real numbers, resulting in $b<-<!--\; -\; -->a+4$ and $b>-<!--\; -\; -->a+1$
How can I solve this?

System of Linear Inequalities to create feasible solution $y=min(x_1,x_2)$
The question is $L_1\le <!--\; \le \; -->x_1\le <!--\; \le \; -->U_1$,...,$L_n\le <!--\; \le \; -->x_n\le <!--\; \le \; -->U_n$. Can we introduce decision variables and define a system of mixed-integer linear inequalities whose feasible solution is $y=min(x_1,...,x_n)$?

Let $Ax\le <!--\; \le \; -->b$ be a system of linear inequalities where $A\in <!--\; \in \; -->R^{m\times <!--\; \times \; -->n}$, $x\in <!--\; \in \; -->R^n$ and $b\in <!--\; \in \; -->R^m$
Suppose $A$ is a matrix with linearly independent rows.
I wonder what would be the geometric or intuitive interpretations of these independent constraints?