Solving System of Inequalities with Examples

Suppose we have some ODE of the form
$\stackrel{\dot{}<!--\; \dot{}\; -->}{x}=f(\mathrm{\alpha <!--\; \alpha \; -->},x),$
where $\mathrm{\alpha <!--\; \alpha \; -->}\in <!--\; \in \; -->{\mathbb{R}}^k$ is some vector of parameters. And for $\mathrm{\alpha <!--\; \alpha \; -->}\in <!--\; \in \; -->A$ we have $x=e_{\mathrm{\alpha <!--\; \alpha \; -->}}$ stable hyperbolic equilibriums for the corresponding equations. Now, lets change $\mathrm{\alpha <!--\; \alpha \; -->}$ for $\mathrm{\alpha <!--\; \alpha \; -->}(t)$ with $\mathrm{\alpha <!--\; \alpha \; -->}(t)\in <!--\; \in \; -->A,\mathrm{\forall <!--\; \forall \; -->}t$. Can we ensure (under some conditions on $f$ and $\mathrm{\alpha <!--\; \alpha \; -->}(t)$ that the equilibrium "will turn" into a trajectory $e(t)$ which attracts solutions (locally) of the non-autonomous equation $\stackrel{\dot{}<!--\; \dot{}\; -->}{x}=f(\mathrm{\alpha <!--\; \alpha \; -->}(t),x)$ (forward or in a pull-back sense)?
In particular, I'm working with a system for which I know the equilibriums for $\mathrm{\alpha <!--\; \alpha \; -->}\in <!--\; \in \; -->A$. And if I change α for a periodic function, I can prove (using degree theory) that a periodic solution exists, provided $\mathrm{\alpha <!--\; \alpha \; -->}(t)\in <!--\; \in \; -->A$ and the amplitude of oscillation is small. Is this solution a "perturbation" of the equilibrium? does it preserve the attractiveness?

Let:
$f(x_1,\cdots <!--\; \cdots \; -->,x_n)=\underset{i}{\prod <!--\; \prod \; -->}{x}_i(1-<!--\; -\; -->x_i)\underset{i<j}{\prod <!--\; \prod \; -->}|x_i-<!--\; -\; -->x_j|$
Suppose all $x_i\in <!--\; \in \; -->(0,1)$ are fixed and $\underset{i}{\sum <!--\; \sum \; -->}{x}_i<\frac{n}{2}$. Show that there is some $i$ and a sufficiently small $\mathrm{ϵ<!--\; ϵ\; -->}$ so that $x_i\mapsto <!--\; \mapsto \; -->x_i+\mathrm{ϵ<!--\; ϵ\; -->}$ doesn't decrease the value of $f$.
That is to say, at least one of the partial derivatives of $f$ is non-negative.
After taking the derivative in each $x_i$ one gets a system of inequalities. I was able to prove the statement for $n=2,3$ this way through basically brute force. This doesn't generalize well though.

Let A and B represent two linear inequalities:
$A:a_1{x}_1+...+a_n{x}_n\ge <!--\; \ge \; -->k_1$
$B:b_1{x}_1+...+b_n{x}_n\ge <!--\; \ge \; -->k_2$
If A and B is unsatisfiable (does not have solution), does the following hold in general (the conjunction of two inequalities implies the summation of them )? If so, I am looking for a formal proof?
$A\wedge <!--\; \wedge \; -->B⟹<!--\; ⟹\; -->A+B$
${𝑎}_1{𝑥}_1+...+{𝑎}_n{𝑥}_n\ge <!--\; \ge \; -->𝑘1\wedge <!--\; \wedge \; -->{𝑏}_1{𝑥}_1+...{𝑏}_n{𝑥}_n\ge <!--\; \ge \; -->{𝑘}_2⟹<!--\; ⟹\; -->{𝑎}_1{𝑥}_1+...+{𝑎}_n{𝑥}_n+{𝑏}_1{𝑥}_1+...{𝑏}_n{𝑥}_n\ge <!--\; \ge \; -->{𝑘}_1+{𝑘}_2$
and then I would like to generalize the above theorem to summation of several inequalities.

Looking for any complex number solutions to the system of equations:
$\begin{array}{rl}|a{|}^2+|b{|}^2+|c{|}^2& =\frac{1}{3}\\ \stackrel{¯<!--\; ¯\; -->}{a}b+a\stackrel{¯<!--\; ¯\; -->}{c}+\stackrel{¯<!--\; ¯\; -->}{b}c& =\frac{1}{6}(2+\sqrt{3}i).\end{array}$

how to solve system of linear inequations involving only $\ne <!--\; \ne \; -->$ inequalities
Specifically, I need to know how to find numbers $x_1,x_2,\dots <!--\; \dots \; -->,x_r$ such that they satisfy all of these inequalities:
$\{\begin{array}{rl}{a}_{11}{x}_1+\dots <!--\; \dots \; -->+a_{1r}{x}_r& \ne <!--\; \ne \; -->b_1\\ a_{21}x+\dots <!--\; \dots \; -->+a_{2r}{x}_r& \ne <!--\; \ne \; -->b_2\\ a_{s1}x+\dots <!--\; \dots \; -->+a_{sr}{x}_r& \ne <!--\; \ne \; -->b_s\end{array}$
Is there any general way to find these numbers?

Find the extrema of $f$ subject to the stated constraints.
$f(x,y)=x-<!--\; -\; -->y$, subject to $x^2-<!--\; -\; -->y^2=2$
I'm solving a problem involving Lagrange's multipliers, and I've got this system of equations to solve:
$1=\mathrm{\lambda <!--\; \lambda \; -->}2x$
$-<!--\; -\; -->1=-<!--\; -\; -->\mathrm{\lambda <!--\; \lambda \; -->}2y$
$x^2-<!--\; -\; -->y^2-<!--\; -\; -->2=0$

Question regarding Monoalphabetic Phi Test
$1\ge <!--\; \ge \; -->\mathrm{\varphi <!--\; \varphi \; -->}(T)\ge <!--\; \ge \; -->\frac{n-<!--\; -\; -->k}{k(n-<!--\; -\; -->1)}$
Where $\mathrm{\varphi <!--\; \varphi \; -->}(T)=\underset{i=1}{\overset{k}{\sum <!--\; \sum \; -->}}\frac{n_i(n_i-<!--\; -\; -->1)}{n(n-<!--\; -\; -->1)}$, $T=$ some text, n= length of text $T$, $k=$ size of the alphabet, and $n_i=$ the frequency of a letter from the alphabet within the text.
Finding the equalities was relatively easy; they exist when the size of the alphabet, $k$, is equal to one.Where I'm having trouble is proving the inequalities. I can pretty much see that, due to $\mathrm{\varphi <!--\; \varphi \; -->}(T)$ being a sum of frequencies, it'll always be less than one, but should I be proving this with induction? And further, I can't really see how to prove that $\mathrm{\varphi <!--\; \varphi \; -->}(T)><!--\; >\; -->\frac{n-<!--\; -\; -->k}{k(n-<!--\; -\; -->1)}$. Again, I can see how it works, and I've done it with a few examples, but I'm not too sure how to translate that into a rigorous proof.

Trying to solve the following inequality numerous times I've been reaching a wrong solution time after time:
$(x^2-<!--\; -\; -->4x+4{)}^x+(2-<!--\; -\; -->x{)}^x<2$
after setting all the demands using systems of inequalities etc, I reached the solution: $2<x<3,x>3$ which is wrong.
how can this be solved?

given a system of inequalities expressed in the following way:
$Ax>\underset{\_<!--\; \_\; -->}{k}$
where $A\in <!--\; \in \; -->{\mathbb{M}}_{n,m(\mathbb{R})}$, with n>m, and $\underset{\_<!--\; \_\; -->}{k}=(k,k,\dots <!--\; \dots \; -->,k)\in <!--\; \in \; -->{\mathbb{R}}^n$
In general, the system might or might not admit solutions. I would like to find a solution $x\in <!--\; \in \; -->{\mathbb{R}}^m$ that minimizes the number of violated inequalities.

Are the following range of values on $x_3$ correct given a system of inequalities with 3 variables?
Given the system of inequalities:
$\begin{array}{rl}{x}_3& \le <!--\; \le \; -->\frac{x_1+x_2-<!--\; -\; -->2}{2}\\ x_3& \ge <!--\; \ge \; -->x_1+3x_2\\ x_3& \ge <!--\; \ge \; -->1-<!--\; -\; -->x_2\\ -<!--\; -\; -->\frac{11}{3}& \le <!--\; \le \; -->x_2\le <!--\; \le \; -->-<!--\; -\; -->3\\ 13& \le <!--\; \le \; -->x_1<\mathrm{\infty <!--\; \infty \; -->}\end{array}$
I know that $x_1\in <!--\; \in \; -->[lowe{r}_{x_1},uppe{r}_{x_1})=[13,\mathrm{\infty <!--\; \infty \; -->})$ and $x_2\in <!--\; \in \; -->[lowe{r}_{x_2},uppe{r}_{x_2}]=[-<!--\; -\; -->11/3,-<!--\; -\; -->3]$.
In order to find $lowe{r}_{x_3},uppe{r}_{x_3}$, I did the following:
$lowe{r}_{x_3}=max\{x_1+3x_2,1-<!--\; -\; -->x_2\}=\{uppe{r}_{x_1}+3uppe{r}_{x_2},1-<!--\; -\; -->uppe{r}_{x_2}\}=\{\mathrm{\infty <!--\; \infty \; -->},4\}=\mathrm{\infty <!--\; \infty \; -->}$
$uppe{r}_{x_3}=\frac{lowe{r}_{x_1}+lowe{r}_{x_2}-<!--\; -\; -->2}{2}=\frac{13-<!--\; -\; -->11/3-<!--\; -\; -->2}{2}=\frac{11}{3}$
Is the above calculation correct? In general, is $uppe{r}_{x_n}=f(lowe{r}_{x_1},lowe{r}_{x_2},\dots <!--\; \dots \; -->,lowe{r}_{x_{n-<!--\; -\; -->1}})$ and is $lowe{r}_{x_n}=g(uppe{r}_{x_1},uppe{r}_{x_2},\dots <!--\; \dots \; -->,uppe{r}_{x_{n-<!--\; -\; -->1}})$, where $f$ and $g$ are some real-valued functions?

$T\{x[n]\}=\underset{k=\text{min}(n,n_0)}{\overset{\text{max}(n,n_0)}{\sum <!--\; \sum \; -->}}x[k]$
for some integer constant $n_0$.
Intuitively, it's unstable, and it can be easily proven by a counterexample if $x[n]$ is the unit step and $n_0=0$.However, the TA tried to use a general proof by invoking the triangle inequality. Assuming $x[n]$ is bounded (or $|x[n]|\le <!--\; \le \; -->M<\mathrm{\infty <!--\; \infty \; -->}$), he said:
$\left|T\{x[n]\}\right|=\left|\underset{k=\text{min}(n,n_0)}{\overset{\text{max}(n,n_0)}{\sum <!--\; \sum \; -->}}x[k]\right|\le <!--\; \le \; -->\underset{k=\text{min}(n,n_0)}{\overset{\text{max}(n,n_0)}{\sum <!--\; \sum \; -->}}|x[k]|\le <!--\; \le \; -->(|n-<!--\; -\; -->n_0|+1)M$
Obviously the right-hand side is unbounded, as it goes to infinity with increasing $n$, but to me it doesn't seem to imply that the system on the left-hand side is unbounded (because of the inequality).
My question is, can his attempt be augmented to show that the left-hand side is also unbounded? Or is a counter-example the only way to prove it?

Solving a system of quadratic inequalities
$\lfloor \sqrt{c^2(x^2+y^2)+2c(x+y)+2}+\frac{1}{2}\rfloor =c$
For a given positive integer c, I am trying to define algebraically, in terms of $x$ & $y$, the region in 2D space for which the above equation and the following conditions hold:
1. $c\in <!--\; \in \; -->\mathbb{N}$
2. $0\le <!--\; \le \; -->x,y\le <!--\; \le \; -->1$
I.E, which constraints/inequalities $x$ & $y$ must fulfill (in terms of $c$)

1. Why is the polyhedron, $P$ considered with the transposes $A^{T}p=0,b^{T}p=1$ as opposed to $p^{T}A=0^T$
2. Why does theorem 4.7 imply that if x is a solution to the subsystem we must have $0^{T}x=0\le <!--\; \le \; -->-<!--\; -\; -->1$? I also don't get how this is implied by ${\stackrel{^<!--\; ^\; -->}{p}}^{T}A=0^T,{\stackrel{^<!--\; ^\; -->}{p}}^{T}b=1,p\ge <!--\; \ge \; -->0$. Theorem 4.7 is given as:
Theorem 4.7 Suppose that the system of linear inequalities $Ax\le <!--\; \le \; -->b$ has at least one solution, and let d be some scalar. Then, the following are equivalent:(a) Every feasible solution to the system $Ax\le <!--\; \le \; -->b$ satisfes $c^{\prime }x\le <!--\; \le \; -->d$.(b) There exists some $p\ge <!--\; \ge \; -->0$ such that $p^{\prime }A=c^{\prime }$ and p'b\leq d.

Let's say I have a system of inequalities: $Ax\le <!--\; \le \; -->g$ for some $A\in <!--\; \in \; -->{\mathbb{R}}^{4\times <!--\; \times \; -->4}$, $x\in <!--\; \in \; -->{\mathbb{R}}^4$, $g\in <!--\; \in \; -->{\mathbb{R}}^4$, and $A$ is full rank. Here, the $\le <!--\; \le \; -->$ denotes element-wise inequality.
Specifically, I know that $x$ lies in a two-dimensional subspace of ${\mathbb{R}}^4$ (determined by the null space of some matrix $N$). What I'm interested in is the dimension of the solution to the above system of inequalities. More succinctly, I'm interested in the dimension of the set
$\{x\in <!--\; \in \; -->{\mathbb{R}}^4|x\in <!--\; \in \; -->\mathrm{N}\mathrm{u}\mathrm{l}\mathrm{l}(N),Ax\le <!--\; \le \; -->g\}$

Given the system
${\mathbf{x}}^{\prime }=A\mathbf{x}$
where
$A=\left[\begin{array}{ccc}a& 0& 4\\ -<!--\; -\; -->1& -<!--\; -\; -->1& 0\\ -<!--\; -\; -->2-<!--\; -\; -->a& 0& -<!--\; -\; -->3\end{array}\right]$
In what interval of a is the system asymptotically stable, and for what value of a is the system stable/unstable if such a case even exists?
Attempt
For a system to be asymptotically stable the real part of all eigenvalues must be negative.
$\mathrm{\Re <!--\; \Re \; -->}(\mathrm{\lambda <!--\; \lambda \; -->})<0,\text{for all}\mathrm{\lambda <!--\; \lambda \; -->}$
Since we are dealing with a $3\times <!--\; \times \; -->3$ matrix, I used maple to find the eigenvalues from the system matrix. I also tried to use Maple to solve the inequality case for the eigenvalues, but I don't think I'm getting correct results.
The results I get from Maple state that the system is asymptotically stable when
$-<!--\; -\; -->8<a\le <!--\; \le \; -->5-<!--\; -\; -->4\sqrt{3}$
I spoke with my peers, and they said that this interval of a is wrong. Can anyone see what is going wrong?

Prove $0$ is an exponentially stable equilibrium of the system $x^{\prime }=f(x)+g(x)$ if $f(0)=g(0)=0$
Besides the conditions in the title, we have:
1. $0$ is an exponential equilibrium of the system $y^{\prime }=f(y)$
2. $|g(x)|\le <!--\; \le \; -->\mathrm{\mu <!--\; \mu \; -->}|x|,\mathrm{\forall <!--\; \forall \; -->}x\in <!--\; \in \; -->{\mathbb{R}}^n$
3. $\mathrm{\mu <!--\; \mu \; -->}$ is sufficiently small!
What I have tried so far is that:
1. $||y(0)||<{\mathrm{\delta <!--\; \delta \; -->}}_1\Rightarrow <!--\; \Rightarrow \; -->||y(t)||<{\mathrm{\alpha <!--\; \alpha \; -->}}_1||y(0)||e^{-<!--\; -\; -->{\mathrm{\beta <!--\; \beta \; -->}}_{1}t},t\ge <!--\; \ge \; -->0$
2. We want to prove: $\mathrm{\exists <!--\; \exists \; -->}\mathrm{\delta <!--\; \delta \; -->},\mathrm{\alpha <!--\; \alpha \; -->},\mathrm{\beta <!--\; \beta \; -->}>0,$ such that
$||x(0)||<\mathrm{\delta <!--\; \delta \; -->}\Rightarrow <!--\; \Rightarrow \; -->||x(t)||<\mathrm{\alpha <!--\; \alpha \; -->}||x(0)||e^{-<!--\; -\; -->\mathrm{\beta <!--\; \beta \; -->}t},t\ge <!--\; \ge \; -->0$
Considering the following system:
1. $x_1^{\prime }(t)=f(x_1)$
2. $x_2^{\prime }(t)=g(x_2)$
Then we will have:
1. $\mathrm{\exists <!--\; \exists \; -->}{\mathrm{\alpha <!--\; \alpha \; -->}}_1,{\mathrm{\beta <!--\; \beta \; -->}}_1,\mathrm{\delta <!--\; \delta \; -->}>0$ such that
$||x_1(0)||<{\mathrm{\delta <!--\; \delta \; -->}}_1\Rightarrow <!--\; \Rightarrow \; -->||x_1(t)||<{\mathrm{\alpha <!--\; \alpha \; -->}}_1||x_1(0)||e^{-<!--\; -\; -->{\mathrm{\beta <!--\; \beta \; -->}}_{1}t},t\ge <!--\; \ge \; -->0$
2. $x_2^{\prime }(t)\le <!--\; \le \; -->\mathrm{\mu <!--\; \mu \; -->}|x_2(t)|$, and by Gronwall inequality $x_2(t)\le <!--\; \le \; -->x_2(0)e^{\mathrm{\mu <!--\; \mu \; -->}t}$
I have a sense that it might be somewhat close to what we want to get. And I am stuck at this stage.

I have a personal problem I want to solve. I have a system of linear inequalities with 97 unknowns and 150000 ineqaulities, I think formal notation of the problem should be something like this
$\left[\begin{array}{ccc}{x}_{00000101}& \cdots <!--\; \cdots \; -->& x_{00000197}\\ ⋮<!--\; ⋮\; -->& \ddots <!--\; \ddots \; -->& ⋮<!--\; ⋮\; -->\\ x_{15000001}& \cdots <!--\; \cdots \; -->& x_{15000097}\end{array}\right]\times <!--\; \times \; -->\left[\begin{array}{c}{y}_1\\ ⋮<!--\; ⋮\; -->\\ y_{97}\end{array}\right]=\left[\begin{array}{c}{z}_1\\ ⋮<!--\; ⋮\; -->\\ z_{150000}\end{array}\right]$
$\mathrm{\forall <!--\; \forall \; -->}z\in <!--\; \in \; -->\{z_i|i\in <!--\; \in \; -->[1,150000]\},z_i>0$
We also know that $\mathrm{\forall <!--\; \forall \; -->}x\in <!--\; \in \; -->$ leftmost matrix $x\in <!--\; \in \; -->\{-<!--\; -\; -->1,0,1\}$ and there is exacly 5 1s and 5 −1's in every row in leftmost matrix. So, each row sums up to 0.
What I want to do is to determine if I can order yi's and if so, I want to find an algorithm to order them.

How to solve the following system of linear inequalities in x, y and z?
$x\ge <!--\; \ge \; -->\frac{y+z}{1.9}$
$y\ge <!--\; \ge \; -->\frac{x+z}{2.25}$
$z\ge <!--\; \ge \; -->\frac{x+y}{1.6}$
$x,y,z\ge <!--\; \ge \; -->0$

For each nonnegative integer $n$, calculate the number of triples $(a,b,c)$ of nonnegative integers which satisfy the system of inequalities below:
$\{\begin{array}{l}a+b\le <!--\; \le \; -->2n\\ a+c\le <!--\; \le \; -->2n\\ c+b\le <!--\; \le \; -->2n\end{array}$

How would one proceed with proving that there exists none $(x,y,z)$ such that all the following hold:
$x+2y+7z=0$
$\mathbb{P}(x+3y+9z\ge <!--\; \ge \; -->0)=1$
$\mathbb{P}(x+y+5z\ge <!--\; \ge \; -->0)=1$
$\mathbb{P}(x+13y+10z\ge <!--\; \ge \; -->0)=1$
$\mathbb{P}(x+3y+9z>0)>0$
$\mathbb{P}(x+y+5z>0)>0$
$\mathbb{P}(x+13y+10z>0)>0$