Solving System of Inequalities with Examples

How to solve systems of polynomial inequalities?

Are convex polytopes Borel sets? Suppose $P\subset <!--\; \subset \; -->{\mathbb{R}}^n$ is a given convex polytope, let's say by a system of linear inequalities $Ax\le <!--\; \le \; -->b$. Is that a Borel set?

Let $f:\mathbb{R}\to <!--\; \to \; -->\mathbb{R}$ be a continuous function and $g:\mathbb{R}\to <!--\; \to \; -->\mathbb{R}$ be a Lipschitz function. Would you help me to prove that the system of differential equation
$x^{\prime }=g(x)$
$y^{\prime }=f(x)y$
with initial value $x(t_0)=x_0$ and $y(t_0)=y_0$ has a unique solution.
Could I prove the uniqueness solution of $x^{\prime }=g(x)$, $x(t_0)=x_0$ by Gronwall Inequality first then use the result to prove the second?

Proof of addition and subtraction rules for systems of inequalities.
$a<b$ and $c<d$
Then $a+c<b+d$ is a valid move, but subtraction isn't. And similarly for a system such as $e>f$ and $g<h$ then $e-<!--\; -\; -->g>f-<!--\; -\; -->h$ is valid and so is $g-<!--\; -\; -->e<h-<!--\; -\; -->f$, but addition isn't valid.

the linear system of inequalities
$(S):Ax\ge <!--\; \ge \; -->0,x\ge <!--\; \ge \; -->0$
has a solution $x\in <!--\; \in \; -->{\mathbb{R}}^n$ that is not zero $x\ne <!--\; \ne \; -->0$. Here $A$ is a $n\times <!--\; \times \; -->n$ square matrix with real entries.
Of course I could combine the two inequalities in (S) into one and write it in the form of $\stackrel{~<!--\; ~\; -->}{A}x\ge <!--\; \ge \; -->0$ where $\stackrel{~<!--\; ~\; -->}{A}$ is $2n\times <!--\; \times \; -->n$.

Prove that the system
$\{\begin{array}{l}a,b,d,e,f,g,h,i>0\\ -<!--\; -\; -->a+e-<!--\; -\; -->i>0\\ -<!--\; -\; -->ae+ai+bd-<!--\; -\; -->ei+fh>0\\ aei-hfa-<!--\; -\; -->bdi-gbf>0\end{array}$
is inconsistent.
I tried using some previously acquired techniques (such as factoring, or multiplying an equality and adding it to another equality) that worked for many families of such systems, but I have only proven it for the case when $i\le <!--\; \le \; -->a$. Here is what I have so far.
From the first inequality, we get
$e>a+i\Rightarrow <!--\; \Rightarrow \; -->e>a.(\ast <!--\; \ast \; -->)$
Multiplying the second inequality by $i$ and adding it to the third gives
$a{i}^2-<!--\; -\; -->e{i}^2+fhi-<!--\; -\; -->hfa-<!--\; -\; -->gbf=i^2(a-<!--\; -\; -->e)+fh(i-<!--\; -\; -->a)-<!--\; -\; -->gbf>0.$
But $a-<!--\; -\; -->e$ is negative by $(\ast <!--\; \ast \; -->)$, and $i-<!--\; -\; -->a\le <!--\; \le \; -->0$ if $i\le <!--\; \le \; -->a$. This leads to a contradiction.
How do I prove this if $a>i$?

System of parametric inequalities
I'm doing some systems of parametric inequalities, but I can not understand how to proceed.This is the system:
$x-<!--\; -\; -->2a<1+a$
$\frac{x}{2}-<!--\; -\; -->2a<2x-<!--\; -\; -->(a-<!--\; -\; -->3)$
$\frac{1}{3}-<!--\; -\; -->x<a$ per $a>0$
These are the solutions I've found:
$x<3a+1$
$x>\frac{-<!--\; -\; -->2a-<!--\; -\; -->6}{3}$
$x>\frac{-<!--\; -\; -->3a+1}{3}$
But now I do not know how to proceed. I thought I could compare the solutions like:
$3a+1>\frac{-<!--\; -\; -->2a-<!--\; -\; -->6}{3}>\frac{-<!--\; -\; -->3a+1}{3}$

What should $a$ be for the system of inequalities to have one solution?
$(1)\frac{3}{x-<!--\; -\; -->a}\ge <!--\; \ge \; -->1(2)|x-<!--\; -\; -->2a-<!--\; -\; -->2|\le <!--\; \le \; -->1$
For them to have one solution, the solution must be an endpoint of an interval. (1) gives me
$x\le <!--\; \le \; -->a+3$
but I don't know how to proceed from here. Any hint would be appreciated.

Notation of a linear inequality system.
$x_1\le <!--\; \le \; -->0$
$x_2\le <!--\; \le \; -->0$
$x_3\le <!--\; \le \; -->0$
is that right?
But what can I get from line 1? Thanks deeply for your help.
$\text{min}\left\{\begin{array}{c}2x_1+4x_2+7x_3:\\ 2x_1+x_2+6x_3\ge <!--\; \ge \; -->5\\ 4x_1-<!--\; -\; -->6x_2+5x_3\ge <!--\; \ge \; -->8\\ x_1,x_2,x_3\ge <!--\; \ge \; -->0\end{array}\right\}$

Systems of equations with one inequality and exponents.
The equation is this:
$8000<<!--\; <\; -->x{y}^5<<!--\; <\; -->8100$
$x{y}^4-<!--\; -\; -->x{y}^3><!--\; >\; -->5000$
I have no idea how to solve for the greater then or less then part of the equation. I also have no idea how to solve for exponents and finally if it even can be solved!

How to solve this complex inequality system
$1\le <!--\; \le \; -->z\stackrel{¯<!--\; ¯\; -->}{z}\le <!--\; \le \; -->4$ and $|\mathrm{\Im <!--\; \Im \; -->}(z)|<\mathrm{\Re <!--\; \Re \; -->}(z)$

Find the smallest possible value of $\mathrm{\omega <!--\; \omega \; -->}$ such that the system of inequalities
$\begin{array}{r}(1009\cdot <!--\; \cdot \; -->1010-<!--\; -\; -->2018)x+2018\le <!--\; \le \; -->2018\mathrm{\omega <!--\; \omega \; -->}\\ (1008\cdot <!--\; \cdot \; -->1009-<!--\; -\; -->2018)x+2018\le <!--\; \le \; -->2016\mathrm{\omega <!--\; \omega \; -->}\\ (1007\cdot <!--\; \cdot \; -->1008-<!--\; -\; -->2018)x+2018\le <!--\; \le \; -->2014\mathrm{\omega <!--\; \omega \; -->}\\ ....................................\\ (1\cdot <!--\; \cdot \; -->2-<!--\; -\; -->2018)x+2018\le <!--\; \le \; -->2\mathrm{\omega <!--\; \omega \; -->}\end{array}$
has a real solution in $x$.
Subtracting the first inequality from the second inequality, we get that $x=\frac{1009}{2017}$. This is also correct when we subtract the second inequality from the third one. When I plugged in $x$, I got an unruly fractional value for $\mathrm{\omega <!--\; \omega \; -->}$. Can someone guide me through the solution? Help is much appreciated.

Suppose that we have following interval (−5,2),we should find such a, which takes all possible values from this interval,creates following inequality systems
$5+a-<!--\; -\; -->|2y|\ge <!--\; \ge \; -->0$
$|x|\le <!--\; \le \; -->\frac{|a-<!--\; -\; -->2|}{2}$

Say there is a cat that in three days eat 12 fishes.
Say that each day the cat eats more than the day before.
Say that the last day the cat has eaten less than the addition of the two previous days.
$\{\begin{array}{l}x+y+z=12\\ z>y>x\\ z<x+y\end{array}$
How am I supposed to solve this?

Solve a system of three inequalities in three unknowns
$\begin{array}{}\text{(1)}& X><!--\; >\; -->1\end{array}$
$\begin{array}{}\text{(2)}& Y<<!--\; <\; -->Z\cdot <!--\; \cdot \; -->(1+X)\end{array}$
$\begin{array}{}\text{(3)}& Z<<!--\; <\; -->\frac{1-<!--\; -\; -->Y}{X}\end{array}$

How to check if a system of linear inequalities is solvable or not?
I have the variables $x_1,x_2,...,x_n$ and the following relations:
$x_1{\mathrm{\rho <!--\; \rho \; -->}}_1{x}_2$
$x_2{\mathrm{\rho <!--\; \rho \; -->}}_2{x}_3$
...
$x_{n-<!--\; -\; -->1}{\mathrm{\rho <!--\; \rho \; -->}}_{n-<!--\; -\; -->1}{x}_n,$
where the relations ${\mathrm{\rho <!--\; \rho \; -->}}_i\in <!--\; \in \; -->[=,<,>]$ are known. How to check if the system of inequalities above are consistent or not?

Quadratic inequalities - why does the solution seem different for $(x-<!--\; -\; -->a)(x-<!--\; -\; -->b)<0$ and $(x-<!--\; -\; -->a)(x-<!--\; -\; -->b)>0$
$3^{2x}-<!--\; -\; -->10\cdot <!--\; \cdot \; -->3^x+9>0$
$(3^x-<!--\; -\; -->9)(3^x-<!--\; -\; -->1)>0$
$3^x\in <!--\; \in \; -->(-<!--\; -\; -->\mathrm{\infty <!--\; \infty \; -->};1)\cup <!--\; \cup \; -->(9;+\mathrm{\infty <!--\; \infty \; -->})$
So to find x we need to solve:
$3^x>9$ and $3^x>1$ (1)
and we get:
$x\in <!--\; \in \; -->(-<!--\; -\; -->\mathrm{\infty <!--\; \infty \; -->};0)\cup <!--\; \cup \; -->(2;+\mathrm{\infty <!--\; \infty \; -->})$
Second example:
$5^{2x}-<!--\; -\; -->6\cdot <!--\; \cdot \; -->5^x+5<0$
$(5^x-<!--\; -\; -->5)(5^x-<!--\; -\; -->1)<0$
$5^x\in <!--\; \in \; -->(1;5)$
So to find x we need to solve:
$\{\begin{array}{l}{5}^x<5\\ 5^x>1\end{array}$ (2)
and we get:
$x\in <!--\; \in \; -->(0;1)$
My question is why in first example I get answer by solving 2 equations (1) seperately but in second example I get answer by solving system of equations (2)

Are the following systems of inequalities the same?
Suppose $x,y\in <!--\; \in \; -->\mathbb{R}$, and ${\mathcal{S}}_{\mathcal{1}}$ is a system of inequalities:
$\begin{array}{rl}{\mathcal{S}}_{\mathcal{1}}& =\left\{\begin{array}{c}x-<!--\; -\; -->y\ge <!--\; \ge \; -->1\\ -<!--\; -\; -->x+2y\ge <!--\; \ge \; -->1\\ 3x-<!--\; -\; -->5y\ge <!--\; \ge \; -->2\end{array}\right\}\\ & =\left\{\begin{array}{c}x\ge <!--\; \ge \; -->1+y\\ x\le <!--\; \le \; -->2y-<!--\; -\; -->1\\ x\ge <!--\; \ge \; -->\frac{2+5y}{3}\end{array}\right\}\end{array}$
I eliminate $x$ from $S_1$ to obtain $S_2$:
$\begin{array}{rl}{\mathcal{S}}_{\mathcal{2}}& =\left\{\begin{array}{c}1+y\le <!--\; \le \; -->2y-<!--\; -\; -->1\\ \frac{2+5y}{3}\le <!--\; \le \; -->2y-<!--\; -\; -->1\end{array}\right\}\\ & =\left\{\begin{array}{c}2\le <!--\; \le \; -->y\\ 2+5y\le <!--\; \le \; -->6y-<!--\; -\; -->3\end{array}\right\}\\ & =\left\{\begin{array}{c}2\le <!--\; \le \; -->y\\ 5\le <!--\; \le \; -->y\end{array}\right\}\end{array}$
Therefore, I know that $5\le <!--\; \le \; -->y<\mathrm{\infty <!--\; \infty \; -->}$. Let $S_3$ denote the following system of inequalities
$\begin{array}{rl}{\mathcal{S}}_{\mathcal{3}}& =\left\{\begin{array}{c}x-<!--\; -\; -->y\ge <!--\; \ge \; -->1\\ -<!--\; -\; -->x+2y\ge <!--\; \ge \; -->1\\ 3x-<!--\; -\; -->5y\ge <!--\; \ge \; -->2\\ 5\le <!--\; \le \; -->y<\mathrm{\infty <!--\; \infty \; -->}\end{array}\right\}\end{array}$
My question is, is $S_3=S_1$?

Suppose $P$ is a polyhedron whose representation as a system of linear inequalities is given to us:
$P=\{x|Ax\le <!--\; \le \; -->b\}$
Define $P^{\prime }$ be the image of $P$ under the linear transformation which maps $x$ to $Cx$:
$P^{\prime }=\{Cx|x\in <!--\; \in \; -->P\}.$
Given $A,b,C$, our goal is to compute a representation of $P^{\prime }$ as a system of linear inequalities, i.e., to compute $D,e$ satisfying
$P^{\prime }=\{x|Dx\le <!--\; \le \; -->e\}.$
What is the complexity of this problem (i.e., how many arithmetic operations or bit operations are required)?