1
vote
1answer
52 views

Order of $\{x\in\mathbb {Z}, |x|+|3x-1|<5\}$

There is a multiple choices which says what is the order of $\{x\in\mathbb {Z}, |x|+|3x-1|<5\}$? a. 1 b. 3 c. 2 d. empty I know that by considering certain cases, for example when $x<0$ or ...
1
vote
1answer
84 views

Inequality: $\left|x^3-y^3\right|<|x|^3+|y|^3$

Could anyone show me why $$\left|x^3-y^3\right|<|x|^3+|y|^3$$ for all real numbers (x,y) except 0? I'm thinking of whether of how to remove the modulus sign on the left hand side of the ...
3
votes
2answers
88 views

Does my proof of $|x+y| \le |x| + |y|$ make sense? How do I conclude a proof?

Thank you for reading it. I know I made a lot of mistakes. This is my first ever proof that I have attempted. Another note is that I only have been studying proofs for about a week. Any advice will be ...
2
votes
2answers
29 views

Taking root from absolute expression

Why is the following true? (Where all terms are positive) $$|x-y| < \epsilon^2 \implies |\sqrt x - \sqrt y| < \epsilon$$
2
votes
3answers
350 views

How to solve inequalities with absolute values on both sides?

If you have an inequality that has two absolute value bars like $|4x+1|<|3x|$, how do you go about doing this? I know that if $4x+1<3x$, then those $x$'s will work but what else do I do? I think ...
0
votes
1answer
48 views

$\left | -(x+2)^2+6(x+2) \right |>13$

I did $-(x+2)^2+6(x+2)>13$ and $-(x+2)^2+6(x+2)< -13$. The first inequality had complex solutions and therefore can be disregarded but the second one has two real solutions, $x \approx -3.7$ and ...
0
votes
3answers
70 views

Absolute value quadratic inequalities not the usual?

$ | -x^2 + 6x | \gt 13 $,for example. I would start off solving $ -x^2 + 6x = \pm 13 $ and either get 4 solutions, 3 solutions or two simply do the the nature of the graph. Without knowing if the two ...
1
vote
1answer
32 views

Don't understand inequality in order to prove Algebraic Limit Theorem

I'm self-studying from the book Understanding Analysis by Stephen Abbott and I'm stuck on Theorem 2.3.3 on page 45, i.e., the Algebraic Limit Theorem. In particular, letting $\lim a_n = a$ and $\lim ...
2
votes
1answer
30 views

If $w_1=a_1+ib_1$ and $w_2=a_2+ib_2$ are complex numbers, then $|e^{w_1}-e^{w_2}|\geq e^{a_1}-e^{a_2}$

Let $w_1=a_1+ib_1$ and $w_2=a_2+ib_2$ be two complex numbers. Ahlfors says that $|e^{w_1}-e^{w_2}|\geq e^{a_1}-e^{a_2}$. I don't understand why that is. Any help would be greatly appreciated.
2
votes
2answers
121 views

solving the inequality

I'm looking for hints on how to efficiently solve this inequality: $$\left( \frac {|x|-|1-x|}{|x|} \right)^{2x-1} \gt \left(\frac {|x|-|1-x|}{|x|} \right)^{8-x} $$
1
vote
2answers
27 views

Where did I go wrong with this inequality involving absolute value function?

Question: Find all $x \in \mathbb R$ such that the inequality $4<|x+2|+ |x-1|<5$ is satisfied. This is my attempt at solving the problem: Case (i): If $x+2 \geq 0 $ and $ x-1\geq0$, then ...
0
votes
4answers
58 views

How to solve Absolute Value Inequality: |x-1| ≥ 3-x

I am learning the topic of solving absolute value inequality question. I had mostly understood the steps in order to solve for an inequality. However, I'm still clueless of a step to solve the ...
1
vote
2answers
33 views

Spivak Absolute Value Problem (Prologue 9-v)

I'm working on the following problem Express the following with at least one less pair of absolute value signs $$|(| \sqrt2 + \sqrt3| - |\sqrt5 - \sqrt7|)|$$ Now I can see that the ...
3
votes
4answers
245 views

inequality method of solution

Im looking for an efficent method of solving the following inequality: $$\left(\frac{x-3}{x+1}\right)^2-7 \left|\frac{x-3}{x+1}\right|+ 10 <0$$ I've tried first determining when the absolute value ...
0
votes
1answer
41 views

Infimum of absolute values versus absolute value of infimum

Let $A\subseteq\mathbb R$. Is there a nice proof of the inequality $\displaystyle\inf_{a\in A} |a|\le|\inf_{a\in A} a|$? The only proof I know is, though not very difficult, annoying because it ...
3
votes
4answers
164 views

Show that $|z+1|\le|z+1|^2 +|z|$ for all $z \in \mathbb{C}$

Question: Show that $|z+1|\le|z+1|^2 +|z|$ for all $z \in \mathbb{C}$ So far I have, Suppose $1\le|z+1|$ $|z+1|\le|z+1|^2$ $|z+1|\le|z+1|^2+|z|$ Now I must show $|z+1|<1$ but this is where ...
2
votes
1answer
91 views

Solving absolute inequality

I have the following inequality: $$|4 - k^2| > |10 + 13k|$$ So how to solve this ?
7
votes
4answers
287 views

Inequality for absolute values

How do you show either of the equivalent inequalities: $$2(|a|+|b|+|c|)\leq |a+b+c|+|a+b-c|+|a-b+c|+|a-b-c|$$ or $$|x+y|+|x+z|+|y+z|\leq |x|+|y|+|z|+|x+y+z|$$ Hold for complex numbers or in $n$ ...
3
votes
2answers
29 views

$5-3|x-6|\leq 3x -7$

I have this inequation: $$5-3|x-6|\leq 3x -7$$ i solved this this way: i said, for $x\geq6$ is the modulus positive, so I made 2 cases in which the modulus gives + or - : 1) for $x\geq6$ ...
1
vote
1answer
32 views

Quadratic inequality with absolute values

I've decided to study calculus on my own, and I've started working on "A First Course in Calculus" by Serge Lang, 5th edition. Now I'm just reading the chapter on preliminaries, and there is a section ...
1
vote
2answers
48 views

Prove that $|x+y| \leq |x|+|y|$ [duplicate]

How to Prove the triangle inequality which says for all x (no matter how big or small) and for all y (no matter its size) in the set of irrational+rational numbers, this holds: $|x+y| \leq |x|+|y|$
2
votes
1answer
52 views

Solve $2\sqrt{(x-1)(x+2)}\ge|x+1|-2$

Can you please show me how can I solve this inequality. I would like to see how it can be done without the graph of the functions. Thank you! $$2\sqrt{(x-1)(x+2)}\ge|x+1|-2$$
0
votes
1answer
48 views

Solve the following inequality…

Can you please verify if I've done this exercise correctly, and if you have a better solution, please, show it to me. Thank you! (The exercise is in the left top corner.)
1
vote
1answer
57 views

Triangle inequality frobenius norm

I'm trying to show that the frobenius norm is a norm. however it appears as if triangle inequality isnt met. $$||A+B||_F = \sqrt{\sum_{i,j=1}^n |a_{ij}+b_{ij}|^2} \leq \sqrt{\sum_{i,j=1}^n ...
0
votes
0answers
41 views

Simple proof explanation - Possibly triangle inequality involved

I'd like some help with understanding the following statements...I saw it on the internet while searching for a proof, and I'd like to understand why its true: let $A$ be a diagonally dominant matrix ...
2
votes
2answers
107 views

Solving inequalities with absolute values

This is the question: $$ \left| \frac{x+2}{3(x-1)} \right| \leq \frac{2}{3} $$ And this is my working out, first I squared both the numerator and denominator, then solved it as if it was a normal ...
1
vote
4answers
126 views

Proving the inequality $|a-b| \leq |a-c| + |c-b|$ for real $a,b,c$

Let $a,b,c$ real numbers. Prove the inequality $|a-b| \leq |a-c| + |c-b|$. Prove that equality holds if and only if $a \leq c \leq b$ or $b \leq c \leq a$. I've tried starting with just $a \leq ...
0
votes
1answer
47 views

How do I prove that $|x+y| \ge \big||x|-|y|\big|$?

I don't know where to start with proving this. Any help will be greatly appreciated.
0
votes
1answer
21 views

finding an absolute value inequality

The question asks, "find an absolute value inequality whose solution's are x>2 and x<-12". I have no idea where to start and was wondering if anyone could help
0
votes
2answers
60 views

Proving absolute value inequality by contradiction

Prove that for $|x|, |y|, |z| \geq 2$ the following holds: $|x^2 + y| + |y^2 + z| + |z^2 + x| \geq |x| + |y| + |z|$ So I thought about a simple proof by contradiction but am not sure whether it's a ...
0
votes
3answers
111 views

Proof for absolute value inequality of three variables: $|x-z| \leq |x-y|+|y-z|$ [duplicate]

$|x-z| \leq |x-y|+|y-z|$ We know that both LHS and RHS are non negatives. So, I thought of proving this by comparing the squares of both sides but can't advance beyond that step. Any help would be ...
2
votes
2answers
64 views

Strategy to solve absolute value inequality

I was wondering if there is any strategy to solve absolute value On both sides inequalities, for example, $$| x^2 -3x + 2 | < | x + 2|$$ Thanks, Eli
0
votes
1answer
60 views

Problem with absolute value

Say that $|\sqrt{x}-1| < \epsilon$. I am having a problem with handling this inequality. I want to exclude x. I. $|\sqrt{x}-1| < \epsilon$ $|\sqrt{x}| - |1| \leq |\sqrt{x}-1| < \epsilon$ ...
2
votes
1answer
48 views

Solving inequation with two absoulte values

I need to solve the following inequation: $$ |x| \cdot |x-1|-1>-x\\ $$ I cant get the correct result. I tried to solve it like this: $$ |x| \cdot |x-1|-1>-x $$ I know that I can write $|x ...
1
vote
2answers
51 views

I want to check that $\left|\left|a+b\right|-\left|a\right|-\left|b\right|\right|\leq2\left|b\right|\forall a,b\in\mathbb{R} $.

I want to check that $\left|\left|a+b\right|-\left|a\right|-\left|b\right|\right|\leq2\left|b\right|\forall a,b\in\mathbb{R} $. It 's equivalent to ...
-5
votes
3answers
79 views

How to solve this: $|3-x|\ge2$ [closed]

How to solve $|3-x|\ge2$ ? I know that if $|x| < y$, then $-y < x < y$. But in this case what to do? Thanks. Here, $|x|$ is the absolute value of $x$.
0
votes
1answer
27 views

Absolute Value Inequality - Precision

So I was writing a computer program, which is supposed to check whether $x$, an approximation of $\sqrt{a}$, is close enough to $\sqrt{x}$. Since these definitions aren't very precise, I defined ...
14
votes
4answers
532 views

How find this inequality $\max{\left(\min{\left(|a-b|,|b-c|,|c-d|,|d-e|,|e-a|\right)}\right)}$

let $a,b,c,d,e\in R$,and such $$a^2+b^2+c^2+d^2+e^2=1$$ find this value $$A=\max{\left(\min{\left(|a-b|,|b-c|,|c-d|,|d-e|,|e-a|\right)}\right)}$$ I use computer have this $$A=\dfrac{2}{\sqrt{10}}$$ ...
3
votes
2answers
102 views

How does the triangle inequality work for $|x-y|$?

I know that $|x+y|\leq |x|+|y|$... But is it similar for $|x-y|$? That is, is $|x-y|\leq |x|+|y|$? I ask because of the following: $x-y=x+(-y)$, so $|x+(-y)|\leq |x|+|-y|=|x|+|y|$ Is it possible ...
5
votes
3answers
131 views

How would I prove $|x + y| \le |x| + |y|$?

How would I write a detailed structured proof for: for all real numbers $x$ and $y$, $|x + y| \le |x| + |y|$ I'm planning on breaking it up into four cases, where both $x,y < 0$, $x \ge 0$ ...
3
votes
3answers
119 views

Absolute Value inequality help: $|x+1| \geq 3$

Find the solutions to the inequality: $$|x+1| \geq 3$$ I translate this as: which numbers are at least $3$ units from $1$? So, picturing a number line, I would place a filled in circle at the ...
2
votes
3answers
91 views

absolute value inequalities

When answer this kind of inequality $|2x^2-5x+2| < |x+1|$ I am testing the four combinations when both side are +, one is + and the other is - and the opposite and when they are both -. When I ...
3
votes
2answers
99 views

Inequality proof with absolute values

How do you prove the following: $$ \varepsilon > 0 \\ \left | y-b \right | < \varepsilon\\ \left | x-a \right | < \varepsilon \\ \Longrightarrow \left | xy - ab \right | < ...
0
votes
2answers
63 views

Solving an equation with absolute values: $ | 2x - 5| + | 2x - 3 | = m $

Given that the following equation does not have solutions in $\mathbb{R}$, find the value of $m$: $$| 2x - 5| + | 2x - 3 | = m $$ I try to resolve this equation on cases, when $| 2x - ...
6
votes
3answers
514 views

How prove this inequality: $\sum_{i,j=1}^{n}|x_{i}+x_{j}|\ge n\sum_{i=1}^{n}|x_{i}|$? [duplicate]

Let $x_{1},x_{2},\cdots,x_{n}$ be real numbers. Show that $$\sum_{i,j=1}^{n}|x_{i}+x_{j}|\ge n\sum_{i=1}^{n}|x_{i}|.$$ I think this problem may be solved using nice methods, but I can't find ...
1
vote
3answers
132 views

Prove that: $\left | 2x-y-4 \right |\geq 4\sqrt{2}+4$

Let $x,y\in \mathbb{R}$ know that $4x^2-9y^2=36$ Prove that: $$\left | 2x-y-4 \right |\geq 4\sqrt{2}+4$$
0
votes
0answers
83 views

Finding the number of integral values of a for which the inequality $3- |{x-a}|>x^2$ is satisfied by at least one negative value of $x$.

Finding the number of integral values of a for which the inequality$$ 3- |{x-a}|>x^2$$ is satisfied by at least one negative value of $x$.Here's a duplicate post ->{{Determine all the values of ...
0
votes
0answers
26 views

Questions about $|f(1+a+bi)|<|f(1+a)|$

Let $a,b >0$ and $|*|$ denote the absolute value. Let $f(z)$ be a realvalued analytic function defined for $Re(z)>1.$ For any $a,b$ we have $|f(1+a+bi)|<|f(1+a)|$. Some questions : $1)$ If ...
0
votes
1answer
66 views

A simpler proof for an equality regarding sums of absolute values

Question Let $I$ be a set of indices, and for all $i \in I$, assume $$0 \le {a_i},{b_i},{c_i},{u_i},{v_i} \le 1 \;,$$ with $u_i+v_i=1$. Assume that $L$ is an upper bound on both ${\sum\nolimits_i ...
2
votes
2answers
112 views

An Inequality Involving $\min(x, y)$

The following problem is from Spivak's Calculus (4th ed., pg. 18): Prove that if: $|x-x_0| < \min(\frac{\epsilon}{2(|y_0|+1)}, 1)$ and $|y - y_0| < \frac{\epsilon}{2(|x_0|+1)}$ then $|xy-x_0 ...