3
votes
4answers
230 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
39 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
159 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
89 views

Solving absolute inequality

I have the following inequality: $$|4 - k^2| > |10 + 13k|$$ So how to solve this ?
6
votes
4answers
257 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
28 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
23 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
43 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: $|x+y| \leq |x|+|y|$
1
vote
1answer
45 views

Solve the inequality…

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
49 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
40 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
92 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
115 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
58 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
96 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
61 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
57 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
46 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
44 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 ...
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
522 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
97 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
126 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
108 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
89 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 | < ...
6
votes
3answers
509 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
128 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
74 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
65 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
111 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 ...
0
votes
2answers
101 views

How to solve $|2x +1|< 1/4$?

How do you solve $$|2x +1|< \frac{1}4$$
0
votes
2answers
82 views

Prove $|x+1|\leq 4$ implies that $-4\leq x\leq 2$.

How do I prove that if $x$ is a real number, then $\lvert x+1 \rvert\leq 3$ implies that $-4\leq x\leq 2$. EDIT: $\lvert x+1 \rvert\leq 4$ should be $\lvert x+1 \rvert\leq 3$
2
votes
4answers
135 views

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

$||x|-|y|| \leq |x-y|$ when $(x,y \in R^k)$ In Principles of MA(Rudin), the author said one sees easily that $||x|-|y|| \leq |x-y|$ when $(x,y \in R^k)$ (p.88, Rudin) from the triangle ...
0
votes
1answer
96 views

Determine all the values of the parameter $a$ for which the inequality $3-|x-a|>x^2$ is satisfied by at least one negative $x$.

I wanted to know, how can I determine all the values of the parameter $a$ for which the inequality $3 - |x-a| > x^2$ is satisfied by at least one negative $x$. I tried for $x<a, |x-a|=-(x-a)$ ...
1
vote
3answers
121 views

Inequalities - Absolute Value

$$|2x-1| \leq |x-3|$$ Answer is $$-2 \leq x \leq \frac43$$ My Question is HOW?
1
vote
2answers
56 views

Finding $x$ from inequality: $\left | \frac{3^n + 2}{3^n + 1} - 1 \right | \le \frac{1}{28}$

Find $x$ in $\mathbb{Z}$ satisfying this inequality: $$\left | \frac{3^n + 2}{3^n + 1} - 1 \right | \le \frac{1}{28}.$$ I tried something, but I don't think it's correct. $$-\frac{1}{28} ...
2
votes
5answers
181 views

Prove the triangle inequality [duplicate]

I want to porve the triangle inequality: $x, y \in \mathbb{R} \text { Then } |x+y| \leq |x| + |y|$ I figured out that probably the cases: $x\geq0$ and $y \geq 0$ $x<0$ and $y < 0$ $x\geq0$ ...
1
vote
2answers
155 views

Graphing - Absolute Value and Circle

The diagram Shows The Graphs of $y = |x + 2|$ and $y = \sqrt{4 - x^2}$ Write down the solution for $\sqrt{4 - x^2}$ is equal to or less than $y = |x + 2|$.
4
votes
5answers
336 views

Prove:$|x-1|+|x-2|+|x-3|+\cdots+|x-n|\geq n-1$

Prove:$|x-1|+|x-2|+|x-3|+\cdots+|x-n|\geq n-1$ example1: $|x-1|+|x-2|\geq 1$ my solution:(substitution) $x-1=t,x-2=t-1,|t|+|t-1|\geq 1,|t-1|\geq 1-|t|,$ square, $t^2-2t+1\geq ...
3
votes
2answers
80 views

How prove this $|x_{p}-y_{q}|>0$

let $$x_{1}=\dfrac{1}{8},x_{n+1}=x_{n}+x^2_{n},y_{1}=\dfrac{1}{10},y_{n+1}=y_{n}+y^2_{n}$$ show that: for any $p,q\in N^{+}$ we have $$|x_{p}-y_{q}|>0$$
1
vote
4answers
278 views

Please help me to prove this inequality: $|x|+|y|≥|x+y|$

Please help me to prove the following inequality: $|x|+|y|\geq|x+y|$ in which $x$ and $y$ are real numbers. Any help or hint would be appreciated. Thanks :)
3
votes
1answer
122 views

Question based on Triangle Inequality $\displaystyle |x+y|\leq |x|+|y|$

If $x,y,z\in \mathbb{R}-\left\{0\right\}$. Then prove that $\displaystyle 1\leq \frac{|x+y|}{|x|+|y|}+\frac{| y+z|}{| y |+| z |}+\frac{| z+x|}{| z |+| x |}\leq 3$ My Try:: Using Triangle Inequality ...
2
votes
3answers
95 views

Question about absolute value in inequalities

My book presents the following: $$7 \le x \le 9 $$ so $$ -1 \le x - 8 \le 1 $$ and $$ |x-8| \le 1$$ I usually get confused with the way that taking the absolute value of an expression works. Could ...
1
vote
2answers
98 views

How to evaluate the inequality $|x+1|<-1$?

Okay perhaps the title isn't specific enough, I didn't know how to word it exactly. I'm finding the interval of convergence for a power series and i know the answer to be (-2,0] I end up with the ...
2
votes
2answers
64 views

proving $|x - 1| < {1\over4} \Rightarrow |2x - 1| \geq {1\over 2}$

I tried solving the above, consider that: ($x \in R)$, I know it's not a complicated problem to solve though I struggle getting on with this question, What I've done far is: $|x-1|<{1\over4} ...