For questions about or involving the absolute value function.

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2
votes
1answer
9k views

Reverse Triangle Inequality Proof

I've seen the full proof of the Triangle Inequality $|x+y|\le|x|+|y| $ However, I haven't seen the proof of the reverse triangle inequality: $||x|-|y||\le|x-y|$ Could you please prove this using ...
11
votes
5answers
765 views

Show that the $\max{ \{ x,y \} }= \dfrac{x+y+|x-y|}{2}$.

Show that the $\max{ \{ x,y \} }= \dfrac{x+y+|x-y|}{2}$. I do not understand how to go about completing this problem or even where to start.
14
votes
4answers
498 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 $$A=\dfrac{2}{\sqrt{10}}$$ ...
6
votes
2answers
2k views

How to use triangle inequality to establish Reverse triangle inequality

I need to use $|a+b| \leq |a|+|b|$ to show that $||a|-|b|| \leq |a-b|$. I have tried to represent $||a|-|b||$ as $||a|+(-|b|)|$, and then get $||a|+(-|b|)| \leq |a|+|-|b||$, but that isn't leading ...
0
votes
6answers
342 views

How could we solve $x$, in $|x+1|-|1-x|=2$?

How could we solve $x$, in $|x+1|-|1-x|=2$? Please suggest a analytical way that I could use in other problems too like this $ |x+1|+|1-x|=2$ and of this genre. Thank you,
6
votes
3answers
503 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 ...
4
votes
3answers
436 views

Proof for Integral Inequality $|\int f| \le \int |f|$ - is it sufficient enough?

Claim: If f is integrable, $\left|\int_a^bf(x)dx\right|\le\int_a^b|f(x)|dx$ Proof (attempt): We know $-|f|\le f \le|f|$, so $\int-|f| \le \int f \le \int|f|$.* Since, if $-b<a<b$, we say ...
1
vote
3answers
1k views

Proving absolute value inequalities

I need help proving the last two cases for the following inequality: $\bigl|\lvert x\rvert-\lvert y\rvert\bigr| \le \lvert x-y\rvert$. Case 1: $x > 0$ and $y > 0$: the inequality simplifies ...
7
votes
2answers
163 views

Maximum of the difference

What is the maximum value of $f(… f(f(f(x_{1} – x_{2}) – x_{3})-x_{4}) … – x_{2012})$ where $x_{1}, x_{2}, … , x_{2012}$ are distinct integers in the set ${1, 2, 3, …, 2012}$ and $f$ is the absolute ...
5
votes
1answer
902 views

Prove variant of triangle inequality containing p-th power for 0 < p < 1

Sorry if this is a trivial question, but I am kind of stuck with proving the following inequality and have been searching for a while: $\rho \left( \sum\limits_i^n d_i \right) \leq \sum\limits_i^n ...
3
votes
2answers
193 views

Prove the monotonicity of the expectation of a binary random variable function

Consider $R$ independent binary random variables $y^1, \ldots, y^R$ over the space $\{-1, +1\}$ such that $\Pr(y^j = 1) = p^j \geq 0.5$ and $\Pr(y^j = -1) = 1 - p^j$, $\forall j = 1,\ldots,R$. ...
3
votes
2answers
306 views

Why exactly can you take the absolute value of one side of this inequality and assume it is still true?

Exercise: Show that if $(b_n) \to b$, then the sequence of absolute values $\left| b_n \right|$ converges to $\left| b \right|$. Solution (partial): By the triangle inequality, ...
2
votes
5answers
165 views

Prove the triangle inequality

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
4answers
104 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
2answers
40 views

Absolute value question false solution

|x| = 3x – 2 Why does this statement eventually give you a solution that isn't valid. So this equation comes out: x = 3x - 2 2 = 2x x = 1 OR x = -3x + 2 4x = 2 x = 1/2 However 1/2 doesn't ...
0
votes
1answer
164 views

Absolute value of a real number

My question is: Solve: $|x-4|< a$, where $a$ belongs to the real numbers. Solve this by considering various cases depending upon whether $a$ is negative, positive or zero. What I have tried ...
9
votes
7answers
866 views

what does $|x-2| < 1$ mean?

I am studying some inequality properties of absolute values and I bumped into some expressions like $|x-2| < 1$ that I just can't get the meaning of them. Lets say I have this expression $$ ...
4
votes
4answers
103 views

Let $x$ be in the set of real numbers $\mathbb{R}$ and let $f(x)=|2x-1|-3|2x+4|+7$ be a function, write $f(x)$ without the absolute value.

Let $x$ be in the set of real numbers $\mathbb{R}$ and let $f(x)=|2x-1|-3|2x+4|+7$ be a function, write $f(x)$ without the absolute value. I thought of it this way: $$f(x)=\begin{cases}2x-1-3(2x+4)+7 ...
1
vote
3answers
615 views

Proving square root of a square is the same as absolute value

Lets say I have a function defined as $f(x) = \sqrt {x^2}$. Common knowledge of square roots tells you to simplify to $f(x) = x$ (we'll call that $g(x)$) which may be the same problem, but it isn't ...
5
votes
2answers
258 views

Is there a lower-bound version of the triangle inequality for more than two terms?

The triangle inequality $|x+y|\leq|x|+|y|$ can be generalized by induction to $$|x_1+\ldots+ x_n|\leq|x_1|+\ldots+|x_n|.$$ Can we generalize the version $|x+y|\geq||x|-|y||$ to $n$ terms too? I need ...
3
votes
1answer
299 views

Why do definitions of distinct conic sections produce a single equation?

I understand how to get from the definitions of a hyperbola — as the set of all points on a plane such that the absolute value of the difference between the distances to two foci at $(-c,0)$ and ...
2
votes
1answer
506 views

Prove by contradiction or contrapositive? If $|x+y|<|x|+|y|$, then $x<0$ or $y<0$.

Prove: If $|x+y|<|x|+|y|$, then $x<0$ or $y<0$ This looks as though it's true from the start. Take $x=-4, y=4$. $|-4+4|<|-4|+|4|$ $0<8$ is true. The question is asking for a ...
1
vote
2answers
591 views

Proof the maximum function $\max(x,y) = \frac {x +y +|x-y|} {2}$ [duplicate]

I want to prove the maximum function max: $\mathbb{R} \rightarrow \mathbb{R}$, which is defined by $$\max(x,y) = \begin{cases}x, \text { if } x \geq y , \\ y, \text { if } x < y \end{cases}$$ ...
0
votes
1answer
93 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 $x<a, |x-a|=-(x-a)$ ...
7
votes
2answers
269 views

How does one DERIVE the formula for the maximum of two numbers

I want to derive (not prove that this is true) the formula $\max (x,y) = \dfrac{x + y + |y-x|}{2}$ I was reading a proof (which they have the result ahead of time already) that we do cases and then ...
4
votes
3answers
710 views

Proving two integral inequalities

Can anyone help me to prove that these integral inequalities hold? Here $x$ is a real value: $$ \left| \int_a^b\ f(x) dx \right| \leq \int_a^b\ |f(x)| dx $$ Here $z$ is a complex value: $$ \left| ...
4
votes
3answers
163 views

Is there an alternate definition for $\{ z \in \mathbb{C} \colon \vert z \vert \leq 1 \} $.

Is there a method of constructing a subset of a reasonably arbitrary ring so that when the construction is applied the $\mathbb{C}$ the result is $B = \{ z \in \mathbb{C} \colon |z| \leq 1 \} $? My ...
2
votes
7answers
116 views

for $n$ an integer, why is $n^0=1$ ??

This is so going to cost me.... I was wondering why for any integer $n$: $n^0 =1$. Perhaps It's because $n$ is a round number and if $m$ is a non negative integer, also round then: $$n^m = 1 \cdot ...
2
votes
3answers
146 views

Absolute ratios

I'm curious about the following idea: suppose we have two values $P$ and $Q$, and the magnitude of the ratio $\frac{P}{Q}$ is between $0$ and $\infty$. If $P$ is smaller, then it's between $0$ and ...
2
votes
3answers
269 views

Solve an absolute value equation simultaneously

My question is : Solve simultaneously $$\left\{\begin{align*}&|x-1|-|y-2|=1\\&y = 3-|x-1|\end{align*}\right.$$ What I did : $y=3 - |x-1|$ is given. Thus $y = 3-(x-1)$ or $y = ...
1
vote
3answers
72 views

Value and simplify

I want to find the value and simplify square root 36 ? Square root of 36 is 6 But I would know how to find the value and simplify it .
1
vote
2answers
53 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} ...
1
vote
1answer
364 views

double integral of an absolute function

I'm just a little unsure of how to tackle this one. I understand that typically you would separate the integral into two for where x is positive or negative, I'm just unsure of how to separate it for ...
1
vote
0answers
88 views

Fields with their own absolute value

Let $F\hspace{.02 in}$ be a field. $\:$ Let $E\hspace{.02 in}$ be a non-zero subring of $F$. Let $\hspace{.03 in}\leq\hspace{.03 in}$ be a total order on $E\hspace{.02 in}$ that makes $E\hspace{.02 ...
1
vote
1answer
4k views

Solving inequality with two absolute values

Hey, ! In my pre-calculus class the teacher showed the solution of the following example: \begin{align} \vert x-3 \vert \lt \vert x - 4 \vert + x \end{align} He started by stated the domains ...
0
votes
3answers
76 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 ...
0
votes
3answers
361 views

Integrating absolute value function

I'm working on a problem drawing phase plane diagrams in my applied mathematics course. I'm supposed to draw the phase line diagram of $x''+\vert x\vert=0.$ In the process, I get to the differential ...
0
votes
5answers
106 views

Prove That $|a +b| = |a| +|b|$ if $a$ and $b$ Have Same Signs, And $|a +b| < |a| + |b|$ if $a$ and $b$ Have Opposite Signs

My Proof: $|a +b| = |a| +|b|$ ..... $(i)$ $|a +b| < |a| + |b|$ ..... $(i)$ If $'a'$ and $'b'$ have same signs: Let $a$ and $b$ be equal to $-x$. Replacing $a$ and $b$ with $-x$ in the equation ...
0
votes
2answers
106 views

solving absolute value equation 2

My question is : Solve simultaneously- $$\left\{\begin{align*}&|x-1|+|y-2|=1\\&y = 3-|x-1|\end{align*}\right.$$ I tried to solve this question by the method told by Marvis as I had ...
0
votes
2answers
124 views

Solving $|x-2| + |x-5|=3$ [duplicate]

Possible Duplicate: How could we solve $x$, in $|x+1|-|1-x|=2$? How should I solve: $|x-2| + |x-5|=3$ Please suggest a way that I could use in other problems of this genre too Any help ...
0
votes
3answers
109 views

Absolute value on a number line

Solve : |x-4|>a if case1:a>0 and case2:a<0 I am getting answers which look similar in both cases. please i wish to know why it is so and how different both answers are when plotted on a number ...
0
votes
5answers
201 views

Solving an equation with absolute values

The equation I am trying to solve is this : $\newcommand\abs[1]{|#1|}\abs{3y+7}=\abs{2y-1}$. My conventional approach is to split this into three intervals with $1/2$ and $-7/3$ being the two "split" ...