# Absolute Value Inequalities

**Absolute value inequalities** deal with the inequalities $(<, \leq, >, \geq)$ on the expressions with absolute value sign.
For an example,

Solve the inequality for $x$: $|x-9|<4.$

We will use the definition of absolute value to solve this problem, which is$\begin{aligned} -4<x-9&<4 \\ 5<x&<13.\ _\square \end{aligned}$

The simplest guideline when solving problems like these is to consider the piecewise function and split it into relevant cases:

$|x - a| = \begin{cases}-x + a & \text{for }x < a \\ x - a & \text{for }x \geq a . \end{cases}$

However, there exist several techniques to solve such inequalities like using basic properties, considering the cases, graph visualization, etc. This wiki page explains all those techniques in detail along with worked examples and problems to try. These techniques start from solving basic inequalities further extending to the harder absolute value inequalities.

#### Contents

## Introduction and Basic Properties

Recall the definition of absolute value:

For any real number $x$, its

absolute valueis defined as$|x| = \begin{cases} x & \textrm{if } x > 0 \\ 0 & \textrm{if } x = 0 \\ -x & \textrm{if } x < 0. \\ \end{cases}$

The function $f(x) = |x|$ is also called the modulus function. $_\square$

Let $x$ be a variable or an algebraic expression and let $a$ be a real number such that $a > 0$. Then the following inequalities hold:

- $|x| \leq a \Leftrightarrow -a \leq x \leq a$
- $|x| \geq a \Leftrightarrow x \leq -a\$ or $\ x \geq a$
- $|x| < a \Leftrightarrow -a<x<a$
- $|x| > a \Leftrightarrow x <-a\$ or $\ x>a$.

For $f(x)$ and $g(x),$ functions of $x,$ we have

$1.~~ |f(x)| \leq g(x)~~ \Leftrightarrow~~ - g(x) \leq f(x) \leq g(x)\\ 2.~~|f(x)| \geq g(x) ~~ \Leftrightarrow~~ f(x) \leq -g(x) \text{ or }f(x) \geq g(x)\\ 3.~~|f(x)| < g(x) ~~ \Leftrightarrow ~~-~g(x)<f(x)<g(x)\\ 4.~~|f(x)| > g(x) ~~ \Leftrightarrow ~~f(x) <- g(x) \text{ or }f(x)>g(x).$

Here are some other properties of absolute value on inequalities:

- $|a+b|\leq |a|+|b|$ with equality if both have the same sign, i.e. $ab > 0$.
- $|a-b|\leq |a|+|b|$ with equality if they have different signs, i.e. $ab<0$.

Generalizing the two inequalities above, we get

$|\pm a_1 \pm a_2 \pm a_3 \pm \cdots \pm a_n| \leq |a_1| +|a_2|+\cdots +|a_n|.$

Let's look at an example to see how to apply this formula.

Find the minimum value of $|x-1|+|x-2|+\dots + |x-100|$.

Using the inequality above, we try to remove variable $x$ by

$\begin{aligned} |x-1|+|x-2|+\dots + |x-100| &\geq |(x-1)+(x-2)+\dots +(x-50)-(x-51)-(x-52)-\dots -(x-100)| \\ &= 2500. \end{aligned}$

To verify that this is a valid case, we must have $\{(x-1), (x-2), \dots , (x-50)\}\geq 0$ and $\{(x-51), (x-52), \dots , (x-100)\}\leq 0$. This reduces to only two cases:

$\begin{cases} x-50 &\geq 0 \\ x-51 & \leq 0. \end{cases}$

Hence, the minimum value is obtained when $x\in [50,51]$. $_\square$

Here is a problem to try.

## Problem Solving Technique - Case Work

Here, we look at the definition of taking the absolute value of a number. In order to "undo" the absolute value signs, we could either get a positive or negative value, since the absolute value of $- 5$ is the same as the absolute value of $5$, which is $5$. This becomes a method where we have multiple cases.

The main steps (for dealing with linear/multiple linear absolute value inequalities) are

- "Undo" the absolute value signs by making the expressions inside the absolute value sign negative or positive.
- Take all the inequalities obtained (these are all solution sets) and figure out the solution set. To figure out the solution set given a bunch of solution sets from the cases you worked out, consider your original problem as a piecewise function, which is why you have the cases. What values of $x$ make the left hand side negative? Positive? Zero? These are all additional restrictions on $x$, and you will take the intersection between these additional restrictions on $x$ and the "final" inequality obtained from the case work you did. This gives what I call a "restricted final" inequality.
- Finally, you take the union of all the "restricted final" inequalities for your final solution set.

We will explore how to do this in the next 3 examples.

Solve $| x + 3 | < 7$.

Case 1:$x+3$ is non-negative, or $x\geqslant -3$

The property of absolute value tells us that $|a| = a$ for non-negative $a$, so in this case $|x+3|<7 \implies x+3 < 7\implies x<4.$ Now, we have two inequalities here, and the solution for this case is the intersection of both inequalities. This is because $x$ should satisfy both as they are dependent of each other $($only because $x\geq -3$ we have $x<4)$. Hence, the solution for this case is $-3 \leq x < 4$.

Case 2:$x+3$ is negative, or $x < -3$

Again, $|a| = -a$ for a negative value $a$, so $|x+3|<7\implies -(x+3)<7\implies x>-10.$ For the same as the reason above, we should take the intersection of both inequalities, which is $-10<x<-3$.Finally, we take the union of these inequalities, as they are independent of each other: $-10<x<4.\ _\square$

Solve $2|x+2| - |x+5| \leqslant 4$.

Case 1:$x+2>0 \text{ and } x+5>0 \implies x>-2$

In this case, we have $\begin{aligned} 2(x+2) - (x+5) &\leqslant 4 \\ 2x + 4 - x - 5 &\leqslant 4 \\ x -1 &\leqslant 4 \\ x &\leqslant 5\\ \Rightarrow -2<x&\leqslant 5. \end{aligned}$

Case 2:$x+2>0 \text{ and } x+5\leqslant 0$

Since it is always true that $x+5>x+2,$ this case is not possible.

Case 3:$x+2\leqslant 0 \text{ and } x+ 5>0 \implies -5<x\leqslant -2$

In this case, we have $\begin{aligned} -2(x+2) - (x+5) &\leqslant 4 \\ -2x - 4 - x - 5 &\leqslant 4 \\ -3x - 9 &\leqslant 4 \\ -3x &\leqslant 13 \\ x &\geqslant \dfrac{-13}{3}\\ \Rightarrow \dfrac{-13}{3} \leqslant x &\leqslant -2. \end{aligned}$

Case 4:$x+2\leqslant \text{ and } x + 5\leqslant \implies x\leqslant -5$

In this case, we have $\begin{aligned} -2(x+2) + (x+5) &\leqslant 4 \\ -2x - 4 + x + 5 &\leqslant 4 \\ -x +1 &\leqslant 4 \\ -x &\leqslant 3 \\ x &\geqslant 3, \end{aligned}$ which is impossible since $x\leqslant -5.$Therefore, from cases 1 and 3, we have $\dfrac{-13}{3} \leqslant x \leqslant 5.\ _\square$

Solve $|x^2 + 4x + 4| - 2|x+2| + 1 < \frac{1}{2}$.

We got lots and lots of cases here. Or do we?

Rewrite the given inequality as $\left| (x+2)^2 \right| - 2|x+2| + 1 < \dfrac{1}{2}.$ Shouldn't $(x+2)^2$ always be positive? Yes. We don't need the absolute value sign there.

We're left with just two cases--one where $x+ 2$ (in the absolute value sign) is positive, and the other where it is negative.

Case 1:$x+2\ge 0 \implies x \ge -2$

In this case, we have $\begin{aligned} (x+2)^2 - 2(x+2) + 1 &< \dfrac{1}{2} \\ (x+2 - 1)^2 &< \dfrac{1}{2} \\ (x+1)^2 &< \dfrac{1}{2} \\ (x+1)^2 - \dfrac{1}{2} &< 0 \\ \left( x+ 1 + \dfrac{1}{\sqrt{2}} \right) \left( x + 1 - \dfrac{1}{\sqrt{2}}\right) &< 0. \end{aligned}$ To solve this quadratic inequality, we can use a sign analysis chart.We know that the inequality will equal zero when one factor of the whole expression equals zero, namely at $-1 - \frac{1}{\sqrt{2}}$ and $-1 + \frac{1}{\sqrt{2}}$, so we want to know whether each factor is positive or negative at specific values of $x$ that are less than $-1 - \frac{1}{\sqrt{2}}$, greater than $-1 - \frac{1}{\sqrt{2}}$

andless than $-1 + \frac{1}{\sqrt{2}}$, and greater than -1 + $\frac{1}{\sqrt{2}}$. We also use the properties that negative times negative gives positive, positive times positive gives positive, and negative times positive gives negative. To satisfy that the left hand side is less than zero, we look for $x$ that give us negative values of left hand side.

We can see that the left hand side of the quadratic inequality is less than zero when $-1 - \frac{1}{\sqrt{2} } < x < -1 + \frac{1}{\sqrt{2}}. \qquad (1)$

$($Note that these inequalities satisfy $x \ge -2.)$

Case 2:$x + 2<0 \implies x<-2$

In this case, we have $\begin{aligned} (x+2)^2 + 2(x+2) + 1 &< \dfrac{1}{2} \\ (x+2 + 1)^2 &< \dfrac{1}{2} \\ (x+3)^2 &< \dfrac{1}{2} \\ (x+3)^2 - \dfrac{1}{2} &< 0 \\ \left( x+ 3 + \dfrac{1}{\sqrt{2}} \right) \left( x + 3 - \dfrac{1}{\sqrt{2}}\right) &< 0. \end{aligned}$ Again, we do the sign analysis chart:

We can see that the left hand side of the quadratic inequality is less than zero when $-3 - \frac{1}{\sqrt{2} } < x < -3 + \frac{1}{\sqrt{2}}. \qquad (2)$

$($Note that these inequalities satisfy $x < -2.)$We now have two compound inequalities (1) and (2), both of which are solutions; $x$ can lie within either interval, making the final solution set

$-3 - \dfrac{1}{\sqrt{2}} < x < -3 + \dfrac{1}{\sqrt{2}}\ \text{ or }\ -1 - \dfrac{1}{\sqrt{2}} < x < -1 + \dfrac{1}{\sqrt{2}}.\ _\square$

## Problem Solving Technique - Visualization

Another way to solve absolute value inequalities is to graph them.

Solve $| x + 3 | < 7$.

The graph solution is to find out when $y= |x+3|$ is less than $y = 7$.

To sketch the first graph, we know that based on the definition of absolute value, $|x+3| = \begin{cases} x+3 & \textrm{if } x > -3 \\ 0 & \textrm{if } x = -3 \\ -x-3 & \textrm{if } x < -3. \\ \end{cases}$

Here, we can see that when $\boxed{-10 < x < 4}$, $y = |x+3|$ is less than $y = 7$. We use open less than signs because if $x = -10$ or $x = 4$, then $y = |x+3|$ will equal $y = 7$, but the original inequality asked when $|x+3|$ is less than $7$, not less than or equal to $7$. $_\square$

Solve $2|x+2| - |x+5| < 4$.

We can also solve this graphically. But we will need to use the definition of absolute value to write a piecewise function for $y = 2|x+2| - |x+5|$.

By cases, we get the following:

- If $|x+2|$ and $|x+5|$ are both positive, the resulting equation is $y = x-1$.
- If $|x+2|$ is negative and $|x+5|$ is positive, the resulting equation is $y= -3x - 9$.
- If $|x+2|$ is positive and $|x+5|$ is negative, the resulting equation is $y= 3x+9$.
- If $|x+2|$ and $|x+5|$ are both negative, the resulting equation is $y = -x+1$.
Let's then start looking at some sets of $x$ values.

- If $x < -5$, then we know that both $|x+2 |$ and $|x+5|$ will be negative. Thus, when $x < -5$, $y = -x + 1$.
- If $-5 < x < -2$, then we know that $|x+2|$ will be negative but $|x+5|$ will be positive. Thus, when $-5 < x < -2$, $y = -3x - 9$.
- If $x > -2$, then we know that both $|x+2|$ and $|x+5|$ will be positive. Thus, when $x > -2$, $y = x -1$.
- You can calculate the $y$ values at $x = -5$ and $x = -2$, as they are the intersection of $y = -x + 1$ and $y= -3x - 9$, and the intersection of $y = -3x - 9$ and $y = x-1$, respectively.
Then, just graph $y = 4$ and determine when $y = 2|x+2| - |x+5 |$ is less than $y = 4$.

We obtain the following graph:

Note: My graph is simply zoomed out to show the intersections of the two graphs.

It is apparent that the solution set is $\dfrac{-13}{3} < x < 5$. $_\square$

Solve $|x^2 + 4x + 4| - 2|x+2| + 1 < \dfrac{1}{2}$.

First, note that $|x^2 + 4x + 4| = |(x+2)^2 | = (x+2)^2$.

Second, let's try to graph $y = |x^2 + 4x + 4| - 2|x+2| + 1$ or rather $y=(x+2)^2 - 2|x+2| + 1$.

Our two cases are that $|x+2|$ is positive, or negative.

- If $|x+2|$ is positive, we get $y= (x+2)^2 - 2(x+2) + 1$, or $y=(x+2-1)^2$, or $y=(x+1)^2$. $|x+2|$ is positive when $x>-2$.
- If $|x+2|$ is negative, we get $y= (x+2)^2 + 2(x+2) + 1$, or $y= (x+2+1)^2$, or $y=(x+3)^2$. $|x+2|$ is negative when $x<-2$.
- To find out the $y$-value at $x = -2$, simply find the intersection between $y = (x+1)^2$ and $y = (x+3)^2$.
This gives the following graph:

We can see that $|x^2 + 4x + 4| - 2|x+2| + 1$ is less than $\dfrac{1}{2}$ when

$\boxed{ -3 - \dfrac{1}{\sqrt{2}} < x < -3 + \dfrac{1}{\sqrt{2}} \quad \text{or} \quad -1 - \dfrac{1}{\sqrt{2}} < x < -1 + \dfrac{1}{\sqrt{2}}}.$

Actually, we can't really see that, so you will have to solve some equations to get those exact values using the piecewise function we generated and $y = \dfrac{1}{2}$, but it helps to have the graph because you will not have to go through as much trouble as the cases method. $_\square$

## Problem Solving Technique - Miscellaneous

Prove that there is no solution for $|3x+4|+|2x+3|+|x+2| = \frac{1}{2}$.

The $\text{LHS}$ of this equation is a sum of 3 linear terms of absolute values. This implies that $\frac{1}{2}$ is smaller than the minimum value it can achieve, say $k$. This motivates us to find the minimum value of the expression. A nice tool would be the the inequality

$|a|+|b|\geq |\pm a\pm b|.$

There are too many possible ways we can assign signs to the terms. Let's look at the expressions again. To get a better sense of the minimum value, we attempt to remove all the variables such that the result is a constant. It is not hard to see that the two possible ways are

$\begin{aligned} |3x+4|+|2x+3|+|x+2|&\geq |-(3x+4)+(2x+3)+(x+2)|\\ |3x+4|+|2x+3|+|x+2|&\geq |+(3x+4)-(2x+3)-(x+2)|. \end{aligned}$

We try the first way:

$\begin{aligned} |3x+4|+|2x+3|+|x+2| & \geq |-(3x+4)+(2x+3)+(x+2)| \\ & = 1. \end{aligned}$

This implies that the minimum value is $1,$ and since $\frac{1}{2} < 1$ there is no solution for the equation above.

Note that we should always verify the correctness of the inequality. In this case, we have the system of inequality

$\begin{cases} 3x+4 &\leq 0 \\ 2x+3 &\geq 0 \\ x+2 &\geq 0 \end{cases}$

which reduces to $-\frac{3}{2}\leq x \leq -\frac{4}{3}$. Thus this case is valid. In fact, if you attempt it with the second way, you will get an invalid solution. $_\square$

Solve the equation $|x|^2 - 6|x| + 5 \leq 0$ for all $x$.

Let $y = |x|$, then we have $y^2 - 6y + 5 \leq 0$. Factorizing it yields $(y-1)(y-5) \leq 0 \implies 1 \leq y \leq 5$. This means that $1 \leq |x| \leq 5$, and hence the solutions are $-5 \leq x \leq -1$ and $1 \leq x \leq 5$. $_\square$

Find the solutions to the inequality $|x+1| + |x+2| + |x+3| \leq -1.$

Instead of trying cases on this problem, note that the result of taking absolute value always yields a positive value (or zero), and the sum of three non-negative values cannot be negative. Because $|x+1| + |x+2| + |x+3|$ must be non-negative, it will never be less than $-1$. Thus, there is

no solution. $_\square$

When $y \leqslant 7$ is graphed, all areas below $y \leqslant 7$ and on/above the curve $y = |x| + 2|x+1| - 3|x+2| +4 |x+3|$ are shaded. If the area of the shaded region is $\mathscr{E}$, find $\left \lfloor 1000 \mathscr{E} \right \rfloor$.

**Notations**:

$| \cdot |$ denotes the absolute value function.

$\lfloor \cdot \rfloor$ denotes the floor function.

**Cite as:**Absolute Value Inequalities.

*Brilliant.org*. Retrieved from https://brilliant.org/wiki/absolute-value-inequalities/