3.1 Inequalities and absolute value

Inequalities are statements with direction

An equation records equality. An inequality records order. The difference is small in notation but large in algebra: multiplying by a negative number reverses the inequality sign, while multiplying by a positive number preserves it.

The main discipline in this chapter is to know which transformations preserve the solution set and which ones require cases.

Basic order rules

Theorem

Order rules for real numbers

For real numbers a,b,c:

exactly one of $a<b$ , $a=b$ , $a>b$ holds;
if $a<b$ and $b<c$ , then $a<c$ ;
if $a<b$ , then $a+c<b+c$ ;
if $a<b$ and $c>0$ , then $ac<bc$ ;
if $a<b$ and $c<0$ , then $ac>bc$ .

The last two rules explain why division by an expression such as $x-1$ cannot be done blindly. Its sign depends on x.

Worked example

A square gives an inequality

Let $x,y>0$ . Prove

\frac{x}{y}+\frac{y}{x}\ge 2.

Since $(x-y)^2\ge 0$ , we have

x^2-2xy+y^2\ge 0,

x^2+y^2\ge 2xy.

Because $xy>0$ , division by xy preserves the inequality:

\frac{x^2+y^2}{xy}\ge 2.

Thus

\frac{x}{y}+\frac{y}{x}\ge 2.

Equality holds exactly when $x=y$ .

This proof illustrates a common method: start from a quantity known to be nonnegative, then rearrange it into the desired inequality.

Rational inequalities

Common mistake

Do not multiply by an expression of unknown sign

The inequality

\frac{x+1}{x-1}\le 2

cannot be solved by immediately multiplying by $x-1$ , because $x-1$ is positive for $x>1$ , negative for $x<1$ , and zero at $x=1$ .

There are three reliable methods:

split into sign cases;
multiply by a square such as $(x-1)^2$ after excluding the zero point;
move everything to one side and use a sign chart.

Worked example

Solve a rational inequality

Solve

\frac{x+1}{x-1}\le 2.

The expression is undefined at $x=1$ . Move everything to one side:

\frac{x+1}{x-1}-2\le 0.

Simplify:

\frac{x+1-2x+2}{x-1}\le 0, \qquad \frac{3-x}{x-1}\le 0.

Equivalently,

\frac{x-3}{x-1}\ge 0.

The critical points are 1 and 3. A sign chart shows the fraction is positive on $(-\infty,1)$ and $[3,\infty)$ , but $x=1$ is excluded. Therefore

x<1 \quad\text{or}\quad x\ge 3.

Absolute value as distance

Definition

Absolute value

For a real number a,

|a| = \begin{cases} a, & a\ge 0,\\ -a, & a<0. \end{cases}

Geometrically, |a| is the distance from a to 0 on the real line.

Because absolute value is a distance, it is always nonnegative, and

|a|=\sqrt{a^2}.

Theorem

Useful absolute-value facts

For real numbers a,b:

|-a|=|a|, \qquad |ab|=|a||b|, \qquad -|a|\le a\le |a|.

If $b\ge 0$ , then

|a|\le b \quad\Longleftrightarrow\quad -b\le a\le b.

If $b>0$ , then

|a|<b \quad\Longleftrightarrow\quad -b<a<b.

Theorem

Triangle inequality

For all real numbers a,b,

|a+b|\le |a|+|b|.

More generally,

|a_1+\cdots+a_n|\le |a_1|+\cdots+|a_n|.

The triangle inequality says that going directly from one point to another is never longer than taking a detour through intermediate displacements.

Solving absolute-value inequalities

Absolute-value equations and inequalities often require splitting at the points where the expression inside an absolute value changes sign.

Worked example

Split at the breakpoints

Solve

|x-2|+|2x+1|\ge 4.

The expressions change sign at $x=2$ and $x=-1/2$ . Consider the intervals

(-\infty,-1/2),\qquad [-1/2,2),\qquad [2,\infty).

On $x<-1/2$ ,

|x-2|+|2x+1|=-(x-2)-(2x+1)=-3x+1.

The inequality $-3x+1\ge 4$ gives $x\le -1$ , so this case contributes $x\le -1$ .

On $-1/2\le x<2$ ,

|x-2|+|2x+1|=-(x-2)+(2x+1)=x+3.

The inequality $x+3\ge 4$ gives $x\ge 1$ , so this case contributes $1\le x<2$ .

On $x\ge 2$ ,

|x-2|+|2x+1|=(x-2)+(2x+1)=3x-1.

The inequality $3x-1\ge 4$ gives $x\ge 5/3$ ; combined with $x\ge 2$ , this contributes $x\ge 2$ .

Therefore the full solution is

x\le -1 \quad\text{or}\quad x\ge 1.

Classical inequality habit

Several later examples use named inequalities such as Bernoulli's inequality and AM-GM. The common pattern is the same: state the domain, locate the nonnegative expression or monotonic function, and record the equality case.

Theorem

AM-GM for two nonnegative numbers

If $u,v\ge 0$ , then

\frac{u+v}{2}\ge \sqrt{uv}.

Equality holds exactly when $u=v$ .

This follows from

(\sqrt u-\sqrt v)^2\ge 0.

Quick checks

Quick check

Why does multiplying an inequality by $x-1$ require cases?

Think about the sign of $x-1$ .

Solution

Answer

Quick check

Rewrite $|x-5|<2$ as an interval.

Use distance from 5.

Solution

Answer

Quick check

When does equality hold in $x/y + y/x >= 2$ for $x,y>0$ ?

Look at the square used in the proof.

Solution

Answer

Exercises

Solve $((x-2)(x-3))/(x+1) > 0$ .
Prove Bernoulli's inequality: if $x>-1$ , then $(1+x)^n\ge 1+nx$ for every $n\in Z^+$ .
Solve $|x+2|/(x+1)<-1$ .
Prove $||a|-|b||\le |a-b|$ for real numbers a,b.

Guided solutions

The critical points are $-1$ , 2, and 3, with $x=-1$ excluded. A sign chart gives $(-1,2)\cup(3,\infty)$ .
Use induction. The base case is equality. If $(1+x)^k\ge 1+kx$ , multiply by $1+x>0$ and compare with $1+(k+1)x$ ; the remaining term is $kx^2\ge 0$ .
The domain requires $x\ne -1$ . Splitting at $x=-2$ and $x=-1$ gives $-3/2<x<-1$ .
From the triangle inequality, $|a|=|(a-b)+b|\le |a-b|+|b|$ , so $|a|-|b|\le |a-b|$ . Reverse the roles of a and b to get $|b|-|a|\le |a-b|$ .

3.1 Inequalities and absolute value

MATH1025: Preparatory mathematics

Inequalities are statements with direction

Basic order rules

Order rules for real numbers

A square gives an inequality

Rational inequalities

Do not multiply by an expression of unknown sign

Solve a rational inequality

Absolute value as distance

Absolute value

Useful absolute-value facts

Triangle inequality

Solving absolute-value inequalities

Split at the breakpoints

Classical inequality habit

AM-GM for two nonnegative numbers

Quick checks

Why does multiplying an inequality by $x-1$ require cases?

Answer

Rewrite $|x-5|<2$ as an interval.

Answer

When does equality hold in $x/y + y/x >= 2$ for $x,y>0$ ?

Answer

Exercises

Guided solutions

Prerequisites

Key terms in this unit

Premium learning add-ons

More notes in this series

3.1 Inequalities and absolute value

MATH1025: Preparatory mathematics

Inequalities are statements with direction

Basic order rules

Order rules for real numbers

A square gives an inequality

Rational inequalities

Do not multiply by an expression of unknown sign

Solve a rational inequality

Absolute value as distance

Absolute value

Useful absolute-value facts

Triangle inequality

Solving absolute-value inequalities

Split at the breakpoints

Classical inequality habit

AM-GM for two nonnegative numbers

Quick checks

Why does multiplying an inequality by x−1x-1x−1 require cases?

Answer

Rewrite ∣x−5∣<2|x-5|<2∣x−5∣<2 as an interval.

Answer

When does equality hold in x/y+y/x>=2x/y + y/x >= 2x/y+y/x>=2 for x,y>0x,y>0x,y>0?

Answer

Exercises

Guided solutions

Prerequisites

Key terms in this unit

Premium learning add-ons

More notes in this series

Why does multiplying an inequality by $x-1$ require cases?

Rewrite $|x-5|<2$ as an interval.

When does equality hold in $x/y + y/x >= 2$ for $x,y>0$ ?