Evanalysis
3.1Estimated reading time: 8 min

3.1 Inequalities and absolute value

Solve inequalities by preserving order, tracking domains, and using absolute value as distance.

Course contents

MATH1025: Preparatory mathematics

Preparatory mathematics notes.

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All note collectionsSeries overview
0-12-34
Chapter 0-1Foundations and early methods2 sections
0.10.1 Course foundations and notation1.11.1 Equation structure and trigonometric identities
Chapter 2-3Proof and inequalities2 sections
2.12.1 Mathematical induction3.13.1 Inequalities and absolute valueCurrent section
Chapter 4Binomial theorem1 sections
4.14.1 Binomial coefficients and expansions

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<ba<b, a=ba=b, a>ba>b holds;
  • if a<ba<b and b<cb<c, then a<ca<c;
  • if a<ba<b, then a+c<b+ca+c<b+c;
  • if a<ba<b and c>0c>0, then ac<bcac<bc;
  • if a<ba<b and c<0c<0, then ac>bcac>bc.

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

Worked example

A square gives an inequality

Let x,y>0x,y>0. Prove

xy+yx≥2.\frac{x}{y}+\frac{y}{x}\ge 2.

Since (x−y)2≥0(x-y)^2\ge 0, we have

x2−2xy+y2≥0,x^2-2xy+y^2\ge 0,

so

x2+y2≥2xy.x^2+y^2\ge 2xy.

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

x2+y2xy≥2.\frac{x^2+y^2}{xy}\ge 2.

Thus

xy+yx≥2.\frac{x}{y}+\frac{y}{x}\ge 2.

Equality holds exactly when x=yx=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

x+1x−1≤2\frac{x+1}{x-1}\le 2

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

There are three reliable methods:

  • split into sign cases;
  • multiply by a square such as (x−1)2(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

x+1x−1≤2.\frac{x+1}{x-1}\le 2.

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

x+1x−1−2≤0.\frac{x+1}{x-1}-2\le 0.

Simplify:

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

Equivalently,

x−3x−1≥0.\frac{x-3}{x-1}\ge 0.

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

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

Absolute value as distance

Definition

Absolute value

For a real number a,

∣a∣={a,a≥0,−a,a<0.|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∣=a2.|a|=\sqrt{a^2}.

Theorem

Useful absolute-value facts

For real numbers a,b:

∣−a∣=∣a∣,∣ab∣=∣a∣∣b∣,−∣a∣≤a≤∣a∣.|-a|=|a|, \qquad |ab|=|a||b|, \qquad -|a|\le a\le |a|.

If b≥0b\ge 0, then

∣a∣≤b⟺−b≤a≤b.|a|\le b \quad\Longleftrightarrow\quad -b\le a\le b.

If b>0b>0, then

∣a∣<b⟺−b<a<b.|a|<b \quad\Longleftrightarrow\quad -b<a<b.

Theorem

Triangle inequality

For all real numbers a,b,

∣a+b∣≤∣a∣+∣b∣.|a+b|\le |a|+|b|.

More generally,

∣a1+⋯+an∣≤∣a1∣+⋯+∣an∣.|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∣≥4.|x-2|+|2x+1|\ge 4.

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

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

On x<-1/2,

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

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

On -1/2\le x<2,

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

The inequality x+3≥4x+3\ge 4 gives x≥1x\ge 1, so this case contributes 1≤x<21\le x<2.

On x≥2x\ge 2,

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

The inequality 3x−1≥43x-1\ge 4 gives x\ge 5/3; combined with x≥2x\ge 2, this contributes x≥2x\ge 2.

Therefore the full solution is

x≤−1orx≥1.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≥0u,v\ge 0, then

u+v2≥uv.\frac{u+v}{2}\ge \sqrt{uv}.

Equality holds exactly when u=vu=v.

This follows from

(u−v)2≥0.(\sqrt u-\sqrt v)^2\ge 0.

Quick checks

Quick check

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

Think about the sign of x−1x-1.

Solution

Answer

Quick check

Rewrite ∣x−5∣<2|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>0x,y>0?

Look at the square used in the proof.

Solution

Answer

Exercises

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

Guided solutions

  1. The critical points are −1-1, 2, and 3, with x=−1x=-1 excluded. A sign chart gives (−1,2)∪(3,∞)(-1,2)\cup(3,\infty).
  2. Use induction. The base case is equality. If (1+x)k≥1+kx(1+x)^k\ge 1+kx, multiply by 1+x>01+x>0 and compare with 1+(k+1)x1+(k+1)x; the remaining term is kx2≥0kx^2\ge 0.
  3. The domain requires x≠−1x\ne -1. Splitting at x=−2x=-2 and x=−1x=-1 gives -3/2<x<-1.
  4. From the triangle inequality, ∣a∣=∣(a−b)+b∣≤∣a−b∣+∣b∣|a|=|(a-b)+b|\le |a-b|+|b|, so ∣a∣−∣b∣≤∣a−b∣|a|-|b|\le |a-b|. Reverse the roles of a and b to get ∣b∣−∣a∣≤∣a−b∣|b|-|a|\le |a-b|.

Prerequisites

2.1

2.1 Mathematical induction

Key terms in this unit

More notes in this series

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4.1 Binomial coefficients and expansions