McIntyre MathsMathematics Extension 2
Extension 2 › Proof › Lesson 5

Lesson 5 · Algebraic inequalities I

Mathematics Extension 2 · Proof · Outcome ME2-12-01 · Term 4 · Week 3
Learning intentionTo prove inequalities using the definition a>b⇔a−b>0a>b\Leftrightarrow a-b>0, the fact that squares are non-negative, and the basic properties of inequalities.
Success criteria · I can…
  • prove an inequality by showing LHS−RHS≥0\text{LHS}-\text{RHS}\ge 0
  • start from a true statement such as (a−b)2≥0(a-b)^2\ge0 and build the result
  • prove and use the properties of inequalities listed in the syllabus
  • use a given condition, for example a+b=1a+b=1, in a proof

Lesson flow

LaunchWarm-up · Inequality instinctsSlide 2Key ideasTwo tools and two layoutsSlide 3ExamplesExample 2 · Squares are non-negativeSlide 4HSC-styleHSC practice · Squares are non-negativeSlide 5Key ideasProperties of inequalities · Prove and useSlide 6ExamplesExample 3 · Using the propertiesSlide 7HSC-styleHSC practice · Which property fails?Slide 8ExamplesExample 4 · Using a given conditionSlide 9Exam practiceExam-style · Prove a result, then use itSlide 10Exit ticketExit ticketSlide 11Spot the flawSpot the flaw · Starting from the resultSlide 12

Warm-up · Inequality instincts

Slide 2

Answer each. For the true/false parts, find a counterexample if the statement is false.

(a) Solve 3−2x>73-2x>7.
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Answer
x<−2x<-2.
Why
Subtract 3 to get −2x>4-2x>4, then divide by −2-2. Dividing by a negative number reverses the inequality.
(b) Complete the square: x2−6x+11x^2-6x+11.
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Answer
(x−3)2+2(x-3)^2+2, so the expression is always at least 2.
Why
x2−6x+9=(x−3)2x^2-6x+9=(x-3)^2, leaving +2+2. A square is never negative, so the minimum value is 2 at x=3x=3.
(c) True or false: if a>ba>b, then a2>b2a^2>b^2.
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Answer
False. For example a=1a=1, b=−2b=-2.
Why
1>−21>-2 but 12=1<4=(−2)21^2=1<4=(-2)^2. The statement is true if a>b>0a>b>0.
(d) True or false: if a>ba>b, then 1a<1b\dfrac1a<\dfrac1b.
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Answer
False. For example a=1a=1, b=−1b=-1.
Why
1>−11>-1 but 11=1>−1=1−1\dfrac11=1>-1=\dfrac1{-1}. Both (c) and (d) are true under extra conditions, which are today’s properties.

Two tools and two layouts

Slide 3
Definition
The two tools

a>b⇔a−b>0a>b\Leftrightarrow a-b>0.

Squares: x2≥0x^2\ge0 for all real xx. In particular (a±b)2≥0(a\pm b)^2\ge0.

Key idea
Two layouts
  • Layout 1: LHS−RHS=⋯=(something)2≥0\text{LHS}-\text{RHS}=\dots=(\text{something})^2\ge0, so LHS≥RHS\text{LHS}\ge\text{RHS}.
  • Layout 2: start from a true statement, for example (a−b)2≥0(a-b)^2\ge0, and rearrange.
Watch out
Never start from the result you are asked to prove, unless every step is reversible and marked ⇔\Leftrightarrow.

Example 1

Prove that x2+y2≥2xyx^2+y^2\ge2xy for all real xx and yy, using each layout. When does equality hold?
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Answer
Layout 1: x2+y2−2xy=(x−y)2≥0x^2+y^2-2xy=(x-y)^2\ge0. Layout 2: (x−y)2≥0⇒x2−2xy+y2≥0⇒x2+y2≥2xy(x-y)^2\ge0\Rightarrow x^2-2xy+y^2\ge0\Rightarrow x^2+y^2\ge2xy. Equality holds if and only if x=yx=y.
Why
A square is zero only when what is inside it is zero, so (x−y)2=0⇔x=y(x-y)^2=0\Leftrightarrow x=y. This result is used constantly in Lesson 6 and in HSC questions.

Example 2 · Squares are non-negative

Slide 4
(a) Prove that a2+1a2≥2a^2+\dfrac1{a^2}\ge2 for a≠0a\ne0.
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Answer
a2+1a2−2=(a−1a)2≥0a^2+\dfrac1{a^2}-2=\left(a-\dfrac1a\right)^2\ge0.
Why
Expand (a−1a)2=a2−2+1a2\left(a-\frac1a\right)^2=a^2-2+\frac1{a^2}. The restriction a≠0a\ne0 is needed so that 1a2\frac1{a^2} exists.
(b) Prove that (x+y)2≥4xy(x+y)^2\ge4xy for all real x,yx,y.
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Answer
(x+y)2−4xy=x2−2xy+y2=(x−y)2≥0(x+y)^2-4xy=x^2-2xy+y^2=(x-y)^2\ge0.
Why
Subtract the right-hand side and recognise a perfect square.
(c) Prove that a2+b22≥(a+b2)2\dfrac{a^2+b^2}2\ge\left(\dfrac{a+b}2\right)^2.
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Answer
LHS−RHS=2a2+2b2−a2−2ab−b24=(a−b)24≥0\text{LHS}-\text{RHS}=\dfrac{2a^2+2b^2-a^2-2ab-b^2}4=\dfrac{(a-b)^2}4\ge0.
Why
In words: the mean of the squares is at least the square of the mean. Use a common denominator of 4 before simplifying.

HSC practice · Squares are non-negative

Slide 5
2023 HSC Q12(b)Let xx and yy be real numbers with x2+y2≠0x^2+y^2\ne0. By considering (x−y)2≥0(x-y)^2\ge0, show that (x+y)2x2+y2≤2\dfrac{(x+y)^2}{x^2+y^2}\le2.
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Answer
(x−y)2≥0⇒x2+y2≥2xy⇒2(x2+y2)≥x2+2xy+y2=(x+y)2(x-y)^2\ge0\Rightarrow x^2+y^2\ge2xy\Rightarrow2(x^2+y^2)\ge x^2+2xy+y^2=(x+y)^2. Since x2+y2>0x^2+y^2>0, divide both sides by it: (x+y)2x2+y2≤2\dfrac{(x+y)^2}{x^2+y^2}\le2.
Why
State that x2+y2>0x^2+y^2>0 before dividing: it is non-negative and not zero. Dividing by a positive number keeps the inequality sign.

Properties of inequalities · Prove and use

Slide 6

For real a,b,c,da,b,c,d. Each property can be proved from a>b⇔a−b>0a>b\Leftrightarrow a-b>0.

PropertyWhy?
If a>ba>b, then a±c>b±ca\pm c>b\pm c(a±c)−(b±c)=a−b>0(a\pm c)-(b\pm c)=a-b>0
If a>b>0a>b>0, then 1b>1a>0\dfrac1b>\dfrac1a>0 (and vice versa)1b−1a=a−bab>0\dfrac1b-\dfrac1a=\dfrac{a-b}{ab}>0, as ab>0ab>0
∣a∣>∣b∣⇔a2>b2|a|>|b|\Leftrightarrow a^2>b^2a2−b2=(∣a∣−∣b∣)(∣a∣+∣b∣)a^2-b^2=(|a|-|b|)(|a|+|b|), and ∣a∣+∣b∣>0|a|+|b|>0 when ∣a∣>∣b∣|a|>|b|
If a>ba>b and b>cb>c, then a>ca>ca−c=(a−b)+(b−c)>0a-c=(a-b)+(b-c)>0
If a>ba>b and c>dc>d, then a+c>b+da+c>b+d(a+c)−(b+d)=(a−b)+(c−d)>0(a+c)-(b+d)=(a-b)+(c-d)>0
If a>ba>b and c>0c>0, then ac>bcac>bcac−bc=c(a−b)>0ac-bc=c(a-b)>0
If a>ba>b and c<0c<0, then ac<bcac<bcac−bc=c(a−b)<0ac-bc=c(a-b)<0
Warning
We cannot in general subtract or divide inequalities. Counterexample: 5>15>1 and 4>04>0, but 5−4=15-4=1 is not greater than 1−0=11-0=1. (Also a>ba>b does not give a2>b2a^2>b^2 in general; the syllabus result is ∣a∣>∣b∣⇔a2>b2|a|>|b|\Leftrightarrow a^2>b^2.)

Example 3 · Using the properties

Slide 7
(a) Prove that if a>b>0a>b>0, then a2>b2a^2>b^2.
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Answer
a2−b2=(a−b)(a+b)a^2-b^2=(a-b)(a+b), a product of two positive numbers, so a2>b2a^2>b^2.
Why
a−b>0a-b>0 because a>ba>b, and a+b>0a+b>0 because both are positive.
(b) Prove that if a,b>0a,b>0, then ab+ba≥2\dfrac ab+\dfrac ba\ge2.
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Answer
ab+ba−2=a2+b2−2abab=(a−b)2ab≥0\dfrac ab+\dfrac ba-2=\dfrac{a^2+b^2-2ab}{ab}=\dfrac{(a-b)^2}{ab}\ge0, since ab>0ab>0.
Why
Combine over the common denominator abab. A non-negative numerator over a positive denominator is non-negative.
(c) Prove that a3+b3≥a2b+ab2a^3+b^3\ge a^2b+ab^2 for a,b>0a,b>0.
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Answer
a3+b3−a2b−ab2=a2(a−b)−b2(a−b)=(a−b)(a2−b2)=(a−b)2(a+b)≥0a^3+b^3-a^2b-ab^2=a^2(a-b)-b^2(a-b)=(a-b)(a^2-b^2)=(a-b)^2(a+b)\ge0.
Why
Group in pairs and factor out (a−b)(a-b). Then (a−b)2≥0(a-b)^2\ge0 and a+b>0a+b>0, so the product is non-negative.

HSC practice · Which property fails?

Slide 8
Adapted from 2021 HSC Q5Which statement is NOT true for all real numbers a<ba<b (in the domain of each function)?
A. a3<b3a^3<b^3B. e−a>e−be^{-a}>e^{-b}C. ln⁡a<ln⁡b\ln a<\ln bD. 1a>1b\dfrac1a>\dfrac1b
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Answer
D.
Why
Counterexample: a=−1a=-1, b=1b=1 gives 1a=−1<1=1b\dfrac1a=-1<1=\dfrac1b. The property 1b>1a\dfrac1b>\dfrac1a holds only when a,ba,b are both positive (or both negative). A, B and C hold because x3x^3 and ln⁡x\ln x are increasing and e−xe^{-x} is decreasing. (Options reworded from the paper.)

Example 4 · Using a given condition

Slide 9

Let aa and bb be positive real numbers with a+b=1a+b=1.

(a) Prove that ab≤14ab\le\dfrac14.
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Answer
1=(a+b)2≥4ab1=(a+b)^2\ge4ab (Example 2(b)), so ab≤14ab\le\dfrac14.
Why
Square the condition a+b=1a+b=1, then use (x+y)2≥4xy(x+y)^2\ge4xy and divide by 4.
(b) Prove that a2+b2≥12a^2+b^2\ge\dfrac12.
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Answer
a2+b2=(a+b)2−2ab=1−2ab≥1−12=12a^2+b^2=(a+b)^2-2ab=1-2ab\ge1-\dfrac12=\dfrac12.
Why
By (a), 2ab≤122ab\le\dfrac12, and subtracting a smaller number from 1 gives a larger result.
(c) Hence prove that (1+1a)(1+1b)≥9\left(1+\dfrac1a\right)\left(1+\dfrac1b\right)\ge9.
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Answer
(1+1a)(1+1b)=1+a+bab+1ab=1+2ab≥1+2×4=9\left(1+\dfrac1a\right)\left(1+\dfrac1b\right)=1+\dfrac{a+b}{ab}+\dfrac1{ab}=1+\dfrac2{ab}\ge1+2\times4=9.
Why
Expand, then use a+b=1a+b=1. Since 0<ab≤140<ab\le\dfrac14, we have 1ab≥4\dfrac1{ab}\ge4 (taking reciprocals reverses the inequality).

Exam-style · Prove a result, then use it

Slide 10

Exam-style question in three parts.

(i) Prove that x2+y2≥2xyx^2+y^2\ge2xy for all real xx and yy.
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Answer
x2+y2−2xy=(x−y)2≥0x^2+y^2-2xy=(x-y)^2\ge0.
Why
Layout 1 from Example 1.
(ii) Hence prove that a2+b2+c2≥ab+bc+caa^2+b^2+c^2\ge ab+bc+ca for all real a,b,ca,b,c.
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Answer
Add a2+b2≥2aba^2+b^2\ge2ab, b2+c2≥2bcb^2+c^2\ge2bc and c2+a2≥2cac^2+a^2\ge2ca to get 2(a2+b2+c2)≥2(ab+bc+ca)2(a^2+b^2+c^2)\ge2(ab+bc+ca). Halve both sides.
Why
Apply (i) three times, once to each pair of variables, then add the three inequalities.
(iii) Hence, or otherwise, show that if a+b+c=3a+b+c=3, then ab+bc+ca≤3ab+bc+ca\le3.
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Answer
9=(a+b+c)2=a2+b2+c2+2(ab+bc+ca)≥3(ab+bc+ca)9=(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)\ge3(ab+bc+ca), so ab+bc+ca≤3ab+bc+ca\le3.
Why
Square the condition, then replace a2+b2+c2a^2+b^2+c^2 by the smaller quantity ab+bc+caab+bc+ca using (ii). This is the syllabus idea of using previously obtained inequalities.

Exit ticket

Slide 11
(a) Prove that x2−4x+5>0x^2-4x+5>0 for all real xx.
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Answer
x2−4x+5=(x−2)2+1≥1>0x^2-4x+5=(x-2)^2+1\ge1>0.
Why
Complete the square. The square is at least 0, so the whole expression is at least 1.
(b) Given a>b>0a>b>0, prove that a+1b+1<ab\dfrac{a+1}{b+1}<\dfrac ab.
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Answer
ab−a+1b+1=a(b+1)−b(a+1)b(b+1)=a−bb(b+1)>0\dfrac ab-\dfrac{a+1}{b+1}=\dfrac{a(b+1)-b(a+1)}{b(b+1)}=\dfrac{a-b}{b(b+1)}>0.
Why
The numerator a−b>0a-b>0 and the denominator b(b+1)>0b(b+1)>0, so the difference is positive.

Spot the flaw · Starting from the result

Slide 12
Spot the flawA student wrote this. Find the flaw, explain why it matters, then write a correct version.
Claim. a2+b2≥2aba^2+b^2\ge2ab for all real aa and bb.
1. a2+b2≥2aba^2+b^2\ge2ab
2. ⇒a2−2ab+b2≥0\Rightarrow a^2-2ab+b^2\ge0
3. ⇒(a−b)2≥0\Rightarrow(a-b)^2\ge0, which is true. Hence a2+b2≥2aba^2+b^2\ge2ab. □\square
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Answer
The flaw: the student starts from the statement to be proved and ends at something true. A true ending does not make the start true. Correct version: (a−b)2≥0⇒a2−2ab+b2≥0⇒a2+b2≥2ab(a-b)^2\ge0\Rightarrow a^2-2ab+b^2\ge0\Rightarrow a^2+b^2\ge2ab (Layout 2), or write ⇔\Leftrightarrow between the lines.
Why
From −1=1-1=1, squaring gives 1=11=1, which is true, yet −1=1-1=1 is false. Here every step happens to be reversible, so the argument can be rescued by writing ⇔\Leftrightarrow or by reversing the order. Reversing fails when a step is not reversible, for example squaring both sides, or multiplying by something that may be negative or zero.