McIntyre MathsMathematics Extension 2
Extension 2 › Proof › Lesson 6

Lesson 6 · Algebraic inequalities II

Mathematics Extension 2 · Proof · Outcome ME2-12-01 · Term 4 · Week 3
Learning intentionTo prove and use the AM–GM inequality and the triangle inequality, and to build new inequalities from known ones, including in geometry.
Success criteria · I can…
  • prove a+b2≥ab\dfrac{a+b}2\ge\sqrt{ab} for a,b≥0a,b\ge0 and use it
  • substitute into a proven inequality to obtain a new one
  • prove ∣a+b∣≤∣a∣+∣b∣|a+b|\le|a|+|b| and interpret it geometrically
  • prove inequalities involving geometry

Lesson flow

LaunchWarm-upSlide 2Key ideasArithmetic mean ≥ geometric meanSlide 3HSC-styleHSC practice · Substituting into AM–GMSlide 4ExamplesExample 2 · Using AM–GMSlide 5HSC-styleHSC practice · Building on AM–GMSlide 6HSC-styleHSC practice · A condition and AM–GMSlide 7Key ideasThe triangle inequalitySlide 8ExamplesExample 4 · Using the triangle inequalitySlide 9ExamplesExample 5 · Inequalities in geometrySlide 10ExtensionExtension · AM–GM for four and three numbersSlide 11Exit ticketExit ticketSlide 12Spot the flawSpot the flaw · AM–GMSlide 13Mixed practiceWhich method? · Proofs so farSlide 14

Warm-up

Slide 2
(a) Prove p2+q2≥2pqp^2+q^2\ge2pq.
Show answerHide answer
Answer
p2+q2−2pq=(p−q)2≥0p^2+q^2-2pq=(p-q)^2\ge0.
Why
Subtract the right-hand side and recognise a perfect square.
(b) Expand (a−b)2\left(\sqrt a-\sqrt b\right)^2 for a,b≥0a,b\ge0.
Show answerHide answer
Answer
a−2ab+ba-2\sqrt{ab}+b.
Why
(a)2=a(\sqrt a)^2=a, (b)2=b(\sqrt b)^2=b and ab=ab\sqrt a\sqrt b=\sqrt{ab} since a,b≥0a,b\ge0. This will give AM–GM.
(c) Compare ∣−3+5∣|-3+5| with ∣−3∣+∣5∣|-3|+|5|.
Show answerHide answer
Answer
∣−3+5∣=2|-3+5|=2 and ∣−3∣+∣5∣=8|-3|+|5|=8, so 2<82<8.
Why
The two sides are equal only when the numbers have the same sign. This is the idea of the triangle inequality.
(d) Can a triangle have sides 3 cm, 4 cm and 8 cm?
Show answerHide answer
Answer
No.
Why
3+4=7<83+4=7<8: the two shorter sides together are too short to reach across the longest side.

Arithmetic mean ≥ geometric mean

Slide 3
Theorem
AM–GM for two numbers

For all a,b≥0a,b\ge0:

a+b2≥ab,\frac{a+b}2\ge\sqrt{ab},

with equality if and only if a=ba=b.

Equivalent forms: a+b≥2aba+b\ge2\sqrt{ab}; and for real x,yx,y, x2+y2≥2xyx^2+y^2\ge2xy (put a=x2a=x^2, b=y2b=y^2).

Example 1 (a)

2022 HSC Q12(a)Prove the AM–GM inequality.
Show answerHide answer
Answer
(a−b)2≥0⇒a+b−2ab≥0⇒a+b2≥ab\left(\sqrt a-\sqrt b\right)^2\ge0\Rightarrow a+b-2\sqrt{ab}\ge0\Rightarrow\dfrac{a+b}2\ge\sqrt{ab}. Equality holds if and only if a=b\sqrt a=\sqrt b, that is a=ba=b.
Why
This is the expansion from the warm-up. A square is zero only when its base is zero. (A geometric proof also works: in a semicircle of diameter a+ba+b, the half-chord at the join has length ab\sqrt{ab} and cannot exceed the radius a+b2\frac{a+b}2.)

Example 1 (b)

Prove that for x>0x>0, x+9x≥6x+\dfrac9x\ge6. When is there equality?
Show answerHide answer
Answer
x+9x≥2x⋅9x=29=6x+\dfrac9x\ge2\sqrt{x\cdot\dfrac9x}=2\sqrt9=6. Equality holds when x=9xx=\dfrac9x, so x2=9x^2=9 and x=3x=3.
Why
Apply AM–GM to the two positive numbers xx and 9x\dfrac9x. Their product is the constant 9, which is what makes the bound useful.

HSC practice · Substituting into AM–GM

Slide 4
Adapted from 2020 HSC Q13(c)(i) Show that a+b≥2aba+b\ge2\sqrt{ab} for all a,b≥0a,b\ge0.
(ii) Hence show that p2+4q2≥4pqp^2+4q^2\ge4pq for all real pp and qq.
Show answerHide answer
Answer
(i) (a−b)2≥0\left(\sqrt a-\sqrt b\right)^2\ge0, so a+b≥2aba+b\ge2\sqrt{ab}. (ii) Put a=p2a=p^2, b=4q2b=4q^2 (both ≥0\ge0): p2+4q2≥24p2q2=4∣pq∣≥4pqp^2+4q^2\ge2\sqrt{4p^2q^2}=4|pq|\ge4pq.
Why
Substitution is the key skill: choose aa and bb so that the square roots simplify. Check: (p−2q)2≥0(p-2q)^2\ge0 gives the same result directly. (The original part (i) is proved geometrically from a diagram; the wording here is adapted.)

Example 2 · Using AM–GM

Slide 5
(a) Prove that (a+b)(1a+1b)≥4(a+b)\left(\dfrac1a+\dfrac1b\right)\ge4 for a,b>0a,b>0.
Show answerHide answer
Answer
(a+b)(1a+1b)=2+ab+ba≥2+2=4(a+b)\left(\dfrac1a+\dfrac1b\right)=2+\dfrac ab+\dfrac ba\ge2+2=4.
Why
Expand, then use ab+ba≥2\dfrac ab+\dfrac ba\ge2 from Lesson 5. Alternatively apply AM–GM to each bracket: a+b≥2aba+b\ge2\sqrt{ab} and 1a+1b≥2ab\dfrac1a+\dfrac1b\ge\dfrac2{\sqrt{ab}}, then multiply.
(b) Prove that (a+b)(b+c)(c+a)≥8abc(a+b)(b+c)(c+a)\ge8abc for a,b,c>0a,b,c>0.
Show answerHide answer
Answer
a+b≥2aba+b\ge2\sqrt{ab}, b+c≥2bcb+c\ge2\sqrt{bc}, c+a≥2cac+a\ge2\sqrt{ca}. All sides are positive, so multiply: (a+b)(b+c)(c+a)≥8a2b2c2=8abc(a+b)(b+c)(c+a)\ge8\sqrt{a^2b^2c^2}=8abc.
Why
Multiplying inequalities is valid when all the quantities are positive. The three square roots multiply to (abc)2=abc\sqrt{(abc)^2}=abc.

HSC practice · Building on AM–GM

Slide 6

Let aa, bb, cc be positive real numbers.

2021 HSC Q15(a)(i) Using xy≤x+y2\sqrt{xy}\le\dfrac{x+y}2, show that abc≤a2+b2+2c4\sqrt{abc}\le\dfrac{a^2+b^2+2c}4.
(ii) Hence show that abc≤a2+b2+c2+a+b+c6\sqrt{abc}\le\dfrac{a^2+b^2+c^2+a+b+c}6.
Show answerHide answer
Answer
(i) abc=ab⋅c≤ab+c2\sqrt{abc}=\sqrt{ab\cdot c}\le\dfrac{ab+c}2, and ab≤a2+b22ab\le\dfrac{a^2+b^2}2, so abc≤a2+b22+c2=a2+b2+2c4\sqrt{abc}\le\dfrac{\frac{a^2+b^2}2+c}2=\dfrac{a^2+b^2+2c}4. (ii) By symmetry, abc≤b2+c2+2a4\sqrt{abc}\le\dfrac{b^2+c^2+2a}4 and abc≤c2+a2+2b4\sqrt{abc}\le\dfrac{c^2+a^2+2b}4. Add the three: 3abc≤2(a2+b2+c2)+2(a+b+c)43\sqrt{abc}\le\dfrac{2(a^2+b^2+c^2)+2(a+b+c)}4. Divide by 3.
Why
In (i) apply AM–GM with x=abx=ab, y=cy=c, then use ab≤a2+b22ab\le\frac{a^2+b^2}2 (Lesson 5). In (ii) the left side is the same in every case, which is why adding the three symmetric versions works.

HSC practice · A condition and AM–GM

Slide 7
Adapted from 2024 HSC Q13(d)Let aa, bb, cc be positive real numbers with 1a+1b+1c=1\dfrac1a+\dfrac1b+\dfrac1c=1. Prove that abc+bca+cab≤abca\sqrt{bc}+b\sqrt{ca}+c\sqrt{ab}\le abc.
Show answerHide answer
Answer
AM–GM: abc≤a(b+c)2a\sqrt{bc}\le\dfrac{a(b+c)}2, bca≤b(c+a)2b\sqrt{ca}\le\dfrac{b(c+a)}2, cab≤c(a+b)2c\sqrt{ab}\le\dfrac{c(a+b)}2. Add: LHS≤2ab+2bc+2ca2=ab+bc+ca\text{LHS}\le\dfrac{2ab+2bc+2ca}2=ab+bc+ca. Multiplying the condition by abcabc gives bc+ca+ab=abcbc+ca+ab=abc. So LHS≤abc\text{LHS}\le abc.
Why
Multiply each AM–GM bound by the positive factor (aa, bb or cc). Note that bc≤b+c2\sqrt{bc}\le\frac{b+c}2, so abc≤a(b+c)2a\sqrt{bc}\le\frac{a(b+c)}2. The condition is used by clearing the fractions. (Wording adapted from the paper.)

The triangle inequality

Slide 8
Theorem
Triangle inequality∣a+b∣≤∣a∣+∣b∣(a,b real)|a+b|\le|a|+|b|\qquad(a,b\text{ real})

Geometrically: a side of a triangle is never longer than the sum of the other two. For real numbers, equality holds if and only if aa and bb have the same sign (or one is 0).

Tools: ∣x∣2=x2|x|^2=x^2; ab≤∣ab∣=∣a∣∣b∣ab\le|ab|=|a||b|; and for non-negative X,YX,Y: X2≥Y2⇒X≥YX^2\ge Y^2\Rightarrow X\ge Y (Lesson 5).

Example 3

Prove the triangle inequality by showing (∣a∣+∣b∣)2−∣a+b∣2≥0(|a|+|b|)^2-|a+b|^2\ge0.
Show answerHide answer
Answer
(∣a∣+∣b∣)2−∣a+b∣2=a2+2∣a∣∣b∣+b2−(a2+2ab+b2)=2(∣ab∣−ab)≥0(|a|+|b|)^2-|a+b|^2=a^2+2|a||b|+b^2-(a^2+2ab+b^2)=2(|ab|-ab)\ge0. Both ∣a∣+∣b∣|a|+|b| and ∣a+b∣|a+b| are non-negative, so ∣a+b∣≤∣a∣+∣b∣|a+b|\le|a|+|b|.
Why
Since ab≤∣ab∣ab\le|ab|, the bracket ∣ab∣−ab|ab|-ab is non-negative. Squares of non-negative numbers compare in the same order as the numbers themselves. The same inequality holds later for complex numbers and for vectors.

Example 4 · Using the triangle inequality

Slide 9
(a) Prove that ∣a+b+c∣≤∣a∣+∣b∣+∣c∣|a+b+c|\le|a|+|b|+|c|.
Show answerHide answer
Answer
∣a+b+c∣≤∣a+b∣+∣c∣≤∣a∣+∣b∣+∣c∣|a+b+c|\le|a+b|+|c|\le|a|+|b|+|c|.
Why
Apply the triangle inequality twice: first to (a+b)(a+b) and cc, then to aa and bb.
(b) Prove that ∣a−b∣≥∣a∣−∣b∣|a-b|\ge|a|-|b|. (Hint: write a=(a−b)+ba=(a-b)+b.)
Show answerHide answer
Answer
∣a∣=∣(a−b)+b∣≤∣a−b∣+∣b∣|a|=|(a-b)+b|\le|a-b|+|b|, so ∣a−b∣≥∣a∣−∣b∣|a-b|\ge|a|-|b|.
Why
Apply the triangle inequality to (a−b)(a-b) and bb, then subtract ∣b∣|b| from both sides.

Example 5 · Inequalities in geometry

Slide 10
(a) A rectangle has perimeter PP. Prove that its area is at most P216\dfrac{P^2}{16}. When is there equality?
Show answerHide answer
Answer
Let the sides be x,yx,y with 2(x+y)=P2(x+y)=P. By AM–GM, xy≤(x+y2)2=(P4)2=P216xy\le\left(\dfrac{x+y}2\right)^2=\left(\dfrac P4\right)^2=\dfrac{P^2}{16}. Equality holds when x=yx=y, that is a square.
Why
Use xy≤x+y2\sqrt{xy}\le\frac{x+y}2 and square both sides (both are non-negative). The semi-perimeter x+y=P2x+y=\frac P2 is fixed, so x+y2=P4\frac{x+y}2=\frac P4.
(b) A right-angled triangle has legs aa, bb and hypotenuse cc. Prove that a+b≤c2a+b\le c\sqrt2.
Show answerHide answer
Answer
(a+b)2=a2+b2+2ab≤2(a2+b2)=2c2(a+b)^2=a^2+b^2+2ab\le2(a^2+b^2)=2c^2. Since a+b>0a+b>0, a+b≤c2a+b\le c\sqrt2.
Why
Use 2ab≤a2+b22ab\le a^2+b^2 and Pythagoras, a2+b2=c2a^2+b^2=c^2. Taking the square root is valid because a+ba+b and c2c\sqrt2 are both positive.
(c) Two sides of a triangle are 5 cm and 9 cm. Find the possible lengths of the third side.
Show answerHide answer
Answer
4<x<144<x<14.
Why
The triangle inequality must hold for each side: x<5+9=14x<5+9=14, 9<5+x9<5+x gives x>4x>4 and 5<9+x5<9+x is always true. So 9−5<x<9+59-5<x<9+5.

Extension · AM–GM for four and three numbers

Slide 11
(i) Using AM–GM for two numbers twice, show that for positive reals a+b+c+d4≥abcd4\dfrac{a+b+c+d}4\ge\sqrt[4]{abcd}.
Show answerHide answer
Answer
a+b+c+d4=12[a+b2+c+d2]≥12[ab+cd]≥abcd=abcd4\dfrac{a+b+c+d}4=\dfrac12\left[\dfrac{a+b}2+\dfrac{c+d}2\right]\ge\dfrac12\left[\sqrt{ab}+\sqrt{cd}\right]\ge\sqrt{\sqrt{ab}\sqrt{cd}}=\sqrt[4]{abcd}.
Why
The first step uses AM–GM on each pair. The second step uses AM–GM again on the two numbers ab\sqrt{ab} and cd\sqrt{cd}.
(ii) By choosing d=a+b+c3d=\dfrac{a+b+c}3 in (i), deduce that a+b+c3≥abc3\dfrac{a+b+c}3\ge\sqrt[3]{abc}.
Show answerHide answer
Answer
Let m=a+b+c3m=\dfrac{a+b+c}3, so a+b+c=3ma+b+c=3m. Then (i) gives 3m+m4=m≥abcm4\dfrac{3m+m}4=m\ge\sqrt[4]{abcm}. So m4≥abcmm^4\ge abcm, hence m3≥abcm^3\ge abc (as m>0m>0), and m≥abc3m\ge\sqrt[3]{abc}.
Why
Putting d=md=m makes the left side of (i) equal to mm itself. Raise both sides to the fourth power and divide by m>0m>0.

Exit ticket

Slide 12
(a) Prove that x+4x≥4x+\dfrac4x\ge4 for x>0x>0.
Show answerHide answer
Answer
By AM–GM, x+4x≥2x⋅4x=24=4x+\dfrac4x\ge2\sqrt{x\cdot\dfrac4x}=2\sqrt4=4.
Why
The product of the two terms is the constant 4. Equality holds at x=2x=2.
(b) Two sides of a triangle are 6 and 10. How many integer lengths are possible for the third side?
Show answerHide answer
Answer
11.
Why
10−6<x<10+610-6<x<10+6, so 4<x<164<x<16. The integers 5 to 15 give 11 values.

Spot the flaw · AM–GM

Slide 13
Spot the flawA student wrote this. Find the flaw, explain why it matters, then write a correct version.
Claim. x+1x≥2x+\dfrac1x\ge2 for all real x≠0x\ne0.
1. By AM–GM, x+1x≥2x⋅1xx+\dfrac1x\ge2\sqrt{x\cdot\dfrac1x}
2. =21=2=2\sqrt1=2. □\square
Show answerHide answer
Answer
The flaw: AM–GM holds only for non-negative numbers, and xx may be negative. The claim is false: x=−1x=-1 gives x+1x=−2<2x+\dfrac1x=-2<2. Correct version: for x>0x>0, x+1x≥2x⋅1x=2x+\dfrac1x\ge2\sqrt{x\cdot\dfrac1x}=2, with equality at x=1x=1. (For x<0x<0, x+1x≤−2x+\dfrac1x\le-2.)
Why
For x<0x<0 the product x⋅1x=1x\cdot\frac1x=1 is still defined, which hides the problem, but the two terms are negative so AM–GM does not apply. Check the conditions of a theorem before using it, the same habit as dividing only by a positive quantity.

Which method? · Proofs so far

Slide 14

For each statement, name the method you would use and write the first line of the proof. Then prove any two.

(a) There are no integers m,nm,n with 6m+9n=1006m+9n=100.
Show answerHide answer
Answer
Contradiction (or a direct divisibility argument). First line: suppose 6m+9n=1006m+9n=100 for some integers m,nm,n.
Why
6m+9n=3(2m+3n)6m+9n=3(2m+3n) is divisible by 3, but 100 is not, which is a contradiction. The words "there are no" point to contradiction.
(b) For integers nn, if n2+2nn^2+2n is odd, then nn is odd.
Show answerHide answer
Answer
Contrapositive. First line: we prove that if nn is even, then n2+2nn^2+2n is even.
Why
Let n=2kn=2k. Then n2+2n=4k2+4k=2(2k2+2k)n^2+2n=4k^2+4k=2(2k^2+2k), which is even. "If ... odd, then ... odd" points to the contrapositive.
(c) a+b2≥2aba+b\dfrac{a+b}2\ge\dfrac{2ab}{a+b} for a,b>0a,b>0.
Show answerHide answer
Answer
Difference and squares. First line: LHS−RHS=(a+b)2−4ab2(a+b)\text{LHS}-\text{RHS}=\dfrac{(a+b)^2-4ab}{2(a+b)}.
Why
This equals (a−b)22(a+b)≥0\dfrac{(a-b)^2}{2(a+b)}\ge0, since the numerator is a square and a+b>0a+b>0.
(d) n2−nn^2-n is even for every integer nn.
Show answerHide answer
Answer
Direct proof (or cases). First line: n2−n=n(n−1)n^2-n=n(n-1).
Why
This is a product of consecutive integers, so one factor is even and the product is even. Alternatively use cases n=2kn=2k and n=2k+1n=2k+1.
(e) 5\sqrt5 is irrational.
Show answerHide answer
Answer
Contradiction. First line: suppose 5=pq\sqrt5=\dfrac pq with p,qp,q integers having no common factor.
Why
Then p2=5q2p^2=5q^2, so 5∣p25\mid p^2, which gives 5∣p5\mid p (lemma: prove it by cases p=5k+rp=5k+r with r=1,2,3,4r=1,2,3,4, or by prime factorisation). Writing p=5kp=5k gives 5∣q5\mid q, contradicting no common factor.
(f) x+16x≥8x+\dfrac{16}x\ge8 for x>0x>0.
Show answerHide answer
Answer
AM–GM. First line: x+16x≥2x⋅16xx+\dfrac{16}x\ge2\sqrt{x\cdot\dfrac{16}x}.
Why
This equals 216=82\sqrt{16}=8, with equality at x=4x=4. A sum of positive terms with a fixed product points to AM–GM.
Discussion
What in each statement points to the method? "No integers" suggests contradiction; "if ... odd, then ... odd" suggests the contrapositive; a sum with a fixed product and x>0x>0 suggests AM–GM. The HSC does not tell you which method to use.