Your weak spots

Insights load after your first practice round.

Module 11 · L04 of 10 ~45 min ⚡ +90 XP available

The AM–GM Inequality

A rectangle with perimeter $4$ has maximum area when it is a $1 \times 1$ square. A rectangular paddock split into two equal pens has its largest area when both pens are square. These optimisation facts are not coincidences — they are the AM–GM inequality in disguise. Today you prove $(a+b)/2 \geq \sqrt{ab}$ from the single move $(\sqrt{a} - \sqrt{b})^2 \geq 0$, then deploy it in two- and three-variable optimisations.

Today's hook — A positive real $x$ has $x + \dfrac{1}{x}$ as a famous expression. What is its minimum value, and at which $x$ is the minimum attained? Try a few values: $x = 1, 2, \tfrac{1}{2}, 10$. Compare after card 05.

0/5QUESTS

Recall — your gut answer first

+5 XP warm-up

Pick three pairs of positive numbers $(a, b)$ — for instance $(4, 16)$, $(9, 9)$, $(1, 100)$. Compute $(a+b)/2$ and $\sqrt{ab}$ for each, and write down which is larger. Make a guess at when (if ever) the two are equal.

auto-saved

The two moves for AM–GM proofs

+5 XP to read

Almost every AM–GM proof boils down to one of two ideas: a perfect square is non-negative, or apply the two-variable case twice (or use induction). The two-variable proof is the master move — once you own it, the three- and $n$-variable forms follow.

The square-it strategy: (1) write $(\sqrt{a} - \sqrt{b})^2 \geq 0$; (2) expand to $a - 2\sqrt{ab} + b \geq 0$; (3) rearrange to $(a+b)/2 \geq \sqrt{ab}$. Equality occurs iff $\sqrt{a} = \sqrt{b}$, i.e., $a = b$.

Statement (n = 2): for $a, b \geq 0$: $\dfrac{a+b}{2} \geq \sqrt{ab}$, equality iff $a = b$.

$\dfrac{a+b}{2} \geq \sqrt{ab}$

Both variables must be $\geq 0$

AM–GM in the form $(a+b)/2 \geq \sqrt{ab}$ requires $a, b \geq 0$ so that $\sqrt{ab}$ is real and the square-root manipulations are valid. If a problem gives $a, b > 0$ strictly, the inequality may still be strict (unless $a = b$).

Equality iff $a = b$

A $(\sqrt{a} - \sqrt{b})^2 = 0$ argument forces $\sqrt{a} = \sqrt{b}$, hence $a = b$. State this in every proof — the equality condition usually carries one of the marks.

Use it for optimisation

If you can write a target expression as a sum of two non-negative terms whose product is constant, AM–GM nails the minimum. Conversely, fixed sum gives a maximum product. This is the standard route around calculus for HSC optimisation.

What you'll master

Know

Key facts

Two-variable AM–GM: $(a+b)/2 \geq \sqrt{ab}$ for $a, b \geq 0$
Three-variable form: $(a+b+c)/3 \geq \sqrt[3]{abc}$ for $a, b, c \geq 0$
General $n$-variable form: $\dfrac{a_1 + \cdots + a_n}{n} \geq \sqrt[n]{a_1 \cdots a_n}$, equality iff all equal

Understand

Concepts

Why $(\sqrt{a} - \sqrt{b})^2 \geq 0$ is the single algebraic move that drives the two-variable proof
Why equality requires $a = b$ (and, for $n$ variables, all equal)
How AM–GM gives sharp bounds for sums of positive reciprocals

Can do

Skills

Prove the two-variable AM–GM from first principles and state the equality condition
Apply two- and three-variable AM–GM in optimisation problems
Identify when AM–GM is the right tool (sum vs. product structure)

Key terms

Arithmetic mean (AM)$\dfrac{a_1 + a_2 + \cdots + a_n}{n}$ — the "average" of the $n$ numbers.

Geometric mean (GM)$\sqrt[n]{a_1 a_2 \cdots a_n}$ — the $n$-th root of the product. Defined for non-negative reals; equals zero whenever any $a_i = 0$.

AM–GM (n = 2)For $a, b \geq 0$: $\dfrac{a+b}{2} \geq \sqrt{ab}$. Equality iff $a = b$.

AM–GM (n = 3)For $a, b, c \geq 0$: $\dfrac{a+b+c}{3} \geq \sqrt[3]{abc}$. Equality iff $a = b = c$.

Equality conditionIn every form of AM–GM, equality holds if and only if all the variables are equal. State this explicitly whenever the inequality is used in a proof.

NESA MEX-P1NESA outcome: proves results using a variety of techniques and considers the validity of an argument — specifically prove and use the AM–GM inequality in two and three variables.

The two-variable proof in three lines

core concept

The whole two-variable AM–GM proof is a single algebraic identity squeezed into three lines:

Start. For $a, b \geq 0$ the quantity $(\sqrt{a} - \sqrt{b})^2$ is a square, hence non-negative.
Expand. $(\sqrt{a} - \sqrt{b})^2 = a - 2\sqrt{ab} + b \geq 0$.
Rearrange. $a + b \geq 2\sqrt{ab}$, i.e. $\dfrac{a+b}{2} \geq \sqrt{ab}$. Equality iff $\sqrt{a} = \sqrt{b}$, i.e. $a = b$.

Worked through the hook: For $x > 0$ set $a = x, b = \dfrac{1}{x}$. Then $\dfrac{x + 1/x}{2} \geq \sqrt{x \cdot \tfrac{1}{x}} = 1$, so $x + \dfrac{1}{x} \geq 2$. Equality holds iff $x = \dfrac{1}{x}$, i.e. $x = 1$. Therefore the minimum of $x + 1/x$ over $x > 0$ is $2$, attained at $x = 1$ — exactly what the table $x = 1, 2, \tfrac{1}{2}, 10$ suggests.

Why a perfect square is the engine. Every square of a real number is $\geq 0$. By rearranging a square so that one side is the target inequality, you trade an "inequality to prove" for an "expression to recognise as a square" — a much easier task. Spotting which square to write is the only creative step.

AM–GM: $(a+b)/2 \geq \sqrt{ab}$ for $a, b \geq 0$; equality iff $a = b$ · Proof: $(\sqrt{a}-\sqrt{b})^2 \geq 0$ then expand and rearrange · Hook: $x + 1/x \geq 2$ for $x > 0$, equality at $x = 1$ · State the equality condition explicitly — usually one mark

Pause — copy the two-variable AM–GM $(a+b)/2 \geq \sqrt{ab}$, its three-line proof from $(\sqrt{a}-\sqrt{b})^2 \geq 0$, and the equality condition $a = b$ into your book.

Quick check: Which expansion is used to prove the two-variable AM–GM inequality?

Three variables and the general statement

core concept

We just saw that AM–GM for two variables follows in three lines from $(\sqrt{a} - \sqrt{b})^2 \geq 0$, with equality iff $a = b$. That raises a question: does the same idea extend to three variables? This card answers it → the three-variable AM–GM $(a+b+c)/3 \geq \sqrt[3]{abc}$, proved via the identity $x^2+y^2+z^2 - xy - yz - zx = \tfrac{1}{2}\sum(x-y)^2 \geq 0$.

The three-variable form is the next step up. For $a, b, c \geq 0$:

$$\dfrac{a + b + c}{3} \;\geq\; \sqrt[3]{abc}$$

Equality holds if and only if $a = b = c$. One classical proof for $n = 3$ uses the substitution $a = x^3, b = y^3, c = z^3$ and the identity

$$x^3 + y^3 + z^3 - 3xyz \;=\; (x + y + z)\bigl(x^2 + y^2 + z^2 - xy - yz - zx\bigr).$$

The factor $x^2 + y^2 + z^2 - xy - yz - zx = \tfrac{1}{2}\bigl[(x-y)^2 + (y-z)^2 + (z-x)^2\bigr] \geq 0$, and $x + y + z \geq 0$ for non-negative $x, y, z$. Hence $x^3 + y^3 + z^3 \geq 3xyz$, i.e. $\dfrac{a + b + c}{3} \geq \sqrt[3]{abc}$. Equality requires $x = y = z$, equivalently $a = b = c$.

General statement (proof not required in Ext 2 beyond $n = 2, 3$). For non-negative reals $a_1, \ldots, a_n$:

$$\dfrac{a_1 + a_2 + \cdots + a_n}{n} \;\geq\; \sqrt[n]{a_1 a_2 \cdots a_n}$$

with equality iff $a_1 = a_2 = \cdots = a_n$.

When to reach for AM–GM. If you can pair a sum (which you want to minimise) with a product (which is constant), or pair a product (which you want to maximise) with a sum (which is constant), AM–GM gives the bound and the equality case tells you where the optimum is attained.

$n = 3$: $(a+b+c)/3 \geq \sqrt[3]{abc}$; equality iff $a = b = c$ · Identity $x^3+y^3+z^3 - 3xyz = (x+y+z)(x^2+y^2+z^2-xy-yz-zx)$ · $x^2+y^2+z^2-xy-yz-zx = \tfrac{1}{2}[(x-y)^2+(y-z)^2+(z-x)^2] \geq 0$ · General: $(a_1+\cdots+a_n)/n \geq \sqrt[n]{a_1\cdots a_n}$, equality iff all equal

Pause — copy the three-variable AM–GM, the identity $x^2+y^2+z^2-xy-yz-zx = \tfrac{1}{2}[(x-y)^2+(y-z)^2+(z-x)^2]$, and the equality condition $a=b=c$ into your book.

Did you get this? True or false: for all non-negative real $a, b, c$, equality holds in $\dfrac{a+b+c}{3} \geq \sqrt[3]{abc}$ if and only if $a = b = c$.

Worked examples · 3 in a row, reveal as you go

PROBLEM 1 · PROVE THE TWO-VARIABLE AM-GM AND APPLY

(a) Prove that for all $a, b \geq 0$: $\dfrac{a+b}{2} \geq \sqrt{ab}$. (b) Hence show $x + \dfrac{4}{x} \geq 4$ for all $x > 0$, stating when equality holds.

For $a, b \geq 0$ both $\sqrt{a}$ and $\sqrt{b}$ are real. Since any square of a real number is non-negative:
$(\sqrt{a} - \sqrt{b})^2 \geq 0$.
Expand: $a - 2\sqrt{ab} + b \geq 0$, hence $a + b \geq 2\sqrt{ab}$.

The square move is the entire proof. Always state explicitly that $a, b \geq 0$ so that $\sqrt{a}, \sqrt{b}$ are real.

PROBLEM 2 · THREE-VARIABLE AM-GM IN ACTION

Let $a, b, c > 0$ with $abc = 8$. Use AM–GM to find the minimum value of $a + b + c$ and state when it is attained.

Apply the three-variable AM–GM to $a, b, c$:
$\dfrac{a + b + c}{3} \geq \sqrt[3]{abc}$.

Three positive variables, sum to minimise, product fixed — the textbook AM–GM set-up. Cite the inequality and its assumptions ($a, b, c \geq 0$, all $> 0$ here).

PROBLEM 3 · OPTIMISATION VIA AM-GM

A rectangle is inscribed in a fixed length of fencing of $20$ m. Use AM–GM to show that the maximum area is $25$ m$^2$, attained when the rectangle is a square.

Let the side lengths be $a, b > 0$. Perimeter $2a + 2b = 20$, so $a + b = 10$. Area $A = ab$.

Translate the geometry into a sum constraint and a product target. AM–GM links sum to product, so the structure matches.

Fill the gap: Setting $a = x$ and $b = \dfrac{1}{x}$ for $x > 0$ in two-variable AM–GM yields $\dfrac{x + 1/x}{2} \geq \sqrt{x \cdot 1/x} = 1$, so $x + \dfrac{1}{x} \geq $ , with equality when $x = 1$.

Misconceptions to fix · the 3 traps that cost marks

Trap 01

Forgetting the non-negativity hypothesis

AM–GM in this form requires $a, b \geq 0$ (or $a_i \geq 0$ in the general case). For $a = -1, b = 4$: AM $= 1.5$ and $\sqrt{ab} = \sqrt{-4}$ — not real. Always state the hypothesis at the start of any AM–GM proof or application.

Trap 02

Omitting the equality condition

A complete AM–GM answer states the equality condition: "with equality iff $a = b$" (or $a = b = c$, or all $a_i$ equal). In optimisation, the equality case identifies the optimum — leaving it out means you have only shown a bound, not that the bound is attained.

Trap 03

Picking the wrong split of variables

In optimisation, AM–GM only delivers a tight bound when the product (or sum) of the chosen pieces is a constant. For example, to bound $x + \dfrac{4}{x}$ split as $a = x, b = 4/x$ so that $ab = 4$ is fixed. Splitting as $a = x + 1, b = 4/x - 1$ destroys the constancy and the bound is not tight.

Did you get this? True or false: applying two-variable AM–GM to $a = x, b = \dfrac{9}{x}$ for $x > 0$ shows that $x + \dfrac{9}{x} \geq 6$, with equality at $x = 3$.

Work mode · how are you completing this lesson?

Activities · practice with the ideas

Prove that for all $a, b \geq 0$: $\dfrac{a+b}{2} \geq \sqrt{ab}$ using $(\sqrt{a} - \sqrt{b})^2 \geq 0$. State when equality holds.

For $x > 0$, find the minimum value of $f(x) = 9x + \dfrac{1}{x}$ using AM–GM. State the value of $x$ at the minimum.

Use three-variable AM–GM to prove that for all $a, b, c > 0$: $\dfrac{a}{b} + \dfrac{b}{c} + \dfrac{c}{a} \geq 3$. State the equality condition.

A rectangular box (no lid) has volume $32$ m$^3$ and a square base. Use AM–GM (n = 3) to show the minimum surface area is $48$ m$^2$.

For $a, b \geq 0$, prove $a^2 + b^2 \geq 2ab$ in two ways: (i) directly from $(a-b)^2 \geq 0$; (ii) by applying AM–GM to $a^2$ and $b^2$.

Odd one out: Three of these statements are correct consequences of AM–GM (for positive variables). Which one is NOT?

Revisit your thinking

Earlier you tested $(a+b)/2$ vs. $\sqrt{ab}$ on three pairs and conjectured when the two are equal.

The data points $(4, 16)$, $(9, 9)$, $(1, 100)$ all line up with the same rule: AM $\geq$ GM, equality only when the two numbers coincide. The proof is a single line — a perfect square — but the consequences are wide: $x + 1/x \geq 2$, isoperimetric rectangles are squares, fixed-product triples minimise their sum when equal. AM–GM is the algebraic skeleton of countless HSC optimisation problems.

auto-saved

Multiple choice

+5 XP per correct · +25 XP all-correct

Pick your answer, then rate your confidence — that tells the system what to drill next. Each retry pulls a fresh mix from the bank.

Short answer

ApplyBand 32 marks

Q1. Prove that for $x > 0$, $x + \dfrac{1}{x} \geq 2$, stating when equality holds. (2 marks)

auto-saved

ApplyBand 43 marks

Q2. Prove from the definition that for $a, b \geq 0$: $\dfrac{a + b}{2} \geq \sqrt{ab}$, stating the condition for equality. (3 marks)

auto-saved

AnalyseBand 54 marks

Q3. Suppose $a, b, c > 0$ with $a + b + c = 6$. Using three-variable AM–GM, find the maximum value of $abc$ and the values of $a, b, c$ at which it is attained. (4 marks)

auto-saved

Comprehensive answers (click to reveal)

Activity answers:

1. $(\sqrt{a} - \sqrt{b})^2 \geq 0$ expands to $a - 2\sqrt{ab} + b \geq 0$, i.e. $a + b \geq 2\sqrt{ab}$, i.e. $\tfrac{a+b}{2} \geq \sqrt{ab}$. Equality iff $\sqrt{a} = \sqrt{b}$, i.e. $a = b$.

2. AM–GM with $a = 9x, b = 1/x$: $\tfrac{9x + 1/x}{2} \geq \sqrt{9x \cdot \tfrac{1}{x}} = 3$, so $9x + \tfrac{1}{x} \geq 6$. Equality at $9x = 1/x$, i.e. $x^2 = 1/9$, i.e. $x = 1/3$. Minimum $= 6$ at $x = 1/3$.

3. Apply 3-var AM–GM to $\tfrac{a}{b}, \tfrac{b}{c}, \tfrac{c}{a}$: $\tfrac{1}{3}\bigl(\tfrac{a}{b}+\tfrac{b}{c}+\tfrac{c}{a}\bigr) \geq \sqrt[3]{\tfrac{a}{b}\cdot\tfrac{b}{c}\cdot\tfrac{c}{a}} = 1$. Hence $\tfrac{a}{b}+\tfrac{b}{c}+\tfrac{c}{a} \geq 3$. Equality iff $\tfrac{a}{b} = \tfrac{b}{c} = \tfrac{c}{a}$, i.e. $a = b = c$.

4. Volume: $x^2 h = 32$, so $h = 32/x^2$. Surface area (no lid): $S = x^2 + 4xh = x^2 + 4x \cdot \tfrac{32}{x^2} = x^2 + \tfrac{128}{x}$. Write $S = x^2 + \tfrac{64}{x} + \tfrac{64}{x}$ — three positive terms with constant product $x^2 \cdot \tfrac{64}{x} \cdot \tfrac{64}{x} = 64^2 = 4096$. By 3-var AM–GM, $\tfrac{S}{3} \geq \sqrt[3]{4096} = 16$, so $S \geq 48$. Equality requires $x^2 = \tfrac{64}{x}$, i.e. $x^3 = 64$, i.e. $x = 4$, and $h = 32/16 = 2$. Minimum $S = 48$ m$^2$.

5. (i) $(a-b)^2 \geq 0 \Rightarrow a^2 - 2ab + b^2 \geq 0 \Rightarrow a^2 + b^2 \geq 2ab$. (ii) AM–GM on $a^2, b^2$: $\tfrac{a^2+b^2}{2} \geq \sqrt{a^2 b^2} = |ab| \geq ab$, so $a^2 + b^2 \geq 2ab$.

Q1 (2 marks): AM–GM with $a = x > 0, b = 1/x > 0$: $\tfrac{x + 1/x}{2} \geq \sqrt{1} = 1$ [1]. Hence $x + 1/x \geq 2$; equality iff $x = 1/x$, i.e. $x = 1$ [1].

Q2 (3 marks): Since $a, b \geq 0$, $\sqrt{a}, \sqrt{b}$ are real and $(\sqrt{a}-\sqrt{b})^2 \geq 0$ [1]. Expand: $a - 2\sqrt{ab} + b \geq 0$, i.e. $a + b \geq 2\sqrt{ab}$ [1]. Divide by 2: $\tfrac{a+b}{2} \geq \sqrt{ab}$. Equality iff $\sqrt{a} = \sqrt{b}$, i.e. $a = b$ [1].

Q3 (4 marks): 3-var AM–GM gives $\tfrac{a+b+c}{3} \geq \sqrt[3]{abc}$ [1]; substitute $a+b+c = 6$ to get $2 \geq \sqrt[3]{abc}$ [1]; cube (both sides $\geq 0$): $abc \leq 8$ [1]. Equality requires $a = b = c$; with sum $6$, $a = b = c = 2$, and $abc = 8$. Max value $= 8$ at $a = b = c = 2$ [1].

Boss battle · The Mean Hunter

earn bronze · silver · gold

Five timed questions on AM–GM — two-variable proofs, three-variable applications, and optimisation. Beat the boss to bank a tier — gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.

⚔ Enter the arena

Science Jump · platform challenge

Climb platforms by answering AM–GM and inequality questions. Lighter alternative to the boss.

Mark lesson as complete

Tick when you've finished the practice and review.

← Lesson 03 · Triangle Inequality Lesson 05 →

Module overview · Extension 2