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.
Start with the warm-up and record your first reasoning.
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.
Orient yourself to the lesson goals, language, and core approach.
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 $a = b$.
Work through the concept and complete its check.
The whole two-variable AM–GM proof is a single algebraic identity squeezed into three lines:
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.
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?
Build the next proof move and complete its check.
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 $a, b, c \geq 0$:
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
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 $a_1, \ldots, a_n$:
with equality iff $a_1 = a_2 = \cdots = a_n$.
$n = 3$: $(a+b+c)/3 \geq \sqrt[3]{abc}$; 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 $a = b = c$.
Follow the worked examples, then complete the check.
Worked examples · 3 in a row, reveal as you go
(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.
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.
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.
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 $x + \dfrac{1}{x}$ is at least , with equality when $x = 1$.
Correct the common traps and test your understanding.
Misconceptions to fix · the 3 traps that cost marks
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 $x = 3$.
Complete the practice activities and revisit your opening thinking.
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?
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.
Multiple choice, then short answer under exam conditions.
Pick your answer, then rate your confidence. That tells the system what to drill next. Each retry pulls a fresh mix from the bank.
Q1. Prove that for $x > 0$, $x + \dfrac{1}{x} \geq 2$, stating when equality holds. (2 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)
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)
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].
Sit the module quiz, then close the lesson off.
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