Why is the arithmetic mean always at least as large as the geometric mean? Why does $a^2 + b^2 \geq 2ab$ for any reals? Every algebraic inequality on the HSC paper rests on a single, almost embarrassing observation: a real number squared is never negative. In this lesson you will weaponise that fact, $(x-y)^2 \geq 0$, into a complete toolkit for proving multi-variable inequalities, including symmetric and cyclic forms.
Start with the warm-up and record your first reasoning.
For any real number $x$, what can you say about $x^2$? Before reading on write a one-line proof that $a^2 + b^2 \geq 2ab$ for all real $a, b$, using only that observation. When does equality hold?
Orient yourself to the lesson goals, language, and core approach.
Every clean inequality proof rests on two moves: start from something non-negative (usually $(x-y)^2 \geq 0$), and rearrange forwards until you reach the inequality you want. Working backwards from the target is a common scaffolding tool, but the polished proof must run forwards from a known-true statement.
The sum-of-squares strategy: (1) identify a square (or sum of squares) that captures the gap between the two sides, (2) expand it to expose the difference, (3) write a clean forward proof. Always state the equality case.
Start point: $(a-b)^2 \geq 0$ · Key identity: $a^2 + b^2 \geq 2ab$ · Equality iff $a = b$
Work through the concept and complete its check.
Every elementary algebraic inequality on the HSC paper can be traced back to the single fact that the square of a real number is non-negative. The four-step strategy:
Worked through the hook: Prove $\dfrac{a}{b} + \dfrac{b}{a} \geq 2$ for $a, b > 0$.
Master fact: $(x - y)^2 \geq 0$ for all real $x, y$; equality iff $x = y$ · Derived: $a^2 + b^2 \geq 2ab$ for all real $a, b$ · For $a, b > 0$: $\dfrac{a}{b} + \dfrac{b}{a} \geq 2$ (divide previous by $ab$) · Always state the equality case in your final line
Pause, copy the master fact $(x-y)^2 \geq 0$, the derived inequality $a^2+b^2 \geq 2ab$, and the ratio corollary $a/b + b/a \geq 2$ (for $a,b>0$) into your book.
Quick check: Which starting line gives the cleanest proof that $a^2 + b^2 \geq 2ab$ for all real $a, b$?
Build the next proof move and complete its check.
We just saw that $(x-y)^2 \geq 0$ immediately gives $a^2 + b^2 \geq 2ab$ for all real $a, b$, with equality iff $a = b$. That raises a question: how do we bound $a^2 + b^2 + c^2$ from below for three variables? This card answers it → summing pairwise squares $(a-b)^2 + (b-c)^2 + (c-a)^2 \geq 0$ gives $a^2+b^2+c^2 \geq ab+bc+ca$.
For three (or more) variables the trick is to add up the pairwise squares. The key identity:
The LHS is a sum of squares, hence $\geq 0$. Dividing by 2 gives the famous symmetric inequality:
This single identity unlocks dozens of three-variable problems. For example, dividing both sides by $abc$ when $a, b, c > 0$ gives $\dfrac{a}{bc} + \dfrac{b}{ca} + \dfrac{c}{ab} \geq \dfrac{1}{a} + \dfrac{1}{b} + \dfrac{1}{c}$.
Master identity: $(a-b)^2 + (b-c)^2 + (c-a)^2 = 2(a^2+b^2+c^2) - 2(ab+bc+ca) \geq 0$ · Therefore $a^2 + b^2 + c^2 \geq ab + bc + ca$ · Equality iff $a = b = c$ · Pairwise differences, never the sum $(a+b+c)^2$
Pause, copy the pairwise-square identity $(a-b)^2+(b-c)^2+(c-a)^2 = 2(a^2+b^2+c^2)-2(ab+bc+ca)$ and the resulting inequality $a^2+b^2+c^2 \geq ab+bc+ca$ into your book.
Did you get this? True or false: the inequality $a^2 + b^2 + c^2 \geq ab + bc + ca$ holds for all real numbers $a, b, c$ (not just positive ones).
Follow the worked examples, then complete the check.
Worked examples · 3 in a row, reveal as you go
Prove that for all positive real numbers $a$ and $b$, $\;a + b \geq 2\sqrt{ab}$, and state when equality holds.
Prove that for all positive real numbers $a, b, c$, $\;(a + b)(b + c)(c + a) \geq 8abc$, and state the equality case.
Prove that for all positive real numbers $a, b, c$, $\;\dfrac{a}{b} + \dfrac{b}{c} + \dfrac{c}{a} \geq 3$, and state when equality holds.
Fill the gap: The pairwise-squares identity $(a-b)^2 + (b-c)^2 + (c-a)^2 = 2(a^2 + b^2 + c^2) - 2(ab + bc + ca)$ implies $a^2 + b^2 + c^2 \geq ab + bc + ca$, with equality iff $a = b = c$. The minimum value of $(a-b)^2 + (b-c)^2 + (c-a)^2$ over all real $a, b, c$ is .
Correct the common traps and test your understanding.
Misconceptions to fix · the 3 traps that cost marks
Did you get this? True or false: the inequality $\dfrac{a}{b} + \dfrac{b}{a} \geq 2$ holds for all real $a, b$ with $a, b \neq 0$ (positive or negative).
Complete the practice activities and revisit your opening thinking.
Activities · practice with the ideas
Prove that $a^2 + b^2 \geq 2ab$ for all real $a, b$, and state the equality case.
For $a, b > 0$, prove $\dfrac{a^2 + b^2}{2} \geq ab$ and deduce $\dfrac{a^2 + b^2}{a + b} \geq \dfrac{a + b}{2}$.
Prove that for all real $a, b, c$: $a^2 + b^2 + c^2 \geq ab + bc + ca$. Use the pairwise-squares identity.
For positive reals $a, b$, prove $\dfrac{1}{a} + \dfrac{1}{b} \geq \dfrac{4}{a + b}$.
For positive reals $a, b, c$, prove $a + b + c \geq \sqrt{ab} + \sqrt{bc} + \sqrt{ca}$. State the equality case.
Odd one out: Three of these statements about inequality proofs are correct. Which one is NOT?
Earlier you tried to prove $\dfrac{a}{b} + \dfrac{b}{a} \geq 2$ for $a, b > 0$.
The clean proof begins at $(a-b)^2 \geq 0$, expands to $a^2 + b^2 \geq 2ab$, and divides by $ab > 0$ to land on the target. Equality holds iff $a = b$. The crucial sign assumption is $a, b > 0$, without it the division step would flip the inequality.
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 all real $a, b$: $a^2 + b^2 \geq 2ab$, and state the equality case. (2 marks)
Q2. For positive real numbers $a$ and $b$, prove $\dfrac{1}{a} + \dfrac{1}{b} \geq \dfrac{4}{a + b}$. State when equality holds. (3 marks)
Q3. For all positive real numbers $a, b, c$, prove that $(a + b + c)\!\left(\dfrac{1}{a} + \dfrac{1}{b} + \dfrac{1}{c}\right) \geq 9$. (3 marks)
Activity answers:
1. $(a-b)^2 \geq 0 \Rightarrow a^2 - 2ab + b^2 \geq 0 \Rightarrow a^2 + b^2 \geq 2ab$. Equality iff $a = b$.
2. From $a^2 + b^2 \geq 2ab$, divide by 2: $\frac{a^2+b^2}{2} \geq ab$. For the deduction: $\frac{a^2+b^2}{a+b} \geq \frac{2ab}{a+b}$. Use $a+b \geq 2\sqrt{ab}$, so $\frac{2ab}{a+b} \leq \frac{2ab}{2\sqrt{ab}} = \sqrt{ab} \leq \frac{a+b}{2}$, chain not direct; cleaner: cross-multiply $\frac{a^2+b^2}{a+b} \geq \frac{a+b}{2} \Leftrightarrow 2(a^2+b^2) \geq (a+b)^2 = a^2 + 2ab + b^2 \Leftrightarrow a^2+b^2 \geq 2ab$ ✓.
3. $(a-b)^2 + (b-c)^2 + (c-a)^2 = 2(a^2+b^2+c^2) - 2(ab+bc+ca) \geq 0$, so $a^2 + b^2 + c^2 \geq ab + bc + ca$. Equality iff $a = b = c$.
4. $\frac{1}{a} + \frac{1}{b} = \frac{a+b}{ab}$. Want $\frac{a+b}{ab} \geq \frac{4}{a+b}$, i.e., $(a+b)^2 \geq 4ab$, i.e., $a^2 - 2ab + b^2 \geq 0$, i.e., $(a-b)^2 \geq 0$ ✓. Equality iff $a = b$.
5. Adding three pairwise AM-GMs: $(a+b) + (b+c) + (c+a) \geq 2\sqrt{ab} + 2\sqrt{bc} + 2\sqrt{ca}$, so $2(a+b+c) \geq 2(\sqrt{ab}+\sqrt{bc}+\sqrt{ca})$, hence $a+b+c \geq \sqrt{ab}+\sqrt{bc}+\sqrt{ca}$. Equality iff $a = b = c$.
Q1 (2 marks): $(a-b)^2 \geq 0$ [1]. Expand $a^2 - 2ab + b^2 \geq 0$, so $a^2 + b^2 \geq 2ab$. Equality iff $a = b$ [1].
Q2 (3 marks): $\frac{1}{a} + \frac{1}{b} = \frac{a+b}{ab}$ [1]. Multiply $\frac{1}{a}+\frac{1}{b} \geq \frac{4}{a+b}$ by $ab(a+b) > 0$: equivalent to $(a+b)^2 \geq 4ab$ [1]. This is $(a-b)^2 \geq 0$ ✓. Equality iff $a = b$ [1].
Q3 (3 marks): Expand: $(a+b+c)(1/a + 1/b + 1/c) = 3 + \frac{a}{b}+\frac{b}{a} + \frac{b}{c}+\frac{c}{b} + \frac{a}{c}+\frac{c}{a}$ [1]. Apply $x + 1/x \geq 2$ for each positive pair: each bracket $\geq 2$ [1]. Total $\geq 3 + 2 + 2 + 2 = 9$. Equality iff $a = b = c$ [1].
Sit the module quiz, then close the lesson off.
A full module quiz covering every lesson in this module, not just this one. Set aside a decent block of time and treat it like a real assessment.
Start the module quiz →Tick when you've finished the practice and review.
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