Inequalities with Positive Real Numbers
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.
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?
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 $a = b$
Key facts
- $(x - y)^2 \geq 0$ for all real $x, y$, with equality iff $x = y$
- $a^2 + b^2 \geq 2ab$ for all real $a, b$
- For positive reals: $a + b \geq 2\sqrt{ab}$ (AM-GM) and $\dfrac{a}{b} + \dfrac{b}{a} \geq 2$
Concepts
- Why every algebraic inequality reduces to a sum of squares
- How symmetric and cyclic inequalities exploit pairwise $(a-b)^2$ terms
- Why the equality case identifies the tight configuration
Skills
- Prove two- and three-variable inequalities using $(x-y)^2 \geq 0$
- Use sum-of-squares decomposition for cyclic and symmetric expressions
- Identify and state the equality case explicitly
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:
- Identify the gap. Compute LHS $-$ RHS and try to write it as a sum of squares.
- Choose the square. $(a-b)^2$, $(a-b)^2 + (b-c)^2 + (c-a)^2$, or $(\sqrt a - \sqrt b)^2$ are the usual suspects.
- Run the proof forwards. Start from "$(\ldots)^2 \geq 0$" and arrive at the target inequality.
- State equality. Equality holds when each square equals zero, i.e., $a = b$ (or $a = b = c$).
Worked through the hook: Prove $\dfrac{a}{b} + \dfrac{b}{a} \geq 2$ for $a, b > 0$.
- Start: $(a - b)^2 \geq 0$.
- Expand: $a^2 - 2ab + b^2 \geq 0 \;\Rightarrow\; a^2 + b^2 \geq 2ab$.
- Divide by $ab > 0$ (sign preserved): $\dfrac{a}{b} + \dfrac{b}{a} \geq 2$. $\blacksquare$
- Equality iff $a = b$.
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 $a, b$?
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:
- $a^2 + b^2 + c^2 \geq ab + bc + ca$ for all real $a, b, c$.
- Equality iff $a = b = c$ (all three squares vanish simultaneously).
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).
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 .
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).
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.
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)
Comprehensive answers (click to reveal)
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].
Five timed inequality proofs ranging from two-variable AM-GM to three-variable cyclic and symmetric forms. Beat the boss to bank a tier — gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.
⚔ Enter the arenaClimb platforms by answering inequality questions. Lighter alternative to the boss.
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