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Module 11 · L06 of 12 ~40 min ⚡ +90 XP available

Cauchy–Schwarz Inequality

Two vectors and a deceptively short identity unlock a master inequality that powers everything from optics to quantum mechanics. $(a_1 b_1 + a_2 b_2)^2 \leq (a_1^2 + a_2^2)(b_1^2 + b_2^2)$ — at first sight it looks random; you will prove it with one Lagrange identity, then deploy it across two- and three-variable HSC problems where direct expansion would take pages.

Today's hook — Before reading on, write down (no proof yet) what you think the largest possible value of $a_1 b_1 + a_2 b_2$ could be, given $a_1^2 + a_2^2 = 1$ and $b_1^2 + b_2^2 = 1$. Compare with the Cauchy–Schwarz bound after card 05.

0/5QUESTS

Recall — your gut answer first

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You know that $(a_1 b_2 - a_2 b_1)^2 \geq 0$ for all real $a_1, a_2, b_1, b_2$. Before reading on — expand this square fully and write what you notice. Which familiar quantities appear?

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The two moves for Cauchy–Schwarz problems

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Cauchy–Schwarz problems reward two habits: recognise the shape — anything that looks like a sum of products squared, bounded by a product of sums of squares — and choose your $a_i, b_i$ deliberately. The right choice turns a horrible algebra question into a one-line application.

The Cauchy–Schwarz strategy: (1) write the target as $\bigl(\sum a_i b_i\bigr)^2$, (2) identify the $a_i$ and $b_i$ pairings, (3) apply the inequality $\bigl(\sum a_i b_i\bigr)^2 \leq \bigl(\sum a_i^2\bigr)\bigl(\sum b_i^2\bigr)$.

2-var: $(a_1 b_1 + a_2 b_2)^2 \leq (a_1^2 + a_2^2)(b_1^2 + b_2^2)$

$\bigl(\textstyle\sum a_i b_i\bigr)^2 \leq \bigl(\sum a_i^2\bigr)\bigl(\sum b_i^2\bigr)$

Spot the shape

A target like $(x + y + z)^2$ with extra info on $x^2 + y^2 + z^2$ or $\frac{1}{a} + \frac{1}{b}$ paired with $a + b$ usually screams Cauchy–Schwarz.

Engineer the pairing

The freedom is in choosing $a_i$ and $b_i$. For $(\sqrt a + \sqrt b)^2 \leq 2(a+b)$, pair $a_i = (1,1)$ with $b_i = (\sqrt a, \sqrt b)$.

Equality is proportionality

Cauchy–Schwarz equality holds iff $a_i = \lambda b_i$ for some real $\lambda$ — the $a$-vector and $b$-vector are parallel.

What you'll master

Know

Key facts

2-variable: $(a_1 b_1 + a_2 b_2)^2 \leq (a_1^2 + a_2^2)(b_1^2 + b_2^2)$
3-variable: $(a_1 b_1 + a_2 b_2 + a_3 b_3)^2 \leq (a_1^2 + a_2^2 + a_3^2)(b_1^2 + b_2^2 + b_3^2)$
Lagrange identity: $(a_1^2 + a_2^2)(b_1^2 + b_2^2) - (a_1 b_1 + a_2 b_2)^2 = (a_1 b_2 - a_2 b_1)^2$

Understand

Concepts

Why Cauchy–Schwarz follows from a single sum of squares
How the discriminant proof generalises to $n$ variables
Why equality holds when the two "vectors" are proportional

Can do

Skills

State and prove the 2-variable Cauchy–Schwarz inequality
Apply Cauchy–Schwarz to deduce inequalities involving sums and sums of squares
Identify the right pairing $a_i, b_i$ for a given problem

Key terms

Cauchy–Schwarz inequalityFor any reals $a_1,\ldots,a_n$ and $b_1,\ldots,b_n$: $\bigl(\sum a_i b_i\bigr)^2 \leq \bigl(\sum a_i^2\bigr)\bigl(\sum b_i^2\bigr)$.

Lagrange identity (2-var)$(a_1^2 + a_2^2)(b_1^2 + b_2^2) = (a_1 b_1 + a_2 b_2)^2 + (a_1 b_2 - a_2 b_1)^2$. The cleanest 2-variable proof.

Discriminant proofConsider $f(t) = \sum (a_i t - b_i)^2 \geq 0$. As a quadratic in $t$, its discriminant must be $\leq 0$, yielding Cauchy–Schwarz.

Equality (proportionality)Equality holds iff there exists $\lambda \in \mathbb{R}$ with $a_i = \lambda b_i$ for all $i$ (one tuple is a scalar multiple of the other).

Vector formGeometrically: $|\mathbf{a} \cdot \mathbf{b}| \leq |\mathbf{a}|\,|\mathbf{b}|$. Equivalent to $|\cos\theta| \leq 1$ where $\theta$ is the angle between the vectors.

NESA MEX-P1Proof: the nature of proof. Includes algebraic and analytic proofs of classical inequalities such as Cauchy–Schwarz.

Statement and the Lagrange-identity proof

core concept

Statement (2-variable). For all real numbers $a_1, a_2, b_1, b_2$:

$$(a_1 b_1 + a_2 b_2)^2 \leq (a_1^2 + a_2^2)(b_1^2 + b_2^2)$$

Proof. The cleanest argument is the Lagrange identity:

Expand $(a_1 b_2 - a_2 b_1)^2 = a_1^2 b_2^2 - 2 a_1 a_2 b_1 b_2 + a_2^2 b_1^2$.
Expand $(a_1^2 + a_2^2)(b_1^2 + b_2^2) = a_1^2 b_1^2 + a_1^2 b_2^2 + a_2^2 b_1^2 + a_2^2 b_2^2$.
Expand $(a_1 b_1 + a_2 b_2)^2 = a_1^2 b_1^2 + 2 a_1 a_2 b_1 b_2 + a_2^2 b_2^2$.
Subtract: $(a_1^2 + a_2^2)(b_1^2 + b_2^2) - (a_1 b_1 + a_2 b_2)^2 = a_1^2 b_2^2 - 2 a_1 a_2 b_1 b_2 + a_2^2 b_1^2 = (a_1 b_2 - a_2 b_1)^2 \geq 0$.
Therefore $(a_1 b_1 + a_2 b_2)^2 \leq (a_1^2 + a_2^2)(b_1^2 + b_2^2)$. $\blacksquare$

Equality holds iff $a_1 b_2 - a_2 b_1 = 0$, i.e., $a_1 b_2 = a_2 b_1$, i.e., the pairs $(a_1, a_2)$ and $(b_1, b_2)$ are proportional.

Answer to the hook. The maximum of $a_1 b_1 + a_2 b_2$ subject to $a_1^2 + a_2^2 = 1$ and $b_1^2 + b_2^2 = 1$ is $\sqrt{1 \cdot 1} = 1$, achieved when $(a_1, a_2) = (b_1, b_2)$.

$n$-variable statement. The same inequality extends: $\bigl(\sum_{i=1}^{n} a_i b_i\bigr)^2 \leq \bigl(\sum_{i=1}^{n} a_i^2\bigr)\bigl(\sum_{i=1}^{n} b_i^2\bigr)$. The standard HSC proof uses the discriminant: $f(t) = \sum (a_i t - b_i)^2 \geq 0$ is a non-negative quadratic in $t$, so its discriminant $\leq 0$.

Statement (2-var): $(a_1 b_1 + a_2 b_2)^2 \leq (a_1^2 + a_2^2)(b_1^2 + b_2^2)$ · Lagrange identity: RHS $-$ LHS $= (a_1 b_2 - a_2 b_1)^2 \geq 0$ · Equality: iff $(a_1, a_2)$ and $(b_1, b_2)$ are proportional · $n$-var: use discriminant of $f(t) = \sum (a_i t - b_i)^2 \geq 0$

Pause — copy the two-variable Cauchy–Schwarz $(a_1 b_1+a_2 b_2)^2 \leq (a_1^2+a_2^2)(b_1^2+b_2^2)$, the Lagrange identity proof, and the equality condition into your book.

Quick check: Which identity gives the cleanest proof of the 2-variable Cauchy–Schwarz inequality?

The discriminant proof and $n$-variable form

core concept

We just saw that the Lagrange identity RHS $-$ LHS $= (a_1 b_2 - a_2 b_1)^2 \geq 0$ proves Cauchy–Schwarz for two variables with equality iff the vectors are proportional. That raises a question: how does this generalise to $n$ variables without needing an $n$-variable Lagrange identity? This card answers it → the discriminant proof: $f(t) = \sum(a_i t - b_i)^2 \geq 0$ for all $t$ forces $\Delta \leq 0$.

For the $n$-variable Cauchy–Schwarz, the discriminant proof is the cleanest. Consider the function:

$$f(t) = \sum_{i=1}^{n}(a_i t - b_i)^2 \geq 0 \quad \text{for all } t \in \mathbb{R}$$

Expanding: $f(t) = \bigl(\sum a_i^2\bigr) t^2 - 2\bigl(\sum a_i b_i\bigr) t + \bigl(\sum b_i^2\bigr)$. This is a non-negative quadratic in $t$ (with positive leading coefficient, assuming not all $a_i$ are zero). For a non-negative quadratic, the discriminant $\Delta \leq 0$:

$\Delta = \bigl(-2\sum a_i b_i\bigr)^2 - 4\bigl(\sum a_i^2\bigr)\bigl(\sum b_i^2\bigr) \leq 0$.
Divide by 4: $\bigl(\sum a_i b_i\bigr)^2 - \bigl(\sum a_i^2\bigr)\bigl(\sum b_i^2\bigr) \leq 0$.
Rearrange: $\bigl(\sum a_i b_i\bigr)^2 \leq \bigl(\sum a_i^2\bigr)\bigl(\sum b_i^2\bigr)$. $\blacksquare$

Equality holds iff $\Delta = 0$, i.e., $f(t)$ has a (double) real root, i.e., there exists $t$ with $a_i t - b_i = 0$ for all $i$, i.e., $b_i = t a_i$ for all $i$ (proportional).

Common mistake. Students forget the case "all $a_i = 0$", which makes $f(t) = \sum b_i^2$ a constant (not a quadratic). The inequality is trivially $0 \leq 0$ in that case, so it still holds — just note it separately.

Define $f(t) = \sum(a_i t - b_i)^2 \geq 0$ for all real $t$ · Expand: $\bigl(\sum a_i^2\bigr) t^2 - 2\bigl(\sum a_i b_i\bigr) t + \bigl(\sum b_i^2\bigr) \geq 0$ · Discriminant $\leq 0$ $\Rightarrow$ Cauchy–Schwarz · Equality iff one tuple is a scalar multiple of the other

Pause — copy the discriminant proof setup $f(t) = \sum(a_i t - b_i)^2 \geq 0$, the expansion, and the conclusion $\Delta \leq 0 \Rightarrow$ Cauchy–Schwarz into your book.

Did you get this? True or false: in the Cauchy–Schwarz inequality, equality holds iff there exists a real $\lambda$ with $a_i = \lambda b_i$ for every $i$.

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

PROBLEM 1 · 2-VARIABLE APPLICATION

Use the Cauchy–Schwarz inequality to prove that for all real $x, y$: $(x + y)^2 \leq 2(x^2 + y^2)$. State the equality case.

Choose the pairing: $a_1 = 1, a_2 = 1$ and $b_1 = x, b_2 = y$. Then $a_1 b_1 + a_2 b_2 = x + y$.

The target $(x+y)^2$ tells us to write $x + y$ as a "sum of products". The simplest pairing uses $a_i = 1$.

PROBLEM 2 · 3-VARIABLE APPLICATION

For all positive real numbers $a, b, c$, use Cauchy–Schwarz to prove $(a + b + c)\!\left(\dfrac{1}{a} + \dfrac{1}{b} + \dfrac{1}{c}\right) \geq 9$. State the equality case.

Choose the pairing: $a_i = (\sqrt{a}, \sqrt{b}, \sqrt{c})$ and $b_i = (1/\sqrt{a}, 1/\sqrt{b}, 1/\sqrt{c})$. Then $\sum a_i b_i = 1 + 1 + 1 = 3$.

Engineer the pairing so $\sum a_i b_i$ collapses to a constant — that constant becomes the lower bound for the RHS.

PROBLEM 3 · CLASSIC RMS-AM

For all real $a_1, a_2, \ldots, a_n$, use Cauchy–Schwarz to prove $(a_1 + a_2 + \cdots + a_n)^2 \leq n\bigl(a_1^2 + a_2^2 + \cdots + a_n^2\bigr)$. Deduce that the arithmetic mean is bounded by the quadratic (RMS) mean.

Pairing: $a_i = a_i$ (the given numbers) and $b_i = 1$ for $i = 1, \ldots, n$. Then $\sum a_i b_i = a_1 + a_2 + \cdots + a_n$.

Choosing $b_i = 1$ converts $\sum b_i^2$ into a count $n$ — the cleanest way to introduce a factor of $n$.

Fill the gap: Applying Cauchy–Schwarz with $a_i = (1, 1, 1)$ and $b_i = (x, y, z)$ gives $(x + y + z)^2 \leq (x^2 + y^2 + z^2)$.

Misconceptions to fix · the 3 traps that cost marks

Trap 01

Wrong direction of inequality

Cauchy–Schwarz gives $(\sum a_i b_i)^2 \leq (\sum a_i^2)(\sum b_i^2)$ — the square of the cross-sum is $\leq$ the product of the squared sums. Students sometimes write it backwards. Sanity check: $(1 \cdot 1)^2 = 1$ vs $(1)(1) = 1$, so the bound is tight when one term is involved; in general the RHS is bigger.

Trap 02

Forgetting the equality case

Cauchy–Schwarz equality is "the two tuples are proportional" — not "all equal" or "all positive". For $(x+y)^2 \leq 2(x^2+y^2)$, equality is $x = y$ (proportional to $(1,1)$). For $(a+b+c)(1/a+1/b+1/c) \geq 9$, equality is $a = b = c$. Always check what proportionality means in your specific pairing.

Trap 03

Choosing a pairing that doesn't telescope

The whole skill of Cauchy–Schwarz problems is choosing $a_i, b_i$ so that $\sum a_i^2$ and $\sum b_i^2$ become the quantities in the target. If you pick a pairing that leaves messy radicals or unrelated sums, try a different choice. A common rescue: replace $a_i$ with $\sqrt{f_i}$ to introduce $f_i$ into $\sum a_i^2$.

Did you get this? True or false: the Cauchy–Schwarz inequality $(a_1 b_1 + a_2 b_2)^2 \leq (a_1^2 + a_2^2)(b_1^2 + b_2^2)$ requires $a_i$ and $b_i$ to be positive.

Work mode · how are you completing this lesson?

Activities · practice with the ideas

State the 2-variable Cauchy–Schwarz inequality, and prove it using the Lagrange identity.

Use Cauchy–Schwarz to prove $(x + y + z)^2 \leq 3(x^2 + y^2 + z^2)$ for all real $x, y, z$.

For positive reals $a, b$, use C–S to prove $\dfrac{a^2}{b} + \dfrac{b^2}{a} \geq a + b$. (Hint: pair $(\sqrt b, \sqrt a)$ with $(a/\sqrt b, b/\sqrt a)$.)

Use Cauchy–Schwarz with $b_i = 1$ to prove that for any real $a_1, a_2, a_3$ with $a_1 + a_2 + a_3 = 6$, $\;a_1^2 + a_2^2 + a_3^2 \geq 12$.

Verify the Lagrange identity $(a_1^2 + a_2^2)(b_1^2 + b_2^2) = (a_1 b_1 + a_2 b_2)^2 + (a_1 b_2 - a_2 b_1)^2$ by expanding both sides.

Odd one out: Three of these statements about Cauchy–Schwarz are correct. Which one is NOT?

Revisit your thinking

Earlier you expanded $(a_1 b_2 - a_2 b_1)^2$ and looked for familiar quantities.

That single expansion is the proof of Cauchy–Schwarz: it equals $(a_1^2 + a_2^2)(b_1^2 + b_2^2) - (a_1 b_1 + a_2 b_2)^2$. The Lagrange identity packs the entire 2-variable Cauchy–Schwarz into one line. The $n$-variable version requires the discriminant trick — but the heart of every C–S proof is "a sum of squares is non-negative".

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Multiple choice

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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. State the 2-variable Cauchy–Schwarz inequality and prove it using the Lagrange identity. (2 marks)

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ApplyBand 43 marks

Q2. Use the Cauchy–Schwarz inequality to prove that for all real $x, y, z$: $(x + 2y + 3z)^2 \leq 14(x^2 + y^2 + z^2)$. State the equality case. (3 marks)

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AnalyseBand 53 marks

Q3. For positive real numbers $a, b, c$ satisfying $a + b + c = 1$, use Cauchy–Schwarz to prove $\dfrac{1}{a} + \dfrac{1}{b} + \dfrac{1}{c} \geq 9$, and state when equality holds. (3 marks)

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Comprehensive answers (click to reveal)

Activity answers:

1. Statement: $(a_1 b_1 + a_2 b_2)^2 \leq (a_1^2 + a_2^2)(b_1^2 + b_2^2)$. Proof: $(a_1^2 + a_2^2)(b_1^2 + b_2^2) - (a_1 b_1 + a_2 b_2)^2 = (a_1 b_2 - a_2 b_1)^2 \geq 0$, so $(a_1 b_1 + a_2 b_2)^2 \leq (a_1^2 + a_2^2)(b_1^2 + b_2^2)$. Equality iff $a_1 b_2 = a_2 b_1$.

2. With $a_i = (1, 1, 1)$ and $b_i = (x, y, z)$: $(x + y + z)^2 \leq (1+1+1)(x^2 + y^2 + z^2) = 3(x^2 + y^2 + z^2)$. Equality iff $x = y = z$.

3. With $a_i = (\sqrt b, \sqrt a)$ and $b_i = (a/\sqrt b, b/\sqrt a)$: $\sum a_i b_i = a + b$, $\sum a_i^2 = a + b$, $\sum b_i^2 = a^2/b + b^2/a$. C–S: $(a + b)^2 \leq (a + b)(a^2/b + b^2/a)$. Divide by $a + b > 0$: $a + b \leq a^2/b + b^2/a$. Equality iff $a = b$.

4. C–S with $b_i = 1$: $(a_1 + a_2 + a_3)^2 \leq 3(a_1^2 + a_2^2 + a_3^2)$. Given $a_1 + a_2 + a_3 = 6$: $36 \leq 3(a_1^2 + a_2^2 + a_3^2)$, so $a_1^2 + a_2^2 + a_3^2 \geq 12$. Equality iff $a_1 = a_2 = a_3 = 2$.

5. LHS $= a_1^2 b_1^2 + a_1^2 b_2^2 + a_2^2 b_1^2 + a_2^2 b_2^2$. RHS $= (a_1^2 b_1^2 + 2 a_1 a_2 b_1 b_2 + a_2^2 b_2^2) + (a_1^2 b_2^2 - 2 a_1 a_2 b_1 b_2 + a_2^2 b_1^2) = a_1^2 b_1^2 + a_1^2 b_2^2 + a_2^2 b_1^2 + a_2^2 b_2^2$ ✓.

Q1 (2 marks): Statement [1]. Proof using Lagrange identity, showing $(a_1^2 + a_2^2)(b_1^2 + b_2^2) - (a_1 b_1 + a_2 b_2)^2 = (a_1 b_2 - a_2 b_1)^2 \geq 0$ [1].

Q2 (3 marks): Choose $a_i = (1, 2, 3)$, $b_i = (x, y, z)$ [1]. Apply C–S: $(x + 2y + 3z)^2 \leq (1 + 4 + 9)(x^2 + y^2 + z^2) = 14(x^2 + y^2 + z^2)$ [1]. Equality iff $(1, 2, 3) \parallel (x, y, z)$, i.e., $x : y : z = 1 : 2 : 3$ [1].

Q3 (3 marks): Choose $a_i = (\sqrt a, \sqrt b, \sqrt c)$, $b_i = (1/\sqrt a, 1/\sqrt b, 1/\sqrt c)$, so $\sum a_i b_i = 3$, $\sum a_i^2 = a + b + c = 1$, $\sum b_i^2 = 1/a + 1/b + 1/c$ [1]. C–S: $9 = 3^2 \leq 1 \cdot (1/a + 1/b + 1/c)$, so $1/a + 1/b + 1/c \geq 9$ [1]. Equality iff $a = b = c = 1/3$ [1].

Boss battle · The Cauchy Cascade

earn bronze · silver · gold

Five timed Cauchy–Schwarz problems: identify the pairing, apply the inequality, state the equality case. 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 applying Cauchy–Schwarz to short questions. Lighter alternative to the boss.

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Module overview · Extension 2 · Checkpoint 2