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.
Start with the warm-up and record your first reasoning.
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?
Orient yourself to the lesson goals, language, and core approach.
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: $(a_1 b_1 + a_2 b_2)^2 \leq (a_1^2 + a_2^2)(b_1^2 + b_2^2)$
Work through the concept and complete its check.
Statement (2-variable). For all real numbers $a_1, a_2, b_1, b_2$:
Proof. The cleanest argument is the Lagrange identity:
Equality holds iff $a_1 b_2 - a_2 b_1 = 0$, i.e., $a_1 b_2 = a_2 b_1$. This says the pairs $(a_1,a_2)$ and $(b_1,b_2)$ are linearly dependent, including when either pair is zero.
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 $(a_1, a_2) = (b_1, b_2)$.
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 the two tuples are linearly dependent (including if either is zero) · $n$-var: use the discriminant of $f(t) = \sum (a_i t - b_i)^2 \geq 0$, with the all-zero leading tuple handled separately
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?
Build the next proof move and complete its check.
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 two tuples are linearly dependent (including if either is zero). 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:
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 $\Delta \leq 0$:
If every $a_i=0$, both sides of Cauchy–Schwarz are zero, so equality holds for any tuple $(b_i)$. Otherwise, $\sum a_i^2>0$ and equality holds iff $\Delta=0$; then $f(t)$ has a real root, so $a_i t-b_i=0$ for every $i$, and one tuple is a scalar multiple of the other. Thus, in all cases, equality holds iff the two tuples are linearly dependent.
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 $\leq 0$ gives Cauchy–Schwarz · Equality iff the tuples are linearly dependent, including if either tuple is zero
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: for any real tuples $(a_1,\ldots,a_n)$ and $(b_1,\ldots,b_n)$, equality in Cauchy–Schwarz holds iff the two tuples are linearly dependent (including when either tuple is zero).
Follow the worked examples, then complete the check.
Worked examples · 3 in a row, reveal as you go
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.
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.
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.
Fill the gap: Applying Cauchy–Schwarz with $a_i = (1, 1, 1)$ and $b_i = (x, y, z)$ gives $(x + y + z)^2 \leq c\,(x^2 + y^2 + z^2)$, where $c =$ .
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 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.
Complete the practice activities and revisit your opening thinking.
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?
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".
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. State the 2-variable Cauchy–Schwarz inequality and prove it using the Lagrange identity. (2 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)
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)
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 $(x,y,z)=t(1,2,3)$ for some real $t$, including $t=0$ [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].
Sit the module quiz, then close the lesson off.
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