Worked Solutions
Proof — Worked Solutions (HSC Maths Extension 2)
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Worked examples for HSC Maths Extension 2 proof. Each shows where the marks are awarded, the key idea, and the full solution explained by your choice of tutor — Stella, Ella or Cassie.
How to use these
Try each question first, then check your working. Use the tutor tabs to read the full solution in the style that suits you: Stella is direct and challenging, Ella is warm and explains the why, and Cassie is concise and analytical.
Proof questions reward a clear, logical chain of reasoning — every line must follow from the last, and you should state the result you are using before you use it.
Example 1 — Proving an inequality from a square
Question
Let $a$ and $b$ be positive real numbers. Prove that $\dfrac{a}{b} + \dfrac{b}{a} \ge 2$, and state when equality holds.
Solution
A non-negative square is the engine for almost every basic inequality, so start with $(a - b)^2 \ge 0$.
Expand: $a^2 - 2ab + b^2 \ge 0$, so $a^2 + b^2 \ge 2ab$.
Since $a, b > 0$, divide both sides by $ab > 0$ (the inequality direction is safe because $ab$ is positive):
$\dfrac{a^2 + b^2}{ab} \ge 2$, i.e. $\dfrac{a}{b} + \dfrac{b}{a} \ge 2$.
Equality holds exactly when $(a-b)^2 = 0$, that is $a = b$. Always state the equality case — it is worth a mark and shows your proof is tight.
The trick with inequalities like this is to find something we already know is true and build from it. The safest fact in the toolkit is that a real number squared can never be negative, so $(a - b)^2 \ge 0$.
Let's expand that: $a^2 - 2ab + b^2 \ge 0$, which rearranges to $a^2 + b^2 \ge 2ab$.
Now look at what we want — it has $ab$ on the bottom, so let's divide through by $ab$. We're told $a$ and $b$ are positive, so $ab > 0$, and dividing by a positive number keeps the inequality pointing the same way:
$\dfrac{a^2 + b^2}{ab} \ge 2$, and splitting the fraction gives $\dfrac{a}{b} + \dfrac{b}{a} \ge 2$.
When is it an equality? Only when that first square is zero, i.e. when $a - b = 0$, so $a = b$. That makes sense — the two fractions are reciprocals, and reciprocals balance at 1 each only when the numbers match.
Start from a non-negative square.
- $(a - b)^2 \ge 0$
- $a^2 - 2ab + b^2 \ge 0 \Rightarrow a^2 + b^2 \ge 2ab$
- Divide by $ab > 0$ (valid since $a, b > 0$): $\dfrac{a}{b} + \dfrac{b}{a} \ge 2$
Equality: $(a-b)^2 = 0 \Rightarrow a = b$.
Where the marks go
- 1 mark: States and uses a non-negative square, e.g. $(a-b)^2 \ge 0$
- 1 mark: Correctly divides by $ab > 0$ to reach $\frac{a}{b} + \frac{b}{a} \ge 2$
- 1 mark: States equality holds when $a = b$
Key idea
Many basic inequalities follow from $(\text{real})^2 \ge 0$; dividing by a positive quantity preserves the inequality, and equality tracks back to the square being zero.
Example 2 — Inequality with a constraint
Question
Let $x$ and $y$ be positive real numbers with $x + y = 1$. Prove that $\left(1 + \dfrac{1}{x}\right)\left(1 + \dfrac{1}{y}\right) \ge 9$.
Solution
Expand the product first, then bring in the constraint $x + y = 1$.
$\left(1 + \dfrac{1}{x}\right)\left(1 + \dfrac{1}{y}\right) = 1 + \dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{xy}$.
Since $x + y = 1$, we have $\dfrac{1}{x} + \dfrac{1}{y} = \dfrac{x+y}{xy} = \dfrac{1}{xy}$. So the expression equals $1 + \dfrac{2}{xy}$.
Now bound $xy$. By AM–GM (or $(x-y)^2 \ge 0$ giving $(x+y)^2 \ge 4xy$), $xy \le \dfrac{(x+y)^2}{4} = \dfrac14$. So $\dfrac{1}{xy} \ge 4$ and $\dfrac{2}{xy} \ge 8$.
Therefore the expression $\ge 1 + 8 = 9$, with equality when $x = y = \tfrac12$. Reducing everything to $xy$ is what makes this clean.
There are two ideas hiding here: an algebra simplification and a bound. Let's do the algebra first by expanding the brackets:
$\left(1 + \dfrac{1}{x}\right)\left(1 + \dfrac{1}{y}\right) = 1 + \dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{xy}$.
The constraint $x + y = 1$ is our friend. Combining the middle terms, $\dfrac{1}{x} + \dfrac{1}{y} = \dfrac{x + y}{xy} = \dfrac{1}{xy}$ because the numerator is just 1. So everything collapses to $1 + \dfrac{2}{xy}$ — now it all depends on a single quantity, $xy$.
To finish we need to know how big $\dfrac{2}{xy}$ can be, which means how small $xy$ can be. Since $(x - y)^2 \ge 0$, we get $(x+y)^2 \ge 4xy$, so $xy \le \dfrac{(x+y)^2}{4} = \dfrac14$. A smaller bottom makes a bigger fraction, so $\dfrac{1}{xy} \ge 4$, hence $\dfrac{2}{xy} \ge 8$.
Adding the 1 back: the whole thing is at least $9$. Equality needs $x = y$, and with $x + y = 1$ that means $x = y = \tfrac12$ — a nice check that the answer is sensible.
Expand, simplify using the constraint, then bound $xy$.
- $\left(1 + \tfrac{1}{x}\right)\left(1 + \tfrac{1}{y}\right) = 1 + \tfrac{1}{x} + \tfrac{1}{y} + \tfrac{1}{xy}$
- $\tfrac{1}{x} + \tfrac{1}{y} = \tfrac{x+y}{xy} = \tfrac{1}{xy}$ (since $x + y = 1$)
- Expression $= 1 + \tfrac{2}{xy}$
- $(x-y)^2 \ge 0 \Rightarrow xy \le \tfrac{(x+y)^2}{4} = \tfrac14 \Rightarrow \tfrac{1}{xy} \ge 4$
- Expression $\ge 1 + 8 = 9$
Equality: $x = y = \tfrac12$.
Where the marks go
- 1 mark: Expands the product correctly
- 1 mark: Uses $x + y = 1$ to simplify to $1 + \frac{2}{xy}$
- 1 mark: Establishes $xy \le \frac14$ (e.g. via $(x-y)^2 \ge 0$ or AM–GM)
- 1 mark: Concludes $\ge 9$ and states equality at $x = y = \frac12$
Key idea
Use the constraint to reduce an expression to one variable or quantity, then bound that quantity with a known inequality; track the equality case throughout.