Proving inequalities in HSC Mathematics Extension 2: the non-negativity of squares, the AM-GM inequality, and rigorous algebraic argument

What this dot point is asking

NESA wants you to prove inequalities, not merely verify them numerically. You must build arguments from accepted facts, the most important being that a real square is never negative, and you must know and apply the arithmetic mean-geometric mean (AM-GM) inequality. Crucially, you must structure the argument forwards from known truths to the required result, never assuming what you are trying to prove. Extension 2 also expects you to chain and combine given inequalities to reach a new one, and to recognise when a problem is really an AM-GM or Cauchy-Schwarz problem in disguise.

The non-negative square: the central tool

Every real number squared is greater than or equal to zero:

with equality if and only if $x = 0$. This single fact is the engine behind almost every Extension 2 inequality, because it is the one inequality you never have to justify, it is true for all real values with no domain restriction, and you control exactly when it becomes an equality. A sum of squares inherits both properties: $u^2 + v^2 + w^2 \ge 0$ with equality only when $u = v = w = 0$. The whole craft of these proofs is engineering the right square (or squares) to appear.

To see the tool in its simplest form, prove $a^2 + b^2 \ge 2ab$ for all real $a, b$. Begin with the true statement

Expanding gives $a^2 - 2ab + b^2 \ge 0$, and adding $2ab$ to both sides yields $a^2 + b^2 \ge 2ab$. Equality holds exactly when $a - b = 0$, that is $a = b$.

The "consider LHS minus RHS" method

The most dependable opening move for an unfamiliar inequality $L \ge R$ is to form the difference $L - R$ and try to write it as something obviously non-negative, usually a perfect square or a sum of perfect squares. If you succeed, the inequality is proved and the equality case is read off as the condition that makes every square vanish.

For example, to prove $a^2 + b^2 + 1 \ge ab + a + b$ for all real $a, b$, consider

Doubling (to clear the halves that completing the square will produce) gives $2a^2 + 2b^2 + 2 - 2ab - 2a - 2b$, and this rearranges into a sum of three squares:

A quick expansion confirms the three squares sum to $2a^2 + 2b^2 + 2 - 2ab - 2a - 2b$, so the doubled difference is non-negative, hence the difference itself is non-negative, and the inequality holds. Equality needs all three squares to be zero at once: $a = b$, $a = 1$ and $b = 1$, that is $a = b = 1$.

The arithmetic mean-geometric mean inequality

For non-negative reals $a$ and $b$, the arithmetic mean is at least the geometric mean:

Algebraic proof. Since $a, b \ge 0$, both $\sqrt{a}$ and $\sqrt{b}$ are real. Start from the true statement

Expanding gives $a - 2\sqrt{a}\sqrt{b} + b \ge 0$, that is $a + b \ge 2\sqrt{ab}$. Dividing by $2$ gives the result. Equality holds when $\sqrt{a} = \sqrt{b}$, that is $a = b$.

A frequently used corollary is that for any positive real $x$,

obtained by applying AM-GM to $a = x$ and $b = \frac{1}{x}$, since $\sqrt{x \cdot \frac{1}{x}} = 1$. The same trick gives $\frac{p}{q} + \frac{q}{p} \ge 2$ for positive $p, q$.

A geometric picture of AM-GM

The two-variable AM-GM result has a clean geometric meaning that is worth carrying in your head, because it instantly tells you why equality occurs only when $a = b$. Draw a semicircle on a diameter $PQ$ of length $a + b$, and let $D$ split the diameter into $PD = a$ and $DQ = b$. Drop the perpendicular from $D$ up to the circle at $C$.

The radius of the circle is half the diameter, $OM = \frac{a + b}{2}$, the arithmetic mean. The half-chord $DC$ is the geometric mean: since $PQ$ is a diameter, the angle $PCQ$ is a right angle (angle in a semicircle), and the altitude from the right angle to the hypotenuse satisfies $DC^2 = PD \cdot DQ = ab$, so $DC = \sqrt{ab}$. A perpendicular from a point on the diameter up to the circle can never be longer than the radius, so $DC \le OM$, that is $\sqrt{ab} \le \frac{a + b}{2}$. Equality happens only when $D$ coincides with the centre $O$, which is exactly when $a = b$.

The Cauchy-Schwarz inequality at Extension 2 level

A second standard result, almost always built up within a question rather than quoted cold, is the Cauchy-Schwarz inequality. In its two-term form it states that for all real numbers $a, b, x, y$,

Proof by the difference. Expand both sides and consider the difference $\text{RHS} - \text{LHS}$:

Multiplying out, $(a^2 + b^2)(x^2 + y^2) = a^2x^2 + a^2y^2 + b^2x^2 + b^2y^2$, while $(ax + by)^2 = a^2x^2 + 2abxy + b^2y^2$. Subtracting, the $a^2x^2$ and $b^2y^2$ terms cancel and we are left with

So $\text{RHS} - \text{LHS} = (ay - bx)^2 \ge 0$, which gives the inequality. Equality holds exactly when $ay = bx$, that is when the pairs $(a, b)$ and $(x, y)$ are proportional. The same idea in three terms (the Lagrange identity) writes $(a^2 + b^2 + c^2)(x^2 + y^2 + z^2) - (ax + by + cz)^2$ as the sum of squares $(ay - bx)^2 + (bz - cy)^2 + (cx - az)^2$, again non-negative.

The forwards structure of a proof

The single most important discipline is direction. A valid proof of $L \ge R$ starts from a statement already known to be true and arrives at $L \ge R$. It is invalid to write $L \ge R$, manipulate it, reach a true statement, and declare victory: that only shows the target implies something true, not that it is true. The two readings differ because not every algebraic step is reversible; squaring, for instance, can turn a false statement into a true one.

When working backwards from the target is the natural way to discover the proof, do that as scratch work, then reverse the steps for the final write-up, checking that each reversed step is itself a valid implication. If a step involves multiplying by a quantity, check its sign, because multiplying an inequality by a negative number reverses it.

Building new inequalities from old

Often you are handed one or more inequalities and asked to deduce another. The permitted operations are limited and worth knowing exactly:

• Add the same quantity to both sides (direction preserved).
• Multiply both sides by a positive quantity (direction preserved) or a negative quantity (direction reversed).
• Add two inequalities pointing the same way: from $A \ge B$ and $C \ge D$ you may conclude $A + C \ge B + D$.
• Apply a strictly increasing function to both sides (for example take square roots of non-negative quantities).
• Chain inequalities: from $A \ge B$ and $B \ge C$ conclude $A \ge C$.

What you may not do in general is subtract one inequality from another, or multiply two inequalities together, unless every quantity involved is known to be non-negative. A common Extension 2 task is to apply a given two-variable inequality more than once, feeding the output of one application into the next, exactly as in the 2021 HSC question above.

How exam questions ask about inequalities

• "Prove that $L \ge R$ for all real ...": form $L - R$ and show it is a square or a sum of squares; state the equality case.
• "Prove the AM-GM inequality $\frac{a+b}{2} \ge \sqrt{ab}$": start from $(\sqrt{a} - \sqrt{b})^2 \ge 0$. Watch the domain $a, b \ge 0$.
• "Show that $(ax + by)^2 \le (a^2 + b^2)(x^2 + y^2)$": a Cauchy-Schwarz problem; subtract the sides and recognise $(ay - bx)^2$.
• "It is given that [inequality]. Hence show [new inequality]": do not reprove the given result; apply it, possibly more than once, and combine.
• "Prove that $x + \frac{1}{x} \ge 2$ for $x > 0$": a one-line AM-GM corollary, or consider $\frac{(x - 1)^2}{x} \ge 0$.
• "Deduce that ... and state when equality holds": the equality condition is where the marks finish; trace it back to the squares being zero.

Edge cases and what-ifs

• The domain matters for square roots. AM-GM and any step that takes a square root require the quantities to be non-negative. If the variables can be negative, you cannot use $\sqrt{ab}$ and must work with squares directly.
• Strict versus non-strict. $(a - b)^2 \ge 0$ is non-strict; it becomes the strict $> 0$ only when $a \neq b$. If a question asks for a strict inequality, you must argue that the relevant square cannot be zero (for instance because the variables are distinct).
• Equality can be unreachable. Sometimes the equality condition (all squares zero) cannot be met inside the stated domain, in which case the inequality is in fact strict throughout. Note this when it happens.
• Multiplying inequalities needs non-negativity. Multiplying $a + b \ge 2\sqrt{ab}$ by another inequality is only valid because both sides are non-negative. With possibly negative quantities the product step fails.
• Dividing by a variable. To prove $x + \frac{1}{x} \ge 2$ you must know $x > 0$; for $x < 0$ the inequality reverses to $x + \frac{1}{x} \le -2$.