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NSWMaths Extension 2Syllabus dot point

How do we prove inequalities rigorously, and how do foundational results such as the arithmetic-geometric mean inequality follow from elementary algebra?

Prove inequalities using algebraic manipulation, the fact that squares are non-negative, and standard results such as the arithmetic mean-geometric mean inequality

A focused answer to the HSC Maths Extension 2 dot point on proving inequalities. The non-negative square method, the arithmetic mean-geometric mean inequality, manipulation of given inequalities, and rigorous structure, with verified worked examples.

Generated by Claude Opus 4.76 min answer

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  1. What this dot point is asking
  2. The non-negative square: the central tool
  3. The arithmetic mean-geometric mean inequality
  4. The forwards structure of a proof
  5. Manipulating a given inequality

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.

The non-negative square: the central tool

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

x20for all xR,x^2 \ge 0 \quad \text{for all } x \in \mathbb{R},

with equality if and only if x=0x = 0. Almost every Extension 2 inequality proof reduces to spotting the right square. For instance, to prove a2+b22aba^2 + b^2 \ge 2ab, begin with the true statement

(ab)20.(a - b)^2 \ge 0.

Expanding gives a22ab+b20a^2 - 2ab + b^2 \ge 0, and adding 2ab2ab to both sides yields a2+b22aba^2 + b^2 \ge 2ab. Equality holds exactly when a=ba = b.

The arithmetic mean-geometric mean inequality

For non-negative reals aa and bb, the arithmetic mean is at least the geometric mean:

a+b2ab,a,b0.\frac{a + b}{2} \ge \sqrt{ab}, \qquad a, b \ge 0.

Proof: since a,b0a, b \ge 0, both a\sqrt{a} and b\sqrt{b} are real. Start from the true statement

(ab)20.\left(\sqrt{a} - \sqrt{b}\right)^2 \ge 0.

Expanding gives a2ab+b0a - 2\sqrt{a}\sqrt{b} + b \ge 0, that is a+b2aba + b \ge 2\sqrt{ab}. Dividing by 22 gives the result. Equality holds when a=b\sqrt{a} = \sqrt{b}, i.e. a=ba = b.

A frequently used corollary is that for any positive real xx,

x+1x2,x + \frac{1}{x} \ge 2,

obtained by applying AM-GM to a=xa = x and b=1xb = \frac{1}{x}, since x1x=1\sqrt{x \cdot \frac{1}{x}} = 1.

The forwards structure of a proof

The single most important discipline is direction. A valid proof of LRL \ge R starts from a statement already known to be true and arrives at LRL \ge R. It is invalid to write LRL \ge R, manipulate it, reach a true statement, and declare victory: that only shows the target implies something true, not that it is true.

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

Manipulating a given inequality

Often you are handed an inequality and asked to deduce another. Permitted operations include adding the same quantity to both sides, multiplying both sides by a positive quantity (preserving direction) or a negative quantity (reversing direction), and adding two inequalities of the same direction. You may not in general subtract inequalities or multiply two inequalities unless all terms are known to be positive.

Exam-style practice questions

Practice questions written in the style of NESA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

2022 HSC2 marksFor real numbers a, b >= 0 prove that (a + b)/2 >= sqrt(ab).
Show worked answer →

Start from the fact that a square is never negative. Since a, b >= 0, their square roots are real, so

(sqrt(a) - sqrt(b))^2 >= 0.

Expand the left side:

a - 2 sqrt(a) sqrt(b) + b >= 0, that is a + b - 2 sqrt(ab) >= 0.

Rearrange: a + b >= 2 sqrt(ab), and dividing by 2 (positive) gives

(a + b)/2 >= sqrt(ab),

which is the arithmetic mean - geometric mean inequality. Equality holds when sqrt(a) = sqrt(b), i.e. a = b.

Mark notes: 1 mark for starting from (sqrt(a) - sqrt(b))^2 >= 0, 1 mark for expanding and rearranging to the required form.

2023 HSC2 marksProve that for all real numbers x and y, where x^2 + y^2 is not 0, (x + y)^2/(x^2 + y^2) <= 2.
Show worked answer →

Since x^2 + y^2 > 0 (it is non-zero and a sum of squares), we may compare numerator and denominator directly.

Begin with the non-negative square (x - y)^2 >= 0:

x^2 - 2xy + y^2 >= 0, so x^2 + y^2 >= 2xy.

Add x^2 + y^2 to both sides:

2(x^2 + y^2) >= x^2 + 2xy + y^2 = (x + y)^2.

Divide both sides by the positive quantity x^2 + y^2:

2 >= (x + y)^2/(x^2 + y^2),

which is the required inequality (x + y)^2/(x^2 + y^2) <= 2.

Mark notes: 1 mark for using (x - y)^2 >= 0 to get x^2 + y^2 >= 2xy, 1 mark for completing the algebra and dividing by the positive x^2 + y^2.

2021 HSC2 marksFor all non-negative real numbers x and y, sqrt(xy) <= (x + y)/2 (Do NOT prove this). Using this fact, show that for all non-negative real numbers a, b and c, sqrt(abc) <= (a^2 + b^2 + 2c)/4.
Show worked answer →

Apply the given inequality twice.

First, write sqrt(abc) = sqrt((ab)(c)) and use the fact with the two non-negative numbers ab and c:

sqrt(abc) <= (ab + c)/2.

Next, bound ab. Apply the given fact with the non-negative numbers a^2 and b^2: sqrt(a^2 b^2) <= (a^2 + b^2)/2. Since sqrt(a^2 b^2) = ab (as a, b >= 0),

ab <= (a^2 + b^2)/2.

Substitute this into the first inequality:

sqrt(abc) <= (ab + c)/2 <= ((a^2 + b^2)/2 + c)/2 = (a^2 + b^2 + 2c)/4.

Hence sqrt(abc) <= (a^2 + b^2 + 2c)/4.

Mark notes: 1 mark for applying the given inequality to ab and c, 1 mark for bounding ab by (a^2 + b^2)/2 and combining to the required result.