Ultimate Guide to Proof of Inequalities (Part 1): The Foundation of Squares

Introduction

In lower levels of mathematics, solving an inequality is just like solving an equation—you just flip the sign if you multiply by a negative. But in HSC Mathematics Extension 2, you aren't just solving inequalities; you are proving them from scratch. Proof of Inequalities (Part 1) establishes the foundational axioms of inequalities. We will explore how one incredibly simple fact—that the square of any real number is non-negative—forms the bedrock of almost every major algebraic inequality proof.

Executive Summary

This guide covers the fundamental algebraic inequality proofs:

Formal Definition: Defining $x > y$ mathematically as $x - y > 0$ .
The Core Axiom: Using the property $x^2 \ge 0$ (for real $x$ ) as the starting point for proofs.
Manipulation Rules: Proving and using rules like "if $x > y$ , then $x^2 > y^2$ " (and their restrictions).
The Triangle Inequality: Proving $|x + y| \le |x| + |y|$ for real numbers.

What is this about?

If I ask you to prove that $x^2 + y^2 \ge 2xy$ for all real numbers $x$ and $y$ , where do you start? You cannot start by writing the inequality itself and working backwards, because that assumes the statement is already true!

Instead, you must start with a known, undeniable truth. In the real number system, any number multiplied by itself is positive (or zero). Thus, $(x-y)^2 \ge 0$ . By expanding this single, universally accepted truth, you can derive complex inequalities algebraically. This is the essence of deductive proof.

Main Content

1. The Definition of an Inequality

Before proving anything, we must agree on what an inequality actually means mathematically. For two real numbers $x$ and $y$ : $x > y \iff x - y > 0$

If you want to prove $A > B$ , the most direct algebraic method is to calculate $A - B$ , factorize it, and prove that the resulting expression must be strictly positive.

2. The Core Axiom: Squares are Non-Negative

The most powerful tool in your inequality arsenal is this simple fact: For any real number $a$ , $a^2 \ge 0$ .

Equality only holds when $a = 0$ . By strategically choosing what " $a$ " represents, you can build massive inequalities. For example, let $a = (x - y)$ . Then: $(x - y)^2 \ge 0$ $x^2 - 2xy + y^2 \ge 0$ $x^2 + y^2 \ge 2xy$

This derived result ( $x^2 + y^2 \ge 2xy$ ) is used so often it is practically an axiom itself.

3. Key Manipulation Rules

The syllabus requires you to prove and use several fundamental rules. You must know when they apply (especially regarding positive/negative signs):

Reciprocals: If $x > y > 0$ , then $\frac{1}{x} < \frac{1}{y}$ . (Notice the sign flips!).
Squares: If $x > y > 0$ , then $x^2 > y^2$ . (If $x$ and $y$ can be negative, this is false. E.g. $2 > -5$ , but $4 \ngtr 25$ ).
Square Roots: If $x > y > 0$ , then $\sqrt{x} > \sqrt{y}$ .
Addition: If $x > y$ and $a > b$ , then $x+a > y+b$ . (You can add inequalities pointing the same way).
Multiplication: If $x > y > 0$ and $a > b > 0$ , then $ax > by$ . (You can multiply positive inequalities).

4. The Triangle Inequality

You previously learned the Triangle Inequality for complex numbers. It applies to real numbers too, using absolute values: $|x + y| \le |x| + |y|$

Geometric Interpretation: Imagine the real number line. The distance from the origin to $x+y$ can never be greater than the distance to $x$ plus the distance to $y$ . If $x$ and $y$ have the same sign (both positive or both negative), the distances add perfectly ( $|x+y| = |x| + |y|$ ). If they have opposite signs, they "cancel" each other out, making the sum smaller ( $|x+y| < |x| + |y|$ ).

Simple Worked Example

Question: Prove that for all real numbers $a, b, c$ , $a^2 + b^2 + c^2 \ge ab + bc + ca$

Solution: Step 1: Start with known axioms. We know that the square of any real number is non-negative. We need terms like $ab, bc, ca$ , so we will use pairs of variables.

$(a - b)^2 \ge 0$
$(b - c)^2 \ge 0$
$(c - a)^2 \ge 0$

Step 2: Expand the axioms.

$a^2 - 2ab + b^2 \ge 0 \implies a^2 + b^2 \ge 2ab$
$b^2 - 2bc + c^2 \ge 0 \implies b^2 + c^2 \ge 2bc$
$c^2 - 2ca + a^2 \ge 0 \implies c^2 + a^2 \ge 2ca$

Step 3: Combine the inequalities. Because all three inequalities point the same way, we can add them together (Left side + Left side $\ge$ Right side + Right side). $(a^2 + b^2) + (b^2 + c^2) + (c^2 + a^2) \ge 2ab + 2bc + 2ca$

Step 4: Simplify to reach the goal. Combine like terms on the left: $2a^2 + 2b^2 + 2c^2 \ge 2ab + 2bc + 2ca$

Divide the entire inequality by $2$ (since 2 is positive, the sign does not flip): $a^2 + b^2 + c^2 \ge ab + bc + ca$

Conclusion: The statement is proven for all real numbers $a, b, c$ .

mini-FAQ page

Q: Can I just write the inequality and say "True because squares are positive"? A: No! You must physically show the expansion from the axiom $(x-y)^2 \ge 0$ . You cannot just state the final result without proof unless the question says "Hence or otherwise".

Q: Can I subtract inequalities? A: Never. If $10 > 5$ and $3 > 1$ , subtracting them gives $10-3 > 5-1$ , which is $7 > 4$ (true). But if $10 > 5$ and $9 > 1$ , subtracting gives $10-9 > 5-1$ , which is $1 > 4$ (FALSE!). You can only add inequalities safely.

Common mistakes to avoid

Working backwards: Writing $a^2+b^2 \ge 2ab$ , then $a^2-2ab+b^2 \ge 0$ , then $(a-b)^2 \ge 0$ , and concluding "this is true so the start is true" is logically flawed. (It assumes the converse is true). You MUST start at $(a-b)^2 \ge 0$ and work forwards.
Multiplying by variables: If you have $\frac{1}{x} < y$ , do not multiply by $x$ to get $1 < xy$ . You don't know if $x$ is positive or negative! If it's negative, the sign should have flipped. Only multiply if you are given $x > 0$ .

Practice on Vu's Maths Hub

Mastering the $(x-y)^2 \ge 0$ trick is essential for the exam.

Practice building complex inequalities from simple squares in the HSC Proof Booklet.
Test your algebra skills with our interactive quizzes at Vu's Maths Hub.

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