Ultimate Guide to Proof of Inequalities (Part 2): AM-GM and Calculus

Introduction

While the "squares are positive" axiom from Part 1 is powerful, it can be slow and clunky for massive expressions. In Proof of Inequalities (Part 2), we upgrade our toolkit. We will introduce the famous AM-GM inequality, allowing you to bypass tedious algebraic expansions. We will also explore how to use geometry, the limits of the Squeeze Theorem, and derivatives from Calculus to prove inequalities that pure algebra cannot touch.

Executive Summary

This guide covers advanced techniques for proving inequalities:

The AM-GM Inequality: Proving and applying the rule that the Arithmetic Mean is always greater than or equal to the Geometric Mean ( $\frac{a+b}{2} \ge \sqrt{ab}$ ).
Geometry Inequalities: Using areas of shapes (like rectangles under curves) to bound values.
The Squeeze Theorem: Finding the limit of a function by trapping it between two other functions.
Calculus Techniques: Using the first derivative ( $f'(x) > 0$ ) to prove a function is strictly increasing.

What is this about?

Sometimes algebra isn't enough. If asked to prove that $e^x > 1 + x$ for $x > 0$ , expanding squares won't help you because $e^x$ is a transcendental function, not a polynomial. Instead, you need calculus. By looking at the rate of change (the derivative), you can prove that $e^x$ grows faster than $1+x$ , and since they start at the same place when $x=0$ , $e^x$ must always be larger.

Similarly, the AM-GM inequality is a generalized shortcut. It takes the $(x-y)^2 \ge 0$ axiom and packages it into a ready-to-use formula that instantly links addition with multiplication, making it a favorite tool for Extension 2 examiners.

Main Content

1. The AM-GM Inequality

For any two non-negative numbers $a$ and $b$ , the Arithmetic Mean is greater than or equal to the Geometric Mean: $\frac{a+b}{2} \ge \sqrt{ab}$ Equality holds if and only if $a = b$ .

The Proof: We start with our core axiom for real numbers $\sqrt{a}$ and $\sqrt{b}$ : $(\sqrt{a} - \sqrt{b})^2 \ge 0$ $a - 2\sqrt{ab} + b \ge 0$ $a + b \ge 2\sqrt{ab}$ $\frac{a+b}{2} \ge \sqrt{ab}$

The true power of AM-GM is that you can chain it. If you have variables like $x^2$ and $\frac{1}{x^2}$ , applying AM-GM makes the variables cancel out under the square root!

2. Geometric Inequalities

Many inequalities can be proven by visualizing areas on a graph. The most common technique involves drawing a curve $y = f(x)$ and comparing the exact area under the curve (using integration) to the area of rectangles drawn underneath or above the curve.

Because the rectangles are either entirely contained within the curve or stick out above it, their areas are strictly less than (or greater than) the integral. This visually proves inequalities involving sums and integrals.

3. The Squeeze Theorem (Pinching Theorem)

The Squeeze Theorem is used to evaluate limits of tricky functions by "trapping" them. If you know that: $g(x) \le f(x) \le h(x)$ for all $x$ near $c$ , AND you can prove that: $\lim_{x \to c} g(x) = L \text{ and } \lim_{x \to c} h(x) = L$ Then $f(x)$ , trapped in the middle, must also have a limit of $L$ : $\lim_{x \to c} f(x) = L$

This is often used with trigonometric functions like $\sin(x)$ , because we know they are eternally trapped: $-1 \le \sin(x) \le 1$ .

4. Calculus Techniques

To prove $f(x) > g(x)$ for $x > a$ :

Create a Difference Function: Let $h(x) = f(x) - g(x)$ . Our goal is to prove $h(x) > 0$ .
Check the starting point: Show that $h(a) \ge 0$ . (The function starts at zero or above zero).
Find the derivative: Calculate $h'(x)$ .
Prove it is increasing: Show that $h'(x) > 0$ for all $x > a$ .
Conclude: If the function starts at $0$ and is constantly increasing, it MUST be strictly positive ( $h(x) > 0$ ) for all $x > a$ . Thus, $f(x) > g(x)$ .

Simple Worked Example

Question: Using the AM-GM inequality, prove that for any positive real numbers $x, y, z$ : $(x+y)(y+z)(z+x) \ge 8xyz$

Solution: Step 1: Apply AM-GM to pairs. Since $x, y, z$ are positive, we can apply AM-GM: $\frac{a+b}{2} \ge \sqrt{ab} \implies a+b \ge 2\sqrt{ab}$ .

Apply it to $x$ and $y$ :

$x + y \ge 2\sqrt{xy}$

Apply it to $y$ and $z$ : 2) $y + z \ge 2\sqrt{yz}$

Apply it to $z$ and $x$ : 3) $z + x \ge 2\sqrt{zx}$

Step 2: Multiply the inequalities. Because all terms ( $x, y, z$ ) are positive, the expressions on both sides of the inequalities are positive. Therefore, we can safely multiply the three inequalities together without flipping the signs.

Left Side: $(x+y) \times (y+z) \times (z+x)$ Right Side: $2\sqrt{xy} \times 2\sqrt{yz} \times 2\sqrt{zx}$

$(x+y)(y+z)(z+x) \ge 8 \sqrt{xy \cdot yz \cdot zx}$

Step 3: Simplify the Right Side. Combine the terms under the square root: $\sqrt{xy \cdot yz \cdot zx} = \sqrt{x^2 y^2 z^2} = xyz$ (since $x,y,z$ are positive).

Substitute this back: $(x+y)(y+z)(z+x) \ge 8xyz$

Conclusion: The statement is proven.

mini-FAQ page

Q: Can I use AM-GM if the numbers are negative? A: No! AM-GM only applies to non-negative numbers. If you try to use it on negative numbers, you will end up square-rooting a negative (an imaginary number), which breaks the entire concept of inequality (you can't say $3i > 2i$ ).

Q: How many times do I have to differentiate for the calculus method? A: Usually just once ( $h'(x)$ ). But if $h'(x)$ is still too complex to prove it is $>0$ , you can differentiate again to find $h''(x)$ . If $h''(x) > 0$ , then $h'(x)$ is increasing, etc. This is called the 'cascading derivative' technique.

Common mistakes to avoid

Forgetting the starting point in calculus proofs: Just showing $h'(x) > 0$ proves the function is going up. But what if it started at $-10$ ? It could be going up and still be negative! You MUST explicitly evaluate $h(a) \ge 0$ before concluding $h(x) > 0$ .
Not checking for equality conditions in AM-GM: The question might ask "When does equality hold?". For AM-GM, equality only holds when the terms are identical. In the example above, equality holds when $x = y$ , $y = z$ , and $z = x$ , which means $x = y = z$ .

Practice on Vu's Maths Hub

AM-GM and Calculus proofs are the bread and butter of Band 6 students.

Practice chaining AM-GM inequalities in the HSC Proof Booklet.
Test your integration vs. rectangle area geometry in the HSC Integration Booklet.

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Want to master Mathematical Proof and guarantee top marks in HSC Mathematics Extension 2? Visit Vu's Maths Hub for in-depth booklets, rigorous worked solutions, and expert advice to help you ace your exams!