Overview

Solving inequalities is fundamentally different from solving equations. While an equation usually gives you a specific destination (e.g., $x = 5$), an inequality defines a landscape of possibilities. Mastering inequalities for the AMC 8 and AMC 10 requires shifting your mindset from finding "the answer" to finding the boundaries of what is possible.

This module covers the core tools for bounding expressions, visualizing absolute values, recognizing symmetry, and utilizing the strict mathematical hierarchy of averages.

1. Interval and Number Line Reasoning

Before applying advanced theorems, you must know how to safely manipulate boundaries. Inequalities can be viewed as intervals (e.g., $x \in [-3, 4]$ means $-3 \le x \le 4$).

Adding and Subtracting Intervals

• Adding: You can safely add inequalities that face the same direction. If $a \le x \le b$ and $c \le y \le d$, then $a+c \le x+y \le b+d$.
• Subtracting: Never subtract inequalities directly. To find the range of $x - y$, you must first find the bounds of $-y$ (which flips the signs: $-d \le -y \le -c$), and then add that to the bounds of $x$.

Bounding Products (The Four-Corner Check)

Students often assume that if $a \le x \le b$ and $c \le y \le d$, then the minimum of $xy$ is $ac$ and the maximum is $bd$. This is only true if all numbers are positive. If signs are mixed, multiplying endpoints blindly will trap you. To find the true bounds of $xy$, you must calculate all four boundary products:

The minimum of $xy$ is the smallest of these four, and the maximum is the largest.

The "Plugging Endpoints" Strategy

If an expression is strictly linear in a variable (e.g., $3x - 7$), it has no curves, peaks, or valleys. It only goes strictly up or strictly down (monotonicity). Therefore, its absolute minimum and maximum on a given interval must occur exactly at the endpoints of that interval.

2. Absolute Value Inequalities (The Distance Masterkey)

Absolute value is one of the most frequently tested concepts on the AMC. While the algebraic definition is piece-wise ($|x| = x$ if $x \ge 0$, and $-x$ if $x < 0$), the geometric definition is far more powerful: $|x - c|$ is the distance between $x$ and $c$ on the number line.

Core Inequality Forms:

• The "Less Than" Trap (Inside the Radius): $|x| \le a \implies -a \le x \le a$. Geometric meaning: The distance from $x$ to $0$ is at most $a$. The solutions are "trapped" between $-a$ and $a$.

• The "Greater Than" Split (Outside the Radius): $|x| \ge a \implies x \le -a \text{ or } x \ge a$. Geometric meaning: The distance from $x$ to $0$ is at least $a$. The solutions split outward in two opposite directions.

Pro-Tip: $|x - 5| \le 3$ simply means "$x$ is within 3 units of 5." The range is instantly $[5-3, 5+3] \implies [2, 8]$.

3. Quadratics and The Trivial Inequality

The fundamental rule of real numbers is The Trivial Inequality: for any real number $x$, $x^2 \ge 0$. The minimum is strictly $0$, occurring when $x = 0$.

When dealing with quadratics like $ax^2 + bx + c$, you have two ways to find the extreme value (minimum if $a > 0$, maximum if $a < 0$):

Completing the Square: Group terms to form $a(x-h)^2 + k$. By the Trivial Inequality, the squared term's minimum is $0$, so the expression's minimum is $k$.

The Vertex Shortcut: For contest math, speed is critical. The vertex of a parabola always occurs at $x = -b/(2a)$. Simply calculate this $x$-value and plug it back into the expression to find the maximum or minimum bound.

4. Symmetry, Equalization, and Matching

Nature loves balance, and so do inequalities.

The Equalization Principle

When you have an expression with symmetric variables (meaning you could swap $x$ and $y$ and the expression doesn't change), the extreme values almost always occur when the variables are equal to each other.

• If $x + y = 10$, the product $xy$ is maximized when $x = y = 5$ (Max = 25).
• If $x + y = 10$, the sum of squares $x^2 + y^2$ is minimized when $x = y = 5$ (Min = 50).

Rearrangement-Lite (Matching Intuition)

To maximize the sum of products of two sequences, pair the largest with the largest. To minimize, pair the largest with the smallest. For example, if you have weights $3$ and $2$, and variables $a$ and $b$ where $a \ge b$, the sum $3a + 2b$ is strictly greater than or equal to $2a + 3b$.

5. The Mean Inequalities (QM-AM-GM-HM)

While you are likely familiar with the standard average (the Arithmetic Mean), mathematics actually defines several types of "means" or averages. For any set of $n$ strictly positive real numbers $x_1, x_2, \dots, x_n$, we define four primary means:

Quadratic Mean (QM / RMS): Also called the Root Mean Square. It squares the numbers before averaging them, then takes the square root.

Arithmetic Mean (AM): The standard average.

Geometric Mean (GM): Multiplies the numbers together and takes the $n$-th root.

Harmonic Mean (HM): The reciprocal of the average of the reciprocals.

The Inequality Chain:

There is a strict, unbreakable hierarchy among these averages:

The Equality Condition (Crucial):

Equality across any part of this chain holds if and only if all the variables are exactly equal to each other ($x_1 = x_2 = \dots = x_n$). If even one number is different, the inequalities are strictly greater than ($>$).

Why Does This Happen? The Quadratic Mean squares inputs, heavily exaggerating the impact of the largest numbers, making it the largest average. The Harmonic Mean uses reciprocals, so very small numbers create massive fractions that drag the average down, making it the smallest. AM and GM sit balanced in the middle.

Although we have introduced the entire QM-AM-GM-HM inequality, the core to focus on is AM-GM as they are by far the most frequent use case.

Strategy Playbook: Approaching AMC Inequalities

When you face an inequality problem, run through this checklist:

1. Is it a fixed sum looking for a product? $\implies$ Think Equalization Principle or AM-GM.
2. Is it a quadratic equation? $\implies$ Find the vertex $x = -b/2a$ or complete the square.
3. Is it an absolute value? $\implies$ Translate it into distance on a number line or split it into $\le$ and $\ge$ boundary cases.
4. Is it a linear expression bounded by an interval? $\implies$ Just test the endpoints.
5. Always Check the Equality Condition: If you find a minimum bound of 12, can the variables actually equal the numbers required to hit 12? If not, your bound is unreachable.

Worked Examples

Example 1: Bounding Products with Mixed Signs

Given $-2 \le x \le 5$ and $-3 \le y \le 4$, what is the range of possible values for $xy$?

Solution: We test the four corner products of the intervals:

• $(-2)(-3) = 6$
• $(-2)(4) = -8$
• $(5)(-3) = -15$
• $(5)(4) = 20$

The smallest product is $-15$ and the largest is $20$. Therefore, $-15 \le xy \le 20$.

Example 2: Absolute Value Distance

Solve the inequality $|2x - 3| \le 5$.

Solution: Method 1 (Algebraic): Trap the expression.

Add $3$ to all parts:

Divide by $2$:

Method 2 (Geometric): Rewrite as $|x - 1.5| \le 2.5$. The distance from $x$ to $1.5$ is at most $2.5$. Going left: $1.5 - 2.5 = -1$. Going right: $1.5 + 2.5 = 4$. The range is $[-1, 4]$.

Example 3: The Vertex Shortcut

Find the minimum value of $f(x) = 3x^2 - 12x + 19$.

Solution: Instead of completing the square, we use the vertex formula $x = -b/(2a)$. Here, $a = 3$ and $b = -12$.

The minimum occurs at $x=2$. Plug this back into the expression:

The minimum value is $7$.

Example 4: Classic AM-GM Optimization

If $x > 0$, find the minimum possible value of the expression $x^2 + 16/x^2$.

Solution: We are asked to minimize a sum of strictly positive terms. Notice that if we multiply the two terms, the variables cancel out: $x^2 \cdot (16/x^2) = 16$. A constant product is the universal trigger to use AM-GM.

Let $a = x^2$ and $b = 16/x^2$.

Verify the equality condition: Equality happens if and only if $a = b \implies x^2 = 16/x^2 \implies x^4 = 16$. Since $x > 0$, $x = 2$. Because a valid $x$ exists, the minimum value is confidently $8$.

Example 5: AM-HM for Reciprocal Sums

Let $a$ and $b$ be positive real numbers such that $a + b = 1$. Find the minimum value of $1/a + 1/b$.

Solution: We are given a constant sum and asked to minimize the sum of their reciprocals.

Apply the AM-HM inequality ($AM \ge HM$):

Substitute $a + b = 1$:

Multiply both sides by $(1/a + 1/b)$ and multiply by $2$:

Verify the equality condition: Equality holds when $a = b$. Since $a + b = 1$, this means $a = 0.5$ and $b = 0.5$. Testing this: $1/0.5 + 1/0.5 = 2 + 2 = 4$. The minimum value is $4$.

Common Pitfalls

• Applying AM-GM to negatives: The AM-GM inequality only works for positive numbers. Do not use it if variables can be negative.
• Squaring absolute values carelessly: If you have $|x| < |y|$, you can safely square both sides to get $x^2 < y^2$. But if you have $x < y$ with mixed signs, squaring might flip the truth (for example, $-5 < 2$, but $25$ is not less than $4$).
• Forgetting the middle of the interval when bounding squares: If $-4 < x < 3$, the range of $x^2$ is not $9 < x^2 < 16$. Because $0$ is in the interval, the true minimum is $0$, making the range $0 \le x^2 \le 16$.

Practice Problems