1.3 Inequalities and Absolute Value Equations

1.3 Inequalities and Absolute Value Equations

Inequalities lie at the core of algebraic manipulation on the Gaokao examination. They are essential not only in isolation but also across domain and range derivations, optimization problems, function monotonicity analysis, and calculus bounding. Mastering quadratic inequalities, rational expressions, absolute value partitioning, and the Mean Value (AM-GM) inequality provides a versatile toolkit for solving complex algebraic problems.

1. Fundamental Properties of Inequalities

Algebraic operations on inequalities are governed by strict ordering rules:

1. Transitivity: $a > b \text{ and } b > c \implies a > c$.
2. Addition Property: $a > b \implies a + c > b + c$.
3. Multiplication by positive/negative numbers:
   • If $c > 0$, $a > b \implies ac > bc$.
   • If $c < 0$, $a > b \implies ac < bc$ (direction reverses!).
4. Additive Combination: $a > b \text{ and } c > d \implies a + c > b + d$.
5. Multiplicative Combination (for positive numbers): If $a > b > 0$ and $c > d > 0$, then $ac > bd$.
6. Power and Reciprocal Rules: If $a > b > 0$, then $a^n > b^n > 0$ ($n \in \mathbb{N}^*$) and $\frac{1}{a} < \frac{1}{b}$.

2. Quadratic Inequalities and Discriminant Analysis

A standard quadratic inequality takes the form $ax^2 + bx + c > 0$ or $ax^2 + bx + c < 0$ (assuming $a > 0$ without loss of generality by multiplying by $-1$ if needed).

Let $\Delta = b^2 - 4ac$ be the discriminant of the corresponding quadratic equation $ax^2 + bx + c = 0$:

1. Case 1: $\Delta > 0$ The quadratic equation has two distinct real roots $x_1 < x_2$.
   • $ax^2 + bx + c > 0 \implies x < x_1 \text{ or } x > x_2$ (interval: $(-\infty, x_1) \cup (x_2, +\infty)$).
   • $ax^2 + bx + c < 0 \implies x_1 < x < x_2$ (interval: $(x_1, x_2)$).
2. Case 2: $\Delta = 0$ The quadratic equation has a repeated real root $x_1 = x_2 = -\frac{b}{2a}$.
   • $ax^2 + bx + c > 0 \implies x \neq -\frac{b}{2a}$ (interval: $\mathbb{R} \setminus {-b/(2a)}$).
   • $ax^2 + bx + c < 0 \implies \varnothing$.
3. Case 3: $\Delta < 0$ The parabola lies entirely above the x-axis.
   • $ax^2 + bx + c > 0 \implies \mathbb{R}$ (all real numbers).
   • $ax^2 + bx + c < 0 \implies \varnothing$.

3. Rational and Higher-Degree Polynomial Inequalities

Rational Inequalities (分式不等式)

Rational inequalities can be converted into sign-equivalent polynomial inequalities while enforcing denominator non-zero conditions:

1. $\frac{f(x)}{g(x)} > 0 \iff f(x) \cdot g(x) > 0$
2. $\frac{f(x)}{g(x)} \le 0 \iff f(x) \cdot g(x) \le 0 \text{ and } g(x) \neq 0$

Number-Line Root-Passing Method (穿针引线法 / 序轴标根法)

To solve higher-degree polynomial inequalities $P(x) = a_n(x - x_1)(x - x_2)\dots(x - x_k) > 0$ ($a_n > 0$):

1. Ensure the leading coefficient of every factor $(x - x_i)$ is positive.
2. Mark all roots $x_1 < x_2 < \dots < x_k$ on the number line.
3. Draw a continuous curve starting from the top-right (above the number line for $x > x_k$).
4. Pass through each root: for odd-multiplicity roots, the curve passes through the axis; for even-multiplicity roots, the curve touches the axis and bounces back ("奇过偶不过").
5. Select regions above the line for $> 0$ and below for $< 0$.

4. Absolute Value Equations and Inequalities

Geometric Interpretation

$|x - a|$ represents the distance between $x$ and $a$ on the real number line.

Basic Algebraic Formulations ($a > 0$)

• $|f(x)| < a \iff -a < f(x) < a$
• $|f(x)| > a \iff f(x) > a \text{ or } f(x) < -a$

Zero-Point Interval Division Method (零点分段法)

For expressions containing multiple absolute values, such as $|x - a| + |x - b| \ge c$:

1. Find all critical zero points where absolute value arguments change sign ($x = a, x = b$).
2. Divide the real number line into disjoint sub-intervals.
3. Remove absolute value signs within each sub-interval based on local signs, solve the resulting linear/quadratic inequality, and intersect the solution with the sub-interval.
4. Take the union of solutions across all sub-intervals.

Absolute Value Triangle Inequalities (绝对值三角不等式)

For any real numbers $a$ and $b$: $||a| - |b|| \le |a \pm b| \le |a| + |b|$

• Equality $|a + b| = |a| + |b|$ holds if and only if $ab \ge 0$ (same sign).
• Equality $|a - b| = |a| + |b|$ holds if and only if $ab \le 0$ (opposite sign).

5. The Mean Value (AM-GM) Inequality (基本不等式)

For any positive real numbers $a, b > 0$: $\frac{a + b}{2} \ge \sqrt{ab}$ Equality holds if and only if $a = b$.

Three Essential Application Requirements ("一正二定三相等")

1. One Positive (一正): Both terms $a$ and $b$ must be strictly positive ($a > 0, b > 0$).
2. Two Constant (二定):
   • To find the minimum of $a + b$, their product $ab$ must be constant ($ab = k \implies a + b \ge 2\sqrt{k}$).
   • To find the maximum of $ab$, their sum $a + b$ must be constant ($a + b = s \implies ab \le \frac{s^2}{4}$).
3. Three Equal (三相等): The values of $a$ and $b$ at the boundary point where equality holds must fall within their allowable domains.

Summary of Inequality Solutions and Operational Rules

6. Worked Gaokao Exam Examples

Worked Example 1: Multi-Absolute Value Inequality via Zero-Point Division

Problem: Solve the inequality $|x + 1| + |x - 2| \ge 5$ for $x \in \mathbb{R}$.

Solution:

• Step 1: Locate zero points. Setting $x + 1 = 0 \implies x = -1$, and $x - 2 = 0 \implies x = 2$. The zero points $-1$ and $2$ divide $\mathbb{R}$ into three sub-intervals: $(-\infty, -1)$, $[-1, 2]$, and $(2, +\infty)$.
• Step 2: Solve in each sub-interval.
  • Sub-interval 1: $x < -1$ In this region, $x + 1 < 0$ and $x - 2 < 0$. Stripping absolute values: $-(x + 1) - (x - 2) \ge 5 \implies -x - 1 - x + 2 \ge 5 \implies -2x + 1 \ge 5 \implies -2x \ge 4 \implies x \le -2$ Intersecting $x \le -2$ with $x < -1$ gives $x \in (-\infty, -2]$.
  • Sub-interval 2: $-1 \le x \le 2$ In this region, $x + 1 \ge 0$ and $x - 2 \le 0$. Stripping absolute values: $(x + 1) - (x - 2) \ge 5 \implies x + 1 - x + 2 \ge 5 \implies 3 \ge 5$ This is a contradiction. No solutions exist in this interval.
  • Sub-interval 3: $x > 2$ In this region, $x + 1 > 0$ and $x - 2 > 0$. Stripping absolute values: $(x + 1) + (x - 2) \ge 5 \implies 2x - 1 \ge 5 \implies 2x \ge 6 \implies x \ge 3$ Intersecting $x \ge 3$ with $x > 2$ gives $x \in [3, +\infty)$.
• Step 3: Combine sub-interval solutions. Taking the union of results from all three regions: $x \in (-\infty, -2] \cup [3, +\infty)$

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Worked Example 2: AM-GM Inequality Optimization under Linear Constraint

Problem: Given positive real numbers $x, y > 0$ satisfying $x + 2y = 4$, find the minimum value of $\frac{1}{x} + \frac{2}{y}$.

Solution:

• Step 1: Apply the 'Multiply by 1' technique. Since $x + 2y = 4$, we have $\frac{x + 2y}{4} = 1$. Multiply the target expression by $\frac{x + 2y}{4}$: $\frac{1}{x} + \frac{2}{y} = \left( \frac{1}{x} + \frac{2}{y} \right) \cdot 1 = \left( \frac{1}{x} + \frac{2}{y} \right) \cdot \frac{x + 2y}{4}$
• Step 2: Expand the product. $\frac{1}{4} \left( \frac{x}{x} + \frac{2y}{x} + \frac{2x}{y} + \frac{4y}{y} \right) = \frac{1}{4} \left( 1 + 4 + \frac{2y}{x} + \frac{2x}{y} \right) = \frac{1}{4} \left( 5 + \frac{2y}{x} + \frac{2x}{y} \right)$
• Step 3: Apply AM-GM inequality to the variable terms. Since $x > 0$ and $y > 0$, the terms $\frac{2y}{x}$ and $\frac{2x}{y}$ are strictly positive. By AM-GM: $\frac{2y}{x} + \frac{2x}{y} \ge 2 \sqrt{\frac{2y}{x} \cdot \frac{2x}{y}} = 2 \sqrt{4} = 4$
• Step 4: Substitute back and check equality condition. $\frac{1}{x} + \frac{2}{y} \ge \frac{1}{4} (5 + 4) = \frac{9}{4}$ Equality holds if and only if $\frac{2y}{x} = \frac{2x}{y} \implies x^2 = y^2 \implies x = y$. Substituting $x = y$ into $x + 2y = 4 \implies 3x = 4 \implies x = \frac{4}{3}, y = \frac{4}{3}$. Since $x = \frac{4}{3} > 0$ and $y = \frac{4}{3} > 0$, the equality condition is achievable.
• Conclusion: The minimum value is $\frac{9}{4}$.