📚 A-Level CCEA Mathematics: Inequalities Comprehensive Revision | A-Level CCEA 数学：不等式 考点精讲
Inequalities are statements that compare two expressions using symbols like <, >, ≤, ≥. In CCEA A-Level Mathematics, mastering inequalities is essential not only for solving standalone problems but also for applying them in calculus, coordinate geometry, and optimisation. This article systematically covers every key type of inequality you will encounter, with clear methods, examples, and common pitfalls.
不等式是使用 <、>、≤、≥ 等符号比较两个表达式的陈述。在 CCEA A-Level 数学中，掌握不等式不仅是为了解决独立问题，更是为了在微积分、坐标几何和优化问题中应用它们。本文系统梳理了你将遇到的每一类关键不等式，提供清晰的解法、示例和常见错误。
1. Linear Inequalities | 线性不等式
Linear inequalities involve expressions of the form ax + b < c, where a ≠ 0. Solving them follows similar rules to linear equations, with one critical exception: multiplying or dividing both sides by a negative number reverses the inequality sign.
线性不等式涉及形如 ax + b < c 的表达式，其中 a ≠ 0。解法与线性方程类似，但有一个关键例外：两边乘以或除以负数时，不等号方向要反转。
Example: Solve 5 − 2x ≥ 3x + 10. First collect x terms: −2x − 3x ≥ 10 − 5, giving −5x ≥ 5. Divide by −5 and reverse the sign: x ≤ −1. Always express the solution as a set or on a number line: {x ∈ ℝ | x ≤ −1}.
例如：解 5 − 2x ≥ 3x + 10。先合并含 x 的项：−2x − 3x ≥ 10 − 5，得 −5x ≥ 5。除以 −5 并反转符号：x ≤ −1。通常将解表示为集合或标在数轴上：{x ∈ ℝ | x ≤ −1}。
If a linear inequality contains brackets or fractions, clear them first. For instance, (2x − 1)/3 > (x + 4)/2 becomes 2(2x − 1) > 3(x + 4) after multiplying both sides by 6. Solve to obtain x > 14.
若线性不等式包含括号或分数，先化简。例如 (2x − 1)/3 > (x + 4)/2，两边同乘 6 得 2(2x − 1) > 3(x + 4)，解得 x > 14。
2. Quadratic Inequalities | 二次不等式
Quadratic inequalities have the form ax² + bx + c < 0, ≤ 0, > 0, or ≥ 0 with a ≠ 0. The safe approach involves finding the roots of the corresponding quadratic equation, sketching the graph, and determining where the quadratic is positive or negative.
二次不等式形如 ax² + bx + c < 0、≤ 0、> 0 或 ≥ 0，其中 a ≠ 0。稳妥的方法是求出对应二次方程的根，画出图像草图，确定二次式为正或负的区间。
Step-by-step: (1) Rearrange to 0 on one side. (2) Solve ax² + bx + c = 0 by factorising or using the quadratic formula. (3) Sketch the parabola: if a > 0 it opens upward; if a < 0 it opens downward. (4) Identify where the graph lies above or below the x-axis as required.
分步解法：(1) 移项使一边为 0。(2) 通过因式分解或求根公式解 ax² + bx + c = 0。(3) 画抛物线草图：若 a > 0 开口向上；若 a < 0 开口向下。(4) 根据题意确定图像在 x 轴上方或下方的区间。
Example: Solve x² − 5x + 6 > 0. Factor: (x − 2)(x − 3) > 0. Roots at x = 2 and x = 3. Parabola opens upward, so it is positive outside the roots: x < 2 or x > 3. Set notation: {x < 2} ∪ {x > 3}.
例：解 x² − 5x + 6 > 0。因式分解：(x − 2)(x − 3) > 0。根为 x = 2 和 x = 3。抛物线开口向上，故在根的外侧为正：x < 2 或 x > 3。集合表示：{x < 2} ∪ {x > 3}。
If the quadratic has no real roots (discriminant < 0), then it is either always positive or always negative, depending on the sign of a. For instance, x² + x + 1 > 0 is true for all real x because discriminant = −3 and a = 1 > 0.
若二次式无实根（判别式 < 0），则由 a 的符号决定其恒正或恒负。例如 x² + x + 1 > 0 对所有实数 x 成立，因为判别式 = −3 且 a = 1 > 0。
3. Polynomial Inequalities of Higher Degree | 高次多项式不等式
For cubic or higher-degree inequalities, we build a sign table after factorising the polynomial into linear and/or quadratic factors. The key is to find the critical values (roots) and test the sign of the expression in each interval formed on the number line.
对于三次或更高次的不等式，我们将多项式分解为一次和/或二次因式后构建符号表。关键是找出临界值（根），并在数轴上分成的各区间内检验表达式的符号。
Procedure: (1) Move all terms to one side to get P(x) ≥ 0 or P(x) < 0 etc. (2) Fully factorise P(x). (3) Find all roots, including repeated ones, and mark them on a number line. (4) Determine the sign of P(x) by picking a test value from each interval. (5) Write the solution as a union of intervals where the inequality holds.
步骤：(1) 将所有项移到一边，得到 P(x) ≥ 0 或 P(x) < 0 等形式。(2) 将 P(x) 完全分解。(3) 求出所有根（包括重根），并在数轴上标出。(4) 从每个区间选一个测试值确定 P(x) 的符号。(5) 将解写为满足不等式的区间的并集。
Example: Solve x³ − 4x ≤ 0. Factor as x(x − 2)(x + 2) ≤ 0. Roots: x = −2, 0, 2. Test intervals: (−∞, −2) → negative; (−2, 0) → positive; (0, 2) → negative; (2, ∞) → positive. Since ≤ 0, include intervals where negative or zero: x ≤ −2 or 0 ≤ x ≤ 2.
例：解 x³ − 4x ≤ 0。分解得 x(x − 2)(x + 2) ≤ 0。根：x = −2, 0, 2。测试各区间：(−∞, −2) → 负；( −2, 0) → 正；(0, 2) → 负；(2, ∞) → 正。因为 ≤ 0，取负值及零的区间：x ≤ −2 或 0 ≤ x ≤ 2。
Repeated roots require care: if a factor is raised to an even power, the sign does not change across that root; if odd, it does change.
重根需特别注意：若因式的幂次为偶数，穿过该根时符号不变；若为奇数，符号改变。
4. Rational Inequalities | 分式不等式
A rational inequality involves a fraction with polynomials in the numerator and denominator, like (ax + b)/(cx + d) ≥ k. Never multiply through by the denominator unless you are certain of its sign; instead, bring all terms to one side to form a single fraction and compare to zero.
分式不等式涉及分子和分母中含有多项式的分式，例如 (ax + b)/(cx + d) ≥ k。除非你确定分母的符号，否则切勿直接乘以分母消去分式；应将所有项移到一边，化为单个分式与零比较。
Method: (1) Rearrange to f(x)/g(x) ≥ 0 or ≤ 0, with 0 on the right. (2) Find critical values: where numerator f(x) = 0 and where denominator g(x) = 0 (these are excluded from the solution). (3) Use a sign table to test intervals. (4) Write solution, excluding values that make the denominator zero.
方法：(1) 移项化为 f(x)/g(x) ≥ 0 或 ≤ 0 的形式，右边为 0。(2) 找出临界值：分子 f(x) = 0 的点，以及分母 g(x) = 0 的点（这些点不被包含在解集中）。(3) 使用符号表检验各区间。(4) 写出解，排除使分母为零的值。
Example: Solve (x + 2)/(x − 3) ≤ 0. Critical values: x = −2 (numerator zero) and x = 3 (denominator zero, always excluded). Intervals: (−∞, −2] test x = −3 → positive over negative = negative; [−2, 3) test x = 0 → positive over negative = negative; (3, ∞) test x = 4 → positive over positive = positive. The inequality ≤ 0 requires negative or zero. Thus solution: −2 ≤ x < 3. Note x cannot be 3.
例：解 (x + 2)/(x − 3) ≤ 0。临界值：x = −2（分子为零）和 x = 3（分母为零，总被排除）。各区间：( −∞, −2] 取 x = −3 → 正除以负 = 负；[−2, 3) 取 x = 0 → 正除以负 = 负；(3, ∞) 取 x = 4 → 正除以正 = 正。不等式 ≤ 0 要求负或零。因此解为 −2 ≤ x < 3。注意 x 不能等于 3。
Always check endpoints carefully: numerator zeros that do not cancel with denominator zeros are included when the inequality allows equality (≤, ≥).
务必仔细检查端点：当不等式允许等号（≤, ≥）时，不与分母零点相消的分子零点应包含在解中。
5. Absolute Value Inequalities | 绝对值不等式
Absolute value inequalities involve expressions like |x − a| < b or |x − a| ≥ b. They can be interpreted geometrically as distance on the number line or solved analytically by considering cases based on the definition of absolute value.
绝对值不等式涉及形如 |x − a| < b 或 |x − a| ≥ b 的表达式。可从几何上将其解释为数轴上的距离，或根据绝对值的定义分情况解析求解。
Basic equivalences: |x| < k (k > 0) ⇔ −k < x < k; |x| > k ⇔ x < −k or x > k. For |x − a| ≤ b, it becomes −b ≤ x − a ≤ b, giving a − b ≤ x ≤ a + b.
基本等价关系：|x| < k（k > 0）⇔ −k < x < k；|x| > k ⇔ x < −k 或 x > k。对于 |x − a| ≤ b，可化为 −b ≤ x − a ≤ b，得 a − b ≤ x ≤ a + b。
Example: Solve |2x − 3| > 5. This gives 2x − 3 < −5 or 2x − 3 > 5. Solving: 2x < −2 → x < −1, or 2x > 8 → x > 4. Solution: x < −1 or x > 4.
例：解 |2x − 3| > 5。等价于 2x − 3 < −5 或 2x − 3 > 5。求解得：2x < −2 → x < −1，或 2x > 8 → x > 4。解为 x < −1 或 x > 4。
For inequalities like |x + 1| < |x − 3|, squaring both sides is often simpler because both sides are non-negative. Squaring yields (x+1)² < (x−3)², which simplifies to a linear inequality.
对于 |x + 1| < |x − 3| 这类不等式，两边同时平方通常更简单，因为两边非负。平方得 (x+1)² < (x−3)²，化简后得到一个线性不等式。
More complex absolute value expressions may need a case analysis based on where the expressions inside the absolute value change sign. Draw a sign diagram for the interior functions to define intervals.
更复杂的绝对值表达式可能需要根据内部表达式符号变化的位置进行分段讨论。为内部函数绘制符号图以确定各区间。
6. Graphical Representation of Inequalities | 不等式的图形表示
Inequalities in two variables, such as y > 2x + 1 or x² + y² ≤ 25, describe regions in the Cartesian plane. The boundary line or curve is drawn as a solid line if the inequality includes equality (≤, ≥), or dashed if it is strict (<, >). The required region is then shaded.
涉及两个变量的不等式，如 y > 2x + 1 或 x² + y² ≤ 25，描述的是笛卡尔平面上的区域。若不等式包含等号（≤, ≥），边界线或曲线画为实线；若是严格不等式（<, >），则画为虚线。随后将所求区域涂上阴影。
Procedure: (1) Graph the boundary equation replacing the inequality sign with =. (2) Determine if the line is solid or dashed. (3) Choose a test point not on the boundary (usually (0,0) if possible) and substitute into the inequality. If it satisfies the inequality, shade the side containing the test point; otherwise shade the opposite side.
步骤：(1) 将不等号换成等号，画出边界方程图像。(2) 确定边界为实线还是虚线。(3) 选择一个不在边界上的测试点（如可能，通常用 (0,0)），代入原不等式。若满足，则涂上包含测试点的一侧；否则涂另一侧。
Example: Shade the region satisfying y ≤ −½x + 3, x ≥ 0, y ≥ 0. This is a triangle bounded by the line, the y-axis, and the x-axis in the first quadrant. Shade the area below the line and to the right of the axes.
例：绘出满足 y ≤ −½x + 3、x ≥ 0、y ≥ 0 的区域。这是一个由该直线、y 轴和 x 轴在第一象限围成的三角形。涂上直线下方及坐标轴右侧的区域。
For quadratic inequalities like y > x² − 4, shade the region above the parabola because y is greater than the expression. For circle inequalities, remember x² + y² < r² represents the interior of the circle, not including the boundary.
对于如 y > x² − 4 的二次不等式，由于 y 大于表达式，涂上抛物线上方的区域。对于圆的不等式，x² + y² < r² 表示圆的内部区域，不包括边界。
7. Systems of Linear Inequalities and Linear Programming | 线性不等式组与线性规划
When multiple linear inequalities in two variables are given simultaneously, the feasible region is the intersection of all half-planes defined by the inequalities. This is a convex polygon (if bounded), and its vertices can be found by solving the boundary equations pairwise.
当同时给出多个关于两个变量的线性不等式时，可行区域是所有不等式定义的半平面的交集。它是一个凸多边形（若有界），其顶点可通过两两联立边界方程求得。
Linear programming is an application where we maximise or minimise a linear objective function over the feasible region. The optimal value always occurs at a vertex (corner point) of the feasible region.
线性规划是一种应用，我们在可行区域上对线性目标函数进行最大化或最小化。最优值总出现在可行区域的顶点（角点）上。
Steps for linear programming: (1) Define variables and write constraints as inequalities. (2) Graph the feasible region. (3) Find the coordinates of all vertices. (4) Evaluate the objective function at each vertex. (5) Select the vertex giving the maximum or minimum value as required.
线性规划步骤：(1) 定义变量，写出约束条件（不等式）。(2) 绘出可行区域。(3) 求出所有顶点坐标。(4) 计算目标函数在每个顶点的值。(5) 根据题意选择给出最大值或最小值的顶点。
Example: Maximise P = 3x + 2y subject to x + y ≤ 10, x ≥ 0, y ≥ 0, and y ≤ 2x. Vertices are (0,0), (10,0) (not in feasible due to y ≤ 2x? Check: y=0, 0≤2*10 true, so (10,0) is feasible. Intersection of x+y=10 and y=2x gives (10/3, 20/3). Also (0,0). Evaluate P at each: (0,0)→0; (10,0)→30; (10/3, 20/3)→ 3*10/3+2*20/3 = 10+40/3=70/3≈23.33. So maximum is 30 at (10,0). Ensure all constraints are met. Indeed, (10,0) satisfies y≤2x (0≤20). So maximum 30.
例：在约束 x + y ≤ 10，x ≥ 0，y ≥ 0，y ≤ 2x 下最大化 P = 3x + 2y。顶点有 (0,0)，(10,0)（检查 y≤2x: 0≤20，可行）。解 x+y=10 与 y=2x 得 (10/3, 20/3)。计算 P：(0,0)→0；(10,0)→30；(10/3, 20/3)→ 10+40/3=70/3≈23.33。故最大值在 (10,0) 处为 30。确认所有约束满足。
8. Proofs Involving Inequalities | 涉及不等式的证明
At A-Level, you may be asked to prove simple inequalities, often using algebraic manipulation, completing the square, or known results like the AM–GM inequality. A common technique is to start from something known to be true and derive the desired inequality, or to show that the difference of two expressions is always non-negative.
在 A-Level 考试中，你可能需要证明简单的不等式，常通过代数操作、配方法或已知结果（如 AM–GM 不等式）。常用技巧是从已知为真的事实出发推出目标不等式，或证明两表达式之差恒为非负。
For example, to prove that x² + y² ≥ 2xy for all real x, y, consider (x − y)² ≥ 0. Expanding gives x² − 2xy + y² ≥ 0, which rearranges to x² + y² ≥ 2xy. This is a simple but powerful method.
例如，证明对所有实数 x、y 有 x² + y² ≥ 2xy。考虑 (x − y)² ≥ 0，展开得 x² − 2xy + y² ≥ 0，移项即得 x² + y² ≥ 2xy。这是一个简单却有力的方法。
Another useful inequality: for positive numbers, the arithmetic mean is greater than or equal to the geometric mean (AM ≥ GM). For two numbers a, b > 0, (a + b)/2 ≥ √(ab). This can be proved by squaring both sides: (a+b)² ≥ 4ab → a² + 2ab + b² ≥ 4ab → a² − 2ab + b² ≥ 0 → (a − b)² ≥ 0, which is true.
另一个有用不等式：对正数而言，算术平均值大于或等于几何平均值（AM ≥ GM）。对两正数 a、b，有 (a + b)/2 ≥ √(ab)。可通过两边平方证明：(a+b)² ≥ 4ab → a² + 2ab + b² ≥ 4ab → a² − 2ab + b² ≥ 0 → (a − b)² ≥ 0，此式成立。
When proving, always state clearly the starting assumption or known fact and show logical steps. Ensure any squaring steps preserve the inequality direction (valid when both sides are non-negative).
证明时，应清晰陈述起始假设或已知事实，并展示逻辑推导步骤。确保任何平方步骤保持不等号方向（当两边均非负时成立）。
9. Common Mistakes and Tips | 常见错误与技巧
One of the most frequent errors is forgetting to reverse the inequality sign when multiplying or dividing by a negative number. Always check the sign of the multiplier.
最常见的错误之一是在乘以或除以负数时忘记反转不等号。务必检查乘数的符号。
Another pitfall is mishandling denominators in rational inequalities. Never cross-multiply without considering the sign of the denominator; instead, bring everything to one side and use a sign table.
另一个陷阱是在分式不等式中错误处理分母。切勿在不考虑分母符号的情况下交叉相乘；应将所有项移到一边并使用符号表。
With quadratic inequalities, students sometimes incorrectly write the solution as a single interval when the graph indicates two separate regions. Sketching a quick parabola helps avoid this.
对于二次不等式，有时学生错误地将解写成单个区间，而图像显示的是两个分离的区域。快速画出抛物线草图可避免此错。
When dealing with absolute values, remember that |x − a| < b is equivalent to a compound inequality, while |x − a| > b splits into two separate inequalities joined by ‘or’.
处理绝对值时，记住 |x − a| < b 等价于一个复合不等式，而 |x − a| > b 则拆分为两个用“或”连接的不等式。
In graphical inequalities, test points must lie off the boundary. Using (0,0) is efficient unless the boundary passes through the origin; then choose another simple point like (1,0) or (0,1).
在图形不等式中，测试点必须不在边界上。使用 (0,0) 很高效，除非边界经过原点；此时选择其他简单点，如 (1,0) 或 (0,1)。
When solving systems by linear programming, read the question carefully to determine whether the objective is to maximise or minimise, and verify that vertices satisfy all constraints—including non-negativity.
用线性规划解不等式组时，仔细审题以确定目标是最大化还是最小化，并验证各顶点满足所有约束——包括非负条件。
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