Inequality Properties and Proof Methods | 不等式性质与证明方法

📚 Inequality Properties and Proof Methods | 不等式性质与证明方法

Inequalities are fundamental in A-Level Mathematics. They appear in algebra, calculus, sequences, and geometry. Understanding their properties and mastering proof techniques is essential for solving both pure mathematics and applied problems.

不等式是 A-Level 数学的基础内容。它们出现在代数、微积分、数列和几何中。理解不等式的性质并掌握证明方法，对于解决纯数学与应用数学问题都至关重要。

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1. Basic Properties of Inequalities | 不等式的基本性质

Let a, b, c be real numbers. The following properties hold true when manipulating inequalities.

设 a, b, c 为实数。在处理不等式时，以下性质始终成立。

• If a < b, then a + c < b + c. Adding the same number to both sides preserves the inequality.

• 如果 a < b，则 a + c < b + c。两边加上同一个数，不等式方向不变。

• If a < b and c > 0, then ac < bc. Multiplying by a positive number preserves the direction.

• 如果 a < b 且 c > 0，则 ac < bc。乘以正数，不等号方向不变。

• If a < b and c < 0, then ac > bc. Multiplying by a negative number reverses the inequality.

• 如果 a < b 且 c < 0，则 ac > bc。乘以负数，不等号方向必须反向。

• If 0 < a < b, then 1/a > 1/b. For positive numbers, taking reciprocals reverses the inequality.

• 如果 0 < a < b，则 1/a > 1/b。对于正数，取倒数后不等号方向反向。

a < b and b < c ⇒ a < c (Transitivity)

Transitivity allows us to chain inequalities, a core step in many proofs.

传递性允许我们将多个不等式串联起来，这是许多证明中的核心步骤。

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2. Intervals and Solution Sets | 区间与解集

Inequality solutions are often expressed using interval notation. Mastery of this notation is required for A-Level exams.

不等式的解通常用区间表示。掌握区间记号是 A-Level 考试的基本要求。

Inequality	Interval	Number Line
x ≥ 2	[2, ∞)	closed dot at 2, arrow right
−3 < x < 5	(−3, 5)	open dots at −3 and 5
x ≤ 0	(−∞, 0]	closed dot at 0, arrow left

When solving quadratic inequalities, always sketch the graph or test critical points to determine the correct interval.

求解二次不等式时，务必画图或检验关键点，以确定正确的区间。

(x − 2)(x + 3) > 0 ⇒ x < −3 or x > 2

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3. Comparison by Difference | 比较法：差值法

To prove a > b, a common strategy is to show a − b > 0. This is called the difference method or comparison method.

要证明 a > b，常用的策略是说明 a − b > 0。这就是差值法，也称作比较法。

Example: Prove that for all real x, x² + 1 ≥ 2x.

例：证明对所有实数 x，x² + 1 ≥ 2x。

x² + 1 − 2x = (x − 1)² ≥ 0

Since every square is non-negative, the inequality follows immediately.

因为任何平方数都非负，所以原不等式立即成立。

This method is particularly effective when the expression can be factorized into a perfect square or a sum of squares.

当表达式可以化为完全平方或平方和时，这种方法尤其有效。

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4. Synthetic and Analytic Methods | 综合法与分析法

The synthetic method starts from known facts and derives the desired inequality step by step. The analytic method works backwards: assume the conclusion is true, trace it back to a true statement, then reverse the steps.

综合法是从已知事实出发，逐步推出所需不等式。分析法则是反过来：先假设结论成立，追溯到某个真命题，然后再反向写出证明。

Analytic example: To prove √3 < 7/4, we can square both sides because both sides are positive.

分析示例：要证明 √3 < 7/4，因为两边均为正数，可以两边平方。

3 < 49/16 ⇔ 48 < 49 ✓

Since 48 < 49 is true, the original inequality is true. However, we must verify that every step is reversible.

因为 48 < 49 为真，所以原不等式为真。但必须确保每一步都是可逆的。

In exams, analytic work is often used as scratch thought, while the final answer should be presented synthetically.

在考试中，分析法常用于草稿思考，而最终答案应写成综合法形式。

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5. Proof by Contradiction | 反证法

To prove an inequality, we can assume the opposite inequality is true and show that this leads to an impossible conclusion.

证明不等式时，我们可以假设相反的不等式成立，然后推出一个不可能的结论。

Example: Prove that there is no largest positive real number.

例：证明不存在最大的正实数。

Assume N is the largest positive real number. Then consider N + 1, which is positive and larger than N, a contradiction.

假设 N 是最大的正实数。考虑 N + 1，它是正数且大于 N，矛盾。

For inequalities, a common contradiction setup is assuming a < b and later finding a ≥ b from a chain of valid deductions.

对于不等式，常见的反设是假设 a < b，然后通过一串有效推理最终得到 a ≥ b，从而产生矛盾。

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6. AM-GM Inequality | 均值不等式

The Arithmetic Mean – Geometric Mean (AM-GM) inequality states that for non-negative real numbers x₁, x₂, …, xₙ,

均值不等式（AM-GM）指出：对于非负实数 x₁, x₂, …, xₙ，有

(x₁ + x₂ + … + xₙ)/n ≥ ⁿ√(x₁x₂…xₙ)

with equality if and only if all xᵢ are equal.

当且仅当所有 xᵢ 相等时取等号。

Example: For x > 0, prove x + 1/x ≥ 2.

例：当 x > 0 时，证明 x + 1/x ≥ 2。

(x + 1/x)/2 ≥ √(x × 1/x) = 1 ⇒ x + 1/x ≥ 2

Equality holds when x = 1/x, that is x = 1.

当 x = 1/x，即 x = 1 时取等号。

AM-GM is extremely powerful for optimization problems in A-Level, especially when the sum or product is constant.

均值不等式在 A-Level 优化问题中非常强大，尤其是当和或积为常数时。

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7. Cauchy-Schwarz Inequality | 柯西-施瓦茨不等式

The Cauchy-Schwarz inequality is another essential proof tool:

柯西-施瓦茨不等式是另一个重要的证明工具：

(a₁² + a₂² + … + aₙ²)(b₁² + b₂² + … + bₙ²) ≥ (a₁b₁ + a₂b₂ + … + aₙbₙ)²

Equality holds when the sequences (aᵢ) and (bᵢ) are proportional.

当序列 (aᵢ) 与 (bᵢ) 成比例时取等号。

Example: For real x, y with x² + y² = 1, prove x + y ≤ √2.

例：对于满足 x² + y² = 1 的实数 x, y，证明 x + y ≤ √2。

(1² + 1²)(x² + y²) ≥ (1·x + 1·y)² ⇒ (x + y)² ≤ 2 ⇒ x + y ≤ √2

This method is common in STEP-style questions but also appears in A-Level further mathematics.

这种方法常见于 STEP 风格题目，但也出现在 A-Level 进阶数学中。

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8. Triangle Inequality | 三角不等式

For real numbers or complex numbers, the triangle inequality gives a bound on the absolute value of a sum.

对于实数或复数，三角不等式给出了和的绝对值的一个上界。

|a + b| ≤ |a| + |b|

Similarly, |a − b| ≥ ||a| − |b||. These are often used to prove bounds in sequences and series.

同理，|a − b| ≥ ||a| − |b||。这些不等式常用于数列与级数的放缩。

Application: Prove |x + 3| ≤ |x| + 3, which directly follows from the triangle inequality.

应用：证明 |x + 3| ≤ |x| + 3，直接由三角不等式得到。

When solving absolute value inequalities, remember to split cases according to the sign of the expression inside the absolute value.

解绝对值不等式时，切记按绝对值内部表达式的符号分类讨论。

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9. Mathematical Induction for Inequalities | 数学归纳法证明不等式

Mathematical induction is indispensable for proving inequalities involving positive integers n.

数学归纳法是证明涉及正整数 n 的不等式时不可或缺的方法。

Example: Prove that 2ⁿ > n² for all integers n ≥ 5.

例：证明对所有整数 n ≥ 5，2ⁿ > n²。

Base case: n = 5, 32 > 25 is true.

基础步骤：n = 5 时，32 > 25 成立。

Inductive hypothesis: assume 2ᵏ > k² for some k ≥ 5. Then:

归纳假设：假设对某个 k ≥ 5，2ᵏ > k²。则：

2ᵏ⁺¹ = 2 × 2ᵏ > 2k²

To complete the proof, we need 2k² ≥ (k + 1)² for k ≥ 5. Indeed, k² − 2k − 1 > 0 for k ≥ 3.

要完成证明，需要 2k² ≥ (k + 1)² 对 k ≥ 5 成立。事实上，k² − 2k − 1 > 0 对 k ≥ 3 成立。

Thus the induction step is verified. Always clearly state the base case, inductive hypothesis, and inductive step.

于是归纳步骤得证。务必清晰写出基础步骤、归纳假设和归纳递推。

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10. Common Pitfalls and Exam Tips | 常见陷阱与考试提示

The following mistakes are frequently seen in A-Level exams. Avoiding them will earn you valuable marks.

以下错误在 A-Level 考试中非常常见。避免它们能为你保住宝贵的分数。

• Multiplying or dividing an inequality by a negative number without reversing the sign.

• 乘以或除以一个负数时没有反转不等号方向。

• Squaring both sides of an inequality when one side might be negative.

• 在不等式两边为负或可能为负时擅自平方。

• Dropping the equality condition in AM-GM or Cauchy-Schwarz problems.

• 在使用均值不等式或柯西不等式时忽略取等条件。

• Writing a proof “backwards” without checking reversibility.

• 用分析法倒推但未检查每一步是否可逆。

When working with rational inequalities, always consider the denominator sign. For example, 1/x < 2 cannot be solved by simply multiplying by x.

处理分式不等式时，务必考虑分母的正负。例如，1/x < 2 不能简单地两边乘以 x 来求解。

1/x < 2 ⇒ 1/x − 2 < 0 ⇒ (1 − 2x)/x < 0 ⇒ x < 0 or x > 1/2

Use sign tables to organize intervals for rational and product inequalities.

针对分式不等式或乘积不等式，使用符号表来整理区间。

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