Standard inequalities

This chapter introduces fundamental inequalities essential for advanced mathematical problem-solving. Mastery of these standard techniques, including absolute value, AM-GM, and basic Cauchy applications, is crucial for success in the CMI examinations, as they frequently appear in various contexts.

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Chapter Contents

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| Topic |

|---|-------| | 1 | Absolute value inequalities | | 2 | AM-GM | | 3 | Cauchy style school-level use |

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We begin with Absolute value inequalities.

Part 1: Absolute value inequalities

Absolute Value Inequalities

Overview

Absolute value inequalities measure distance from zero or, more generally, distance between two quantities. This topic is fundamental because many harder inequalities reduce to statements of the form $|x-a|<r$ or $|x-a|\ge r$. In CMI-style questions, the main difficulty is not algebraic expansion, but choosing the correct interpretation and splitting cases correctly. ---

Learning Objectives

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Core Meaning

This interpretation is the fastest way to solve many inequalities. ---

Standard One-Modulus Inequalities

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Solving by Case Split

This is especially useful for expressions like:

• $|x-2|$

• $|2x+1|$

• $|x-1|+|x+3|$

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Triangle Inequality

These are among the most important tools in estimation questions. ---

Useful Consequences

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Compound Absolute Value Expressions

For example, if $a<b$, then $\qquad |x-a|+|x-b| \ge b-a$ with equality for all $x$ in the interval $[a,b]$. ---

Minimal Worked Examples

Example 1 Solve $\qquad |x-3| < 5$ Using the basic rule, $\qquad -5 < x-3 < 5$ So $\qquad -2 < x < 8$ Hence the solution set is $\boxed{(-2,8)}$. --- Example 2 Solve $\qquad |2x-1| \ge 3$ We split: $\qquad 2x-1 \ge 3 \quad \text{or} \quad 2x-1 \le -3$ So $\qquad 2x \ge 4 \Rightarrow x \ge 2$ or $\qquad 2x \le -2 \Rightarrow x \le -1$ Hence the solution set is $\qquad \boxed{(-\infty,-1]\cup[2,\infty)}$ --- Example 3 Find the minimum value of $\qquad |x-1|+|x-5|$ This is the sum of distances from $x$ to $1$ and $5$. The minimum occurs for all $x$ between $1$ and $5$, and equals the distance between the two fixed points: $\qquad 5-1=4$ So the minimum value is $\boxed{4}$. ---

Common Mistakes

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CMI Strategy

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Practice Questions

:::question type="MCQ" question="The solution set of $|x-2|<3$ is" options=["$(-1,5)$","$[-1,5]$","$(-\infty,-1)\cup(5,\infty)$","$[-1,\infty)$"] answer="A" hint="Use the interval form of $|x-a|<r$." solution="We have $\qquad |x-2|<3 \iff -3<x-2<3$ So $\qquad -1<x<5$ Hence the correct option is $\boxed{A}$." ::: :::question type="NAT" question="Find the minimum value of $|x+1|+|x-3|$." answer="4" hint="Interpret as a sum of distances." solution="The expression is the sum of distances from $x$ to $-1$ and $3$. The minimum possible value is the distance between these two fixed points: $\qquad 3-(-1)=4$ Hence the minimum value is $\boxed{4}$." ::: :::question type="MSQ" question="Which of the following statements are true?" options=["$|x-a|<r$ with $r>0$ describes an interval","$|x-a|>r$ with $r>0$ describes two rays","$|x+y|=|x|+|y|$ for all real $x,y$","$|x-a|$ is the distance between $x$ and $a$"] answer="A,B,D" hint="One statement incorrectly turns an inequality into an identity." solution="1. True.

True.

False. In general, only $\qquad |x+y| \le |x|+|y|$ is true.

True.

Hence the correct answer is $\boxed{A,B,D}$." ::: :::question type="SUB" question="Solve the inequality $|2x-5|\le 7$." answer="$-1 \le x \le 6$" hint="Convert the modulus inequality into a double inequality." solution="We write $\qquad |2x-5|\le 7 \iff -7 \le 2x-5 \le 7$ Add $5$ throughout: $\qquad -2 \le 2x \le 12$ Divide by $2$: $\qquad -1 \le x \le 6$ Hence the solution set is $\boxed{[-1,6]}$." ::: ---

Summary

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Part 2: AM-GM

AM-GM

Overview

The Arithmetic Mean–Geometric Mean inequality is one of the most powerful and frequently used inequalities in algebra. It compares the average of nonnegative numbers with their geometric mean, and it is central in optimization, estimation, and olympiad-style inequality problems. CMI-style questions often test not just the formula, but how to choose the right grouping and how to detect the equality case. ---

Learning Objectives

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Core Statement

Equality holds if and only if $\qquad a=b$ ::: Equality holds if and only if $\qquad a_1=a_2=\cdots=a_n$ ::: ---

Why Nonnegativity Matters

This is one of the most common sources of invalid proofs. ---

Standard Forms

These are all direct consequences of AM-GM. ---

Equality Case

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Minimal Worked Examples

Example 1 For $x>0$, find the minimum value of $\qquad x+\dfrac{1}{x}$ By AM-GM, $\qquad x+\dfrac{1}{x} \ge 2\sqrt{x\cdot \dfrac{1}{x}} = 2$ Equality holds when $\qquad x=\dfrac{1}{x}$, that is, $x=1$ So the minimum value is $\boxed{2}$. --- Example 2 If $a,b \ge 0$ and $a+b=10$, find the maximum value of $ab$. By AM-GM, $\qquad \dfrac{a+b}{2} \ge \sqrt{ab}$ So $\qquad 5 \ge \sqrt{ab}$ Hence $\qquad ab \le 25$ Equality holds when $\qquad a=b=5$ So the maximum value is $\boxed{25}$. ---

Useful Extensions

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Product-Sum Strategy

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Common Mistakes

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CMI Strategy

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Practice Questions

:::question type="MCQ" question="For $x>0$, the minimum value of $x+\dfrac{1}{x}$ is" options=["$1$","$2$","$\sqrt{2}$","$0$"] answer="B" hint="Use AM-GM on the two positive terms." solution="By AM-GM, $\qquad x+\dfrac{1}{x} \ge 2\sqrt{x\cdot \dfrac{1}{x}} = 2$ Equality holds at $x=1$. Hence the correct option is $\boxed{B}$." ::: :::question type="NAT" question="If $a,b\ge 0$ and $a+b=12$, find the maximum value of $ab$." answer="36" hint="Use AM-GM or equal-splitting." solution="By AM-GM, $\qquad \dfrac{a+b}{2} \ge \sqrt{ab}$ So $\qquad 6 \ge \sqrt{ab}$ Hence $\qquad ab \le 36$ Equality holds when $\qquad a=b=6$ Thus the maximum value is $\boxed{36}$." ::: :::question type="MSQ" question="Which of the following statements are true?" options=["For $a,b\ge 0$, $a+b\ge 2\sqrt{ab}$","For positive $x$, $x+\dfrac1x\ge 2$","Equality in AM-GM for two variables holds when the two variables are equal","AM-GM can be applied directly to any real numbers, positive or negative"] answer="A,B,C" hint="One statement ignores the nonnegativity requirement." solution="1. True.

True.

True.

False. AM-GM requires nonnegative real numbers in the standard real setting.

Hence the correct answer is $\boxed{A,B,C}$." ::: :::question type="SUB" question="Prove that for $x>0$, $2x+\dfrac{1}{x^2}\ge 3$." answer="$2x+\dfrac1{x^2}\ge 3$" hint="Apply AM-GM to three nonnegative terms." solution="Apply AM-GM to the three nonnegative numbers $\qquad x,\ x,\ \dfrac{1}{x^2}$ Then $\qquad \dfrac{x+x+\dfrac{1}{x^2}}{3} \ge \sqrt[3]{x\cdot x\cdot \dfrac{1}{x^2}} = 1$ So $\qquad x+x+\dfrac{1}{x^2} \ge 3$ Hence $\qquad 2x+\dfrac{1}{x^2}\ge 3$ Equality holds when $\qquad x=x=\dfrac{1}{x^2}$, that is, when $x=1$." ::: ---

Summary

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Part 3: Cauchy style school-level use

Cauchy Style School-Level Use

Overview

At school level, Cauchy-type arguments are often used not in their most abstract form, but as powerful tools for proving lower bounds, comparing symmetric expressions, and estimating sums of squares. In olympiad-style algebra, these arguments appear in deceptively simple forms such as $\qquad (a+b)^2 \le 2(a^2+b^2)$ or $\qquad \dfrac{x^2}{a}+\dfrac{y^2}{b}\ge \dfrac{(x+y)^2}{a+b}$ The main skill is to recognize when a sum of squares or a square of a sum is present. ---

Learning Objectives

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Core Idea

At school level, you will usually use this in direct simplified forms rather than in full abstract notation. ---

Most Useful School-Level Forms

More generally, for positive $a_i$, $\qquad \sum \dfrac{x_i^2}{a_i}\ge \dfrac{(x_1+x_2+\cdots+x_n)^2}{a_1+a_2+\cdots+a_n}$ ::: This is a very common exam-level form. ---

Titu's Lemma

This is just a clean and frequently used form of Cauchy-Schwarz. ::: ---

Standard Consequences

These often appear in lower-bound and minimum-value problems. ---

How to Recognize a Cauchy-Type Situation

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Minimal Worked Examples

Example 1 Prove that $\qquad a^2+b^2\ge \dfrac{(a+b)^2}{2}$ From the two-variable form of Cauchy, $\qquad 2(a^2+b^2)\ge (a+b)^2$ Dividing by $2$ gives the result. --- Example 2 Show that for positive $x,y$, $\qquad \dfrac{x^2}{y}+\dfrac{y^2}{x}\ge x+y$ Using Titu is not direct here, but Cauchy-style thinking suggests comparing with common structure. By AM-GM, $\qquad \dfrac{x^2}{y}+\dfrac{y^2}{x}\ge 2\sqrt{xy}$ But a sharper school-level method is to write $\qquad \dfrac{x^2}{y}+\dfrac{y^2}{x}-x-y = \dfrac{x^3+y^3-x^2y-xy^2}{xy}$ $\qquad = \dfrac{(x-y)^2(x+y)}{xy}\ge 0$ Hence $\qquad \dfrac{x^2}{y}+\dfrac{y^2}{x}\ge x+y$ This example shows that not every square-looking inequality is solved by direct Cauchy, but Cauchy-style structure recognition still helps. ---

Equality Cases Matter

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Common Patterns

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Common Mistakes

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CMI Strategy

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Practice Questions

:::question type="MCQ" question="For real numbers $a,b,c$, which of the following is always true?" options=["$(a+b+c)^2\le a^2+b^2+c^2$","$(a+b+c)^2\le 3(a^2+b^2+c^2)$","$(a+b+c)^2\ge 3(a^2+b^2+c^2)$","$a^2+b^2+c^2\le ab+bc+ca$"] answer="B" hint="Use the three-variable form of Cauchy." solution="By the standard three-variable form of Cauchy, $\qquad (a+b+c)^2\le 3(a^2+b^2+c^2)$ Hence the correct option is $\boxed{B}$." ::: :::question type="NAT" question="Find the minimum value of $a^2+b^2$ if $a+b=10$." answer="50" hint="Use $a^2+b^2\ge \dfrac{(a+b)^2}{2}$." solution="By the two-variable Cauchy form, $\qquad a^2+b^2\ge \dfrac{(a+b)^2}{2}$ Since $a+b=10$, $\qquad a^2+b^2\ge \dfrac{100}{2}=50$ Equality holds when $a=b=5$. Therefore the minimum value is $\boxed{50}$." ::: :::question type="MSQ" question="Which of the following statements are true for positive $a,b$?" options=["$\dfrac{1}{a}+\dfrac{1}{b}\ge \dfrac{4}{a+b}$","$a^2+b^2\ge \dfrac{(a+b)^2}{2}$","$\dfrac{a^2}{b}+\dfrac{b^2}{a}\ge a+b$","$a^2+b^2\le \dfrac{(a+b)^2}{2}$"] answer="A,B,C" hint="Use standard school-level Cauchy forms and one factorisation." solution="1. True. By Titu or by Cauchy, $\qquad \dfrac{1}{a}+\dfrac{1}{b}\ge \dfrac{(1+1)^2}{a+b}=\dfrac{4}{a+b}$

True. This is the standard two-variable form.

True. Because

$\qquad \dfrac{a^2}{b}+\dfrac{b^2}{a}-(a+b)=\dfrac{(a-b)^2(a+b)}{ab}\ge 0$

False. The reverse inequality is true.

Hence the correct answer is $\boxed{A,B,C}$." ::: :::question type="SUB" question="Prove that for positive numbers $x,y,z$, $\dfrac{x^2}{y}+\dfrac{y^2}{z}+\dfrac{z^2}{x}\ge x+y+z$." answer="The inequality is true." hint="Use Titu's lemma or factorisation-style comparison." solution="By Titu's lemma, $\qquad \dfrac{x^2}{y}+\dfrac{y^2}{z}+\dfrac{z^2}{x}\ge \dfrac{(x+y+z)^2}{x+y+z}=x+y+z$ So the required inequality follows immediately. Therefore, $\qquad \boxed{\dfrac{x^2}{y}+\dfrac{y^2}{z}+\dfrac{z^2}{x}\ge x+y+z}$" ::: ---

Summary

Chapter Summary

Chapter Review Questions

:::question type="MCQ" question="How many integer solutions exist for the inequality $|2x-3| < 5$?" options=["3", "4", "5", "6"] answer="4" hint="Transform the absolute value inequality into a compound linear inequality and identify the integers within the resulting range." solution="The inequality $|2x-3| < 5$ can be rewritten as $-5 < 2x-3 < 5$. Adding 3 to all parts yields $-2 < 2x < 8$. Dividing by 2 gives $-1 < x < 4$. The integers satisfying this range are $0, 1, 2, 3$. There are 4 such integer solutions."
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:::question type="NAT" question="For $x > 0$, what is the minimum value of $x + \frac{4}{x}$?" answer="4" hint="Apply the AM-GM inequality to the two positive terms." solution="By the AM-GM inequality, for $x>0$, we have $x + \frac{4}{x} \ge 2\sqrt{x \cdot \frac{4}{x}} = 2\sqrt{4} = 2 \cdot 2 = 4$. Equality holds when $x = \frac{4}{x}$, i.e., $x^2=4$, so $x=2$ (since $x>0$). Thus, the minimum value is 4."
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:::question type="MCQ" question="If $x,y,z$ are real numbers such that $x^2+y^2+z^2=1$, what is the maximum value of $x+2y+3z$?" options=["$\sqrt{14}$", "$14$", "$7$", "$\sqrt{7}$"] answer="$\sqrt{14}$" hint="Use the Cauchy-Schwarz inequality." solution="By the Cauchy-Schwarz inequality, for real numbers $a_1, a_2, a_3$ and $b_1, b_2, b_3$:

Let $a_1=x, a_2=y, a_3=z$ and $b_1=1, b_2=2, b_3=3$.
Then $(x \cdot 1 + y \cdot 2 + z \cdot 3)^2 \le (x^2+y^2+z^2)(1^2+2^2+3^2)$.
Given $x^2+y^2+z^2=1$, we have:
$(x+2y+3z)^2 \le (1)(1+4+9)$
$(x+2y+3z)^2 \le 14$
Taking the square root of both sides, we get $-\sqrt{14} \le x+2y+3z \le \sqrt{14}$.
The maximum value is $\sqrt{14}$. Equality holds when $(x,y,z)$ is proportional to $(1,2,3)$ and $x^2+y^2+z^2=1$."
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:::question type="NAT" question="If $a,b,c$ are positive real numbers such that $a+b+c=3$, what is the minimum value of $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$?" answer="3" hint="Consider the product $(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)$ or apply AM-HM inequality." solution="By the Cauchy-Schwarz inequality, with sequences $(\sqrt{a}, \sqrt{b}, \sqrt{c})$ and $(\frac{1}{\sqrt{a}}, \frac{1}{\sqrt{b}}, \frac{1}{\sqrt{c}})$:

Given $a+b+c=3$, we have:


The minimum value is 3, which occurs when $a=b=c=1$."
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What's Next?