Inequality in sequences and sums

This chapter introduces fundamental techniques for analyzing inequalities within sequences and sums. Mastery of these methods is crucial for establishing bounds, determining convergence, and performing estimations, which are frequently assessed in advanced mathematical examinations.

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Chapter Contents

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| Topic |

|---|-------| | 1 | Term-wise comparison | | 2 | Sum bounds | | 3 | Product-sum comparison | | 4 | Growth comparison |

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We begin with Term-wise comparison.

Part 1: Term-wise comparison

Term-wise Comparison

Overview

Term-wise comparison is one of the cleanest ways to prove inequalities involving sequences and sums. The idea is simple: compare corresponding terms first, then add. In exam problems, this basic principle is often hidden inside fractions, radicals, powers, or recursive-looking expressions. The real skill is identifying a simpler upper or lower bound for each term. ---

Learning Objectives

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Core Principle

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Basic Sum Comparison Rules

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Comparing Fractions Term by Term

This gives a common exam trick: If $\qquad d_k \ge e_k > 0$ then $\qquad \dfrac{1}{d_k} \le \dfrac{1}{e_k}$ and hence $\qquad \sum_{k=1}^{n} \dfrac{1}{d_k} \le \sum_{k=1}^{n} \dfrac{1}{e_k}$ ::: ---

Positive-Term Comparison

---

Comparison for Infinite Series

Even when the topic is framed in finite sums, this idea often appears in advanced problems. ---

Very Common Bounding Patterns

These simple inequalities power many larger estimates. ---

How to Build a Term-wise Estimate

Examples of replacements:

• replace a complicated denominator by a smaller or larger one

• replace $\sqrt{k^2+k}$ by something easier to compare with
  • compare $k^2+1$ with $k^2$
  • compare $k(k+1)$ with $k^2$
  ::: ---

  Minimal Worked Examples

  Example 1 Show that $\qquad \sum_{k=1}^{n} \dfrac{1}{k(k+1)} \le \sum_{k=1}^{n} \dfrac{1}{k^2}$ For each $k\ge1$, $\qquad k(k+1) \ge k^2$ so $\qquad \dfrac{1}{k(k+1)} \le \dfrac{1}{k^2}$ Now add term by term: $\qquad \sum_{k=1}^{n} \dfrac{1}{k(k+1)} \le \sum_{k=1}^{n} \dfrac{1}{k^2}$ --- Example 2 Show that for positive integers $n$, $\qquad \sum_{k=1}^{n} \dfrac{1}{n+1} \le \sum_{k=1}^{n} \dfrac{1}{k}$ For each $k=1,2,\dots,n$, we have $\qquad k \le n+1$ so $\qquad \dfrac{1}{n+1} \le \dfrac{1}{k}$ Now add over all $k$: $\qquad n\cdot \dfrac{1}{n+1} \le \sum_{k=1}^{n} \dfrac{1}{k}$ Hence $\qquad \dfrac{n}{n+1} \le \sum_{k=1}^{n} \dfrac{1}{k}$ ---

  When Term-wise Comparison Is Safe

  ❗ Safe Uses

  Term-wise comparison is safe when:
  

  • you compare corresponding terms

  
  

  • the inequality direction is correct for each term

  
  

  • you only add the inequalities

  
  

  • denominators stay positive when taking reciprocals

  ---

  When It Can Fail

  ⚠️ Avoid These Errors

  • ❌ If $a_k \le b_k$, then $a_k^2 \le b_k^2$ without checking signs

  ✅ This is safe only under extra sign conditions

  • ❌ If $a_k \le b_k$, then $\dfrac{1}{a_k} \le \dfrac{1}{b_k}$

  ✅ False for positive numbers; reciprocals reverse inequality

  • ❌ Comparing sums by comparing “average size” informally

  ✅ Compare each term precisely

  • ❌ Using term-wise comparison for products without justification

  ✅ Addition is naturally compatible with term-wise inequality; products need extra care
  ---

  A Useful Symmetric Estimate

  📐 Average-Type Bound

  If
  
  $\qquad m \le a_k \le M$
  
  for all $k=1,2,\dots,n$, then dividing the sum bound by $n$ gives
  
  $\qquad m \le \dfrac{a_1+\cdots+a_n}{n} \le M$
  
  So the average also lies between the term-wise minimum and maximum.

  This often appears in olympiad-style estimation. ---

  Standard Patterns to Recognize

  📐 High-Value Patterns

  • Replace each denominator by something smaller or larger.

  
  

  • Compare each summand with a constant.

  
  

  • Compare each summand with the first or last term.

  
  

  • Use positivity to preserve inequality under summation.

  
  

  • Use reciprocal reversal carefully when terms are positive.

  ---

  CMI Strategy

  💡 How to Attack Term-wise Comparison Problems

  • First inspect one general term, not the whole sum.

  
  

  • Decide whether you want an upper bound or a lower bound.

  
  

  • Replace each term by a simpler comparable term.

  
  

  • Check sign conditions before taking reciprocals or squaring.

  
  

  • Only after the term-wise inequality is correct should you add.

  ---

  Practice Questions

  :::question type="MCQ" question="If $0<a_k\le b_k$ for each $k=1,2,\dots,n$, then which of the following is always true?" options=["$\sum_{k=1}^{n} a_k \ge \sum_{k=1}^{n} b_k$","$\sum_{k=1}^{n} a_k \le \sum_{k=1}^{n} b_k$","$\prod_{k=1}^{n} a_k \le \sum_{k=1}^{n} b_k$","$\dfrac{1}{a_k}\le\dfrac{1}{b_k}$ for each $k$"] answer="B" hint="Adding preserves the inequality direction term by term." solution="Since $a_k\le b_k$ for every $k$, we may add the inequalities: $\qquad \sum_{k=1}^{n} a_k \le \sum_{k=1}^{n} b_k$ So the correct option is $\boxed{B}$." ::: :::question type="NAT" question="Find a lower bound for $\sum_{k=1}^{5}\dfrac{1}{k}$ using the term-wise comparison $\dfrac{1}{k}\ge\dfrac{1}{5}$ for $k=1,2,3,4,5$." answer="1" hint="Replace each term by $\dfrac{1}{5}$ and add." solution="For each $k=1,2,3,4,5$, $\qquad \dfrac{1}{k}\ge \dfrac{1}{5}$ So $\qquad \sum_{k=1}^{5}\dfrac{1}{k} \ge 5\cdot \dfrac{1}{5}=1$ Hence the required lower bound is $\boxed{1}$." ::: :::question type="MSQ" question="Which of the following are always true for positive numbers $x\le y$?" options=["$\dfrac{1}{x}\ge\dfrac{1}{y}$","$\sqrt{x}\le\sqrt{y}$","$x^2\le y^2$","$\dfrac{1}{x}\le\dfrac{1}{y}$"] answer="A,B,C" hint="Check monotonicity of reciprocal, square root, and squaring on positive numbers." solution="1. True, reciprocals reverse the order for positive numbers.
• True, square root is increasing.
• True, squaring preserves order for positive numbers.
• False, it reverses the wrong way.
Hence the correct answer is $\boxed{A,B,C}$." ::: :::question type="SUB" question="Show that $\sum_{k=1}^{n}\dfrac{1}{k+1}\le \sum_{k=1}^{n}\dfrac{1}{k}$ for every positive integer $n$." answer="True, by term-wise comparison" hint="Compare $\dfrac{1}{k+1}$ and $\dfrac{1}{k}$ for each $k$." solution="For every positive integer $k$, $\qquad k+1 \ge k > 0$ So by reciprocal comparison, $\qquad \dfrac{1}{k+1}\le \dfrac{1}{k}$ Now add these inequalities for $k=1,2,\dots,n$: $\qquad \sum_{k=1}^{n}\dfrac{1}{k+1}\le \sum_{k=1}^{n}\dfrac{1}{k}$ Hence the inequality is proved term by term. Therefore the statement is $\boxed{\text{true}}$." ::: ---

Summary

❗ Key Takeaways for CMI

• If $a_k\le b_k$ term by term, then their sums satisfy the same inequality.




• Bounding each term by a constant gives a fast estimate for the full sum.




• Reciprocal inequalities reverse direction for positive numbers.




• Term-wise comparison is strongest when every term is positive and easy to control.




• The main art is choosing the right simpler comparison term.




• Check signs before using nonlinear operations like reciprocal or square.

---

💡 Next Up

Proceeding to Sum bounds.

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Part 2: Sum bounds

Sum Bounds

Overview

Sum bounds are inequalities that estimate how large or how small a sum can be under given conditions. In sequence and sum problems, this topic appears constantly: bounding a sum from below using product information, bounding a sum of squares using a fixed total, or using termwise estimates and Cauchy-type inequalities. In exam problems, the main challenge is choosing the correct inequality direction. ---

Learning Objectives

❗ By the End of This Topic

After studying this topic, you will be able to:


• Find lower and upper bounds for sums under simple constraints.




• Use AM-GM to bound sums from below.




• Use Cauchy-Schwarz to bound sums and sums of squares.




• Recognise when termwise bounds immediately yield a sum bound.




• Decide when equality is possible in a sum bound problem.

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Core Idea

📖 What is a sum bound?

A sum bound is an inequality of the form

$\qquad S \ge \text{something}$
or
$\qquad S \le \text{something}$

where $S$ is a sum such as

$\qquad a_1+a_2+\cdots+a_n$
or
$\qquad a_1^2+a_2^2+\cdots+a_n^2$

The main tools depend on what information is given:
• fixed product
• fixed sum
• lower/upper bounds on each term
• relation between sum and sum of squares
::: ---

Termwise Bounds

📐 Immediate Bounds

If
$\qquad m \le a_i \le M$
for each $i=1,2,\dots,n$,
then

$\qquad nm \le a_1+a_2+\cdots+a_n \le nM$

This is the quickest way to bound a sum when each term is individually controlled. ::: ---

AM-GM for Sum Lower Bounds

📐 AM-GM Lower Bound

For positive numbers $a_1,a_2,\dots,a_n$,

$\qquad a_1+a_2+\cdots+a_n \ge n\sqrt[n]{a_1a_2\cdots a_n}$

This gives a lower bound on the sum when the product is known.

Examples:
• if $\qquad xy=9$, then
$\qquad x+y\ge 6$
• if $\qquad abc=1$, then
$\qquad a+b+c\ge 3$ ::: ---

Cauchy-Schwarz and Sum of Squares

📐 Cauchy-Schwarz Basic Form

For real numbers $x_1,x_2,\dots,x_n$,

$\qquad (x_1+x_2+\cdots+x_n)^2 \le n(x_1^2+x_2^2+\cdots+x_n^2)$

Equivalently, $\qquad x_1^2+x_2^2+\cdots+x_n^2 \ge \dfrac{(x_1+x_2+\cdots+x_n)^2}{n}$ ::: This is one of the most important lower bounds for a sum of squares. ::: ---

Fixed Sum, Minimum Sum of Squares

❗ Minimum Sum of Squares

If
$\qquad x_1+x_2+\cdots+x_n=S$,
then

$\qquad x_1^2+x_2^2+\cdots+x_n^2 \ge \dfrac{S^2}{n}$

Equality holds when all variables are equal:

$\qquad x_1=x_2=\cdots=x_n=\dfrac{S}{n}$

This is a direct application of Cauchy-Schwarz or QM-AM. ---

Fixed Product, Minimum Sum

📐 Another Important Sum Bound

If
$\qquad x_1x_2\cdots x_n=P$
with all variables positive, then

$\qquad x_1+x_2+\cdots+x_n \ge n\sqrt[n]{P}$

Equality holds when all variables are equal.

So fixed product gives a lower bound on the sum. ---

Useful Special Cases

📐 Quick Forms to Memorise

For positive $x,y$:



• $\qquad x+y\ge 2\sqrt{xy}$
  
  • $\qquad x^2+y^2 \ge \dfrac{(x+y)^2}{2}$
  
  
  For real $x,y,z$:
  
  
  • $\qquad x^2+y^2+z^2 \ge \dfrac{(x+y+z)^2}{3}$
  
  
  For positive $a,b,c$ with $abc=1$:
  
  
  • $\qquad a+b+c\ge 3$

---

Minimal Worked Examples

Example 1 If $x+y+z=12$, find the minimum value of $x^2+y^2+z^2$. By Cauchy-Schwarz, $\qquad x^2+y^2+z^2 \ge \dfrac{(x+y+z)^2}{3} = \dfrac{12^2}{3}=48$ Equality occurs at $\qquad x=y=z=4$ So the minimum value is $\boxed{48}$. --- Example 2 If $xy=16$ and $x,y>0$, find the minimum value of $x+y$. By AM-GM, $\qquad x+y \ge 2\sqrt{xy}=2\sqrt{16}=8$ Equality occurs at $\qquad x=y=4$ So the minimum value is $\boxed{8}$. ---

Equality Cases

❗ When is the Bound Sharp?

In most standard sum-bound problems, equality occurs when all relevant terms are equal.

Examples:


• for $\qquad x+y\ge 2\sqrt{xy}$, equality at $x=y$
  
  • for $\qquad x^2+y^2+z^2\ge \dfrac{(x+y+z)^2}{3}$, equality at $x=y=z$
  
  
  • for $\qquad a_1+\cdots+a_n \ge n\sqrt[n]{a_1\cdots a_n}$, equality when all $a_i$ are equal

Always check whether the problem’s constraints allow equality. ---

Common Mistakes

⚠️ Avoid These Errors
• ❌ Using a lower-bound inequality when the problem asks for an upper bound
✅ First decide the required direction
• ❌ Applying AM-GM to numbers that are not known to be nonnegative
✅ Standard AM-GM needs nonnegative terms
• ❌ Forgetting that the sum of squares is usually minimized, not maximized, under fixed sum
✅ Cauchy gives a lower bound
• ❌ Stopping after getting a bound without checking equality
✅ A sharp answer needs the equality case
---

CMI Strategy

💡 How to Attack Sum-Bound Problems

• Identify the target sum clearly.




• Check what is fixed: sum, product, or termwise bounds.




• If product is fixed and terms are positive, try AM-GM.




• If sum is fixed and you see squares, try Cauchy-Schwarz.




• Check whether the bound is intended to be sharp, and identify equality.

---

Practice Questions

:::question type="MCQ" question="If $x+y=10$, then the least possible value of $x^2+y^2$ is" options=["$25$","$50$","$100$","$20$"] answer="B" hint="Use the lower bound for sum of squares." solution="We use $\qquad x^2+y^2 \ge \dfrac{(x+y)^2}{2}$ So $\qquad x^2+y^2 \ge \dfrac{10^2}{2}=50$ Equality occurs when $x=y=5$. Hence the correct option is $\boxed{B}$." ::: :::question type="NAT" question="If $abc=8$ and $a,b,c>0$, find the minimum possible value of $a+b+c$." answer="6" hint="Use AM-GM." solution="By AM-GM, $\qquad a+b+c \ge 3\sqrt[3]{abc}=3\sqrt[3]{8}=6$ Equality occurs when $a=b=c=2$. Hence the minimum value is $\boxed{6}$." ::: :::question type="MSQ" question="Which of the following are always true?" options=["For positive $x,y$, $x+y\ge 2\sqrt{xy}$","For real $x,y,z$, $x^2+y^2+z^2\ge \dfrac{(x+y+z)^2}{3}$","If each $a_i\le 4$ for $i=1,\dots,n$, then $\sum a_i\le 4n$","If $xy=1$, then $x+y\le 2$ for positive $x,y$"] answer="A,B,C" hint="Check the direction of each inequality." solution="1. True by AM-GM.
• True by Cauchy-Schwarz.
• True by termwise upper bound.
• False. By AM-GM, if $xy=1$, then $x+y\ge 2$.
Hence the correct answer is $\boxed{A,B,C}$." ::: :::question type="SUB" question="Prove that for real numbers $x_1,x_2,\dots,x_n$, one has $x_1^2+x_2^2+\cdots+x_n^2 \ge \dfrac{(x_1+x_2+\cdots+x_n)^2}{n}$." answer="Use Cauchy-Schwarz." hint="Apply Cauchy to $(x_1,\dots,x_n)$ and $(1,\dots,1)$." solution="By the Cauchy-Schwarz inequality, $\qquad (x_1^2+x_2^2+\cdots+x_n^2)(1^2+1^2+\cdots+1^2)\ge (x_1+x_2+\cdots+x_n)^2$ Since there are $n$ ones, $\qquad 1^2+1^2+\cdots+1^2=n$ So $\qquad n(x_1^2+x_2^2+\cdots+x_n^2)\ge (x_1+x_2+\cdots+x_n)^2$ Dividing by $n$, we get $\qquad x_1^2+x_2^2+\cdots+x_n^2 \ge \dfrac{(x_1+x_2+\cdots+x_n)^2}{n}$ This proves the result." ::: ---

Summary

❗ Key Takeaways for CMI

• Sum bounds depend on what information is given.




• AM-GM gives lower bounds for sums when the product is fixed.




• Cauchy-Schwarz gives lower bounds for sums of squares when the sum is fixed.




• Termwise bounds immediately give bounds on the whole sum.




• Equality cases usually occur when the relevant variables are equal.

---

💡 Next Up

Proceeding to Product-sum comparison.

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Part 3: Product-sum comparison

Product-Sum Comparison

Overview

Product-sum comparison is a central theme in inequalities. Many exam problems ask whether a product is at most a certain sum expression, or whether a sum is at least a certain product-driven quantity. The main tool is usually AM-GM, but the real skill is recognising the direction of comparison and the condition under which equality occurs. ---

Learning Objectives

❗ By the End of This Topic

After studying this topic, you will be able to:


• Compare sums and products of positive quantities using AM-GM.




• Solve fixed-sum maximum-product problems.




• Solve fixed-product minimum-sum problems.




• Recognise equality cases quickly.




• Use algebraic identities to prove sharp two-variable product-sum inequalities.

---

Core Idea

📖 What is product-sum comparison?

A product-sum comparison problem studies expressions such as

$\qquad x+y \quad \text{and} \quad xy$

or, more generally,

$\qquad x_1+x_2+\cdots+x_n \quad \text{and} \quad x_1x_2\cdots x_n$

under assumptions like positivity, fixed sum, or fixed product.

The most important general principle is:



• for fixed positive sum, the product is largest when the variables are equal




• for fixed positive product, the sum is smallest when the variables are equal

---

AM-GM: The Main Tool

📐 Arithmetic Mean - Geometric Mean Inequality

For positive real numbers $x_1,x_2,\dots,x_n$,

$\qquad \dfrac{x_1+x_2+\cdots+x_n}{n} \ge \sqrt[n]{x_1x_2\cdots x_n}$

Equivalently,

$\qquad x_1+x_2+\cdots+x_n \ge n\sqrt[n]{x_1x_2\cdots x_n}$

Equality holds if and only if $\qquad x_1=x_2=\cdots=x_n$ ::: ---

Two-Variable Product-Sum Relations

📐 Most Used Two-Variable Forms

For positive $x,y$,



• $\qquad x+y \ge 2\sqrt{xy}$
  
  • $\qquad xy \le \left(\dfrac{x+y}{2}\right)^2$
  
  
  • $\qquad (x-y)^2 \ge 0 \implies (x+y)^2 \ge 4xy$
  
  
  All three statements are equivalent.

These are the standard sharp comparisons between sum and product of two positive numbers. ---

Fixed Sum, Maximum Product

❗ Maximum Product Under Fixed Sum

If positive numbers $x_1,x_2,\dots,x_n$ satisfy

$\qquad x_1+x_2+\cdots+x_n=S$

then

$\qquad x_1x_2\cdots x_n \le \left(\dfrac{S}{n}\right)^n$

Equality holds when all variables are equal:

$\qquad x_1=x_2=\cdots=x_n=\dfrac{S}{n}$

For $n=2$, if $x+y=S$, then $\qquad xy \le \dfrac{S^2}{4}$ ::: ---

Fixed Product, Minimum Sum

❗ Minimum Sum Under Fixed Product

If positive numbers $x_1,x_2,\dots,x_n$ satisfy

$\qquad x_1x_2\cdots x_n=P$

then

$\qquad x_1+x_2+\cdots+x_n \ge n\sqrt[n]{P}$

Equality holds when all variables are equal.

For $n=2$, if $xy=P$, then $\qquad x+y \ge 2\sqrt{P}$ ::: ---

Standard Consequences

📐 Frequently Used Consequences

For positive $x,y,z$:



• if $\qquad xyz=1$, then


$\qquad x+y+z \ge 3$



• if $\qquad xyz=k$, then


$\qquad x+y+z \ge 3\sqrt[3]{k}$



• if $\qquad x+y+z=S$, then


$\qquad xyz \le \left(\dfrac{S}{3}\right)^3$



• if $\qquad x+y=S$, then


$\qquad xy \le \dfrac{S^2}{4}$



• if $\qquad xy=P$, then


$\qquad x+y \ge 2\sqrt{P}$

---

Minimal Worked Examples

Example 1 If $x+y=10$ and $x,y>0$, find the maximum value of $xy$. Using $\qquad xy \le \left(\dfrac{x+y}{2}\right)^2$ we get $\qquad xy \le \left(\dfrac{10}{2}\right)^2 = 25$ Equality occurs at $\qquad x=y=5$ So the maximum value is $\boxed{25}$. --- Example 2 If $xyz=8$ and $x,y,z>0$, find the minimum value of $x+y+z$. By AM-GM, $\qquad x+y+z \ge 3\sqrt[3]{xyz}=3\sqrt[3]{8}=6$ Equality occurs when $\qquad x=y=z=2$ So the minimum value is $\boxed{6}$. ---

Equality Case: Why It Matters

💡 Equality is Half the Problem

In olympiad-style and CMI-style inequality questions, the answer is often not complete until you identify when equality occurs.

For product-sum comparison using AM-GM, the equality condition is usually:

$\qquad \text{all compared positive terms are equal}$

So whenever you see a sharp bound, always ask:
• when is it attained?
• is equality allowed under the given conditions?
::: ---

Common Mistakes

⚠️ Avoid These Errors
• ❌ Applying AM-GM to numbers that are not known to be positive
✅ Standard AM-GM requires nonnegative numbers, and most sharp forms are used for positive numbers
• ❌ Forgetting the equality case
✅ Equality usually occurs when the relevant variables are equal
• ❌ Claiming that product is always less than sum without conditions
✅ The comparison depends on positivity and constraints such as fixed sum or fixed product
• ❌ Using a fixed-sum idea in a fixed-product problem
✅ First decide what is being held constant
---

CMI Strategy

💡 How to Attack Product-Sum Problems

• Ask whether the problem fixes the sum or fixes the product.




• If the sum is fixed, think “product is maximized when equal.”




• If the product is fixed, think “sum is minimized when equal.”




• For two variables, also test $(x-y)^2\ge0$.




• Check the equality case before finalising the answer.

---

Practice Questions

:::question type="MCQ" question="If $x,y>0$ and $x+y=14$, the maximum value of $xy$ is" options=["$24$","$36$","$49$","$98$"] answer="C" hint="Use the two-variable fixed-sum product bound." solution="For positive $x,y$ with fixed sum, $\qquad xy \le \left(\dfrac{x+y}{2}\right)^2$ So $\qquad xy \le \left(\dfrac{14}{2}\right)^2 = 49$ Equality occurs at $x=y=7$. Hence the correct option is $\boxed{C}$." ::: :::question type="NAT" question="If $x,y,z>0$ and $xyz=27$, find the minimum value of $x+y+z$." answer="9" hint="Use AM-GM for three variables." solution="By AM-GM, $\qquad x+y+z \ge 3\sqrt[3]{xyz}=3\sqrt[3]{27}=9$ Equality occurs when $x=y=z=3$. Hence the minimum value is $\boxed{9}$." ::: :::question type="MSQ" question="Which of the following are true for positive numbers?" options=["$x+y\ge 2\sqrt{xy}$","If $x+y$ is fixed, then $xy$ is maximized when $x=y$","If $xy$ is fixed, then $x+y$ is maximized when $x=y$","If $xyz=1$, then $x+y+z\ge 3$"] answer="A,B,D" hint="Separate fixed-sum and fixed-product statements carefully." solution="1. True by AM-GM.
• True.
• False. For fixed product, the sum is minimized, not maximized, when the variables are equal.
• True by AM-GM.
Hence the correct answer is $\boxed{A,B,D}$." ::: :::question type="SUB" question="Prove that for positive real numbers $x$ and $y$ with $x+y=S$, one has $xy\le \dfrac{S^2}{4}$, and determine when equality holds." answer="Use $(x-y)^2\ge 0$ or AM-GM." hint="Start from a square that is always nonnegative." solution="Since $\qquad (x-y)^2 \ge 0$, we get $\qquad x^2-2xy+y^2 \ge 0$ So $\qquad x^2+2xy+y^2 \ge 4xy$ That is, $\qquad (x+y)^2 \ge 4xy$ Now $x+y=S$, so $\qquad S^2 \ge 4xy$ Hence $\qquad xy \le \dfrac{S^2}{4}$ Equality holds if and only if $\qquad (x-y)^2=0$, that is, when $\qquad x=y=\dfrac{S}{2}$. Thus the product is at most $\dfrac{S^2}{4}$, with equality when the two numbers are equal." ::: ---

Summary

❗ Key Takeaways for CMI

• Product-sum comparison is mainly governed by AM-GM.




• Fixed sum gives maximum product at equality.




• Fixed product gives minimum sum at equality.




• For two variables, $(x-y)^2\ge0$ is often the fastest route.




• Equality conditions are essential in sharp inequality problems.

---

💡 Next Up

Proceeding to Growth comparison.

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Part 4: Growth comparison

Growth Comparison

Overview

Growth comparison is the study of deciding which of two expressions is larger without always computing exact values. In CMI-style questions, this often involves comparing powers, factorials, logarithms, radicals, and sums by using structure rather than direct expansion. The main skill is to choose the right comparison tool. ---

Learning Objectives

❗ By the End of This Topic

After studying this topic, you will be able to:


• Compare expressions involving powers, factorials, logarithms, and radicals.




• Use monotonicity, convexity, and concavity to compare nearby values.




• Apply AM-GM and simple product bounds to factorial-type comparisons.




• Compare large powers by scaling or normalising.




• Detect when direct computation is unnecessary and a structural argument is enough.

---

Core Idea

📖 What is growth comparison?

Growth comparison means deciding whether one expression is less than, equal to, or greater than another by using known behaviour of functions and sequences.

Typical examples:


• comparing $\sqrt{5+5\sqrt5}$ with an integer
  
  • comparing $\log_2 11$ with an average of two logarithmic values
  
  
  • comparing $n^n$ with $n!$ or $(n!)^2$
  
  
  • comparing sums like $a^m+b^m$ with $c^m$

---

First Tool: Monotonicity

📐 Increasing and Decreasing Functions

If $f$ is increasing, then

$\qquad a<b \implies f(a)<f(b)$

If $f$ is decreasing, then

$\qquad a<b \implies f(a)>f(b)$

This is the simplest comparison tool. It is often enough for:
• square roots
• logarithms
• exponentials
• reciprocal functions on positive intervals
Example To compare $\sqrt{5+5\sqrt5}$ with $4$, square both sides because the square root function is increasing on nonnegative numbers. $\qquad \sqrt{5+5\sqrt5} > 4 \iff 5+5\sqrt5 > 16$ $\qquad 5\sqrt5 > 11$ Since $\sqrt5 > 2$, the right side is true. Hence the expression is greater than $4$. ---

Second Tool: Concavity and Convexity

📖 Concavity and Convexity

A function with $f''(x)<0$ on an interval is concave there.

A function with $f''(x)>0$ on an interval is convex there.

📐 Midpoint Comparison

If $f$ is concave, then

$\qquad f\left(\dfrac{x+y}{2}\right) \ge \dfrac{f(x)+f(y)}{2}$

If $f$ is convex, then

$\qquad f\left(\dfrac{x+y}{2}\right) \le \dfrac{f(x)+f(y)}{2}$

This is extremely useful for growth comparison.

Important examples

• $\log x$ is concave on $x>0$
• $\sqrt{x}$ is concave on $x>0$
  • $x^r$ is convex on $x>0$ for $r>1$
  ---

  Logarithmic Comparison

  📐 Concavity of $\log x$

  Since $\log x$ is concave on positive numbers,
  
  $\qquad \log\left(\dfrac{x+y}{2}\right) \ge \dfrac{\log x + \log y}{2}$

  Equivalently, $\qquad \log x \le \dfrac{\log a + \log b}{2}$ whenever $x \le \sqrt{ab}$ This is often the cleanest way to compare one logarithm with the average of two others. Example To compare $\qquad \log_2 11 \quad \text{and} \quad \dfrac{1+\log_2 61}{2}$ note that $1=\log_2 2$, so the right-hand side is $\qquad \dfrac{\log_2 2 + \log_2 61}{2}$ Using log rules, $\qquad \dfrac{\log_2 2 + \log_2 61}{2} = \dfrac{\log_2(122)}{2} = \log_2 \sqrt{122}$ So the comparison becomes $\qquad \log_2 11 \stackrel{?}{<} \log_2 \sqrt{122}$ Since $\log_2 x$ is increasing, this is equivalent to $\qquad 11 < \sqrt{122}$ which is false because $121<122$, so actually $11<\sqrt{122}$ is true. Hence $\qquad \log_2 11 < \dfrac{1+\log_2 61}{2}$ ---

  Third Tool: AM-GM

  📐 AM-GM

  For positive real numbers $a_1,a_2,\dots,a_n$,
  
  $\qquad \dfrac{a_1+a_2+\cdots+a_n}{n} \ge \sqrt[n]{a_1a_2\cdots a_n}$
  
  Equality holds if and only if all the terms are equal.

  AM-GM is one of the most effective tools for growth comparison.

  Common uses

  • comparing products with powers
  • estimating factorials
  • comparing symmetric expressions
  ---

  Factorial vs Power Comparisons

  📐 A Useful Pairing Trick

  To compare $(n!)^2$ with $n^n$, write
  
  $\qquad (n!)^2 = \prod_{k=1}^{n} k(n+1-k)$
  
  Now compare each factor $k(n+1-k)$ with $n$.

  For $1 \le k \le n$, $\qquad k(n+1-k) \ge n$ because $\qquad k(n+1-k)-n = (k-1)(n-k) \ge 0$ Hence every factor is at least $n$, so $\qquad (n!)^2 = \prod_{k=1}^{n} k(n+1-k) \ge n^n$ This is a very elegant comparison. Example With $n=2023$, $\qquad (2023!)^2 \ge (2023)^{2023}$ and in fact the inequality is strict because not all factors are exactly $2023$. So $\qquad (2023)^{2023} < (2023!)^2$ ---

  Fourth Tool: Normalisation

  📐 Compare by Dividing by a Common Large Quantity

  When comparing large powers like
  
  $\qquad a^m+b^m \quad \text{and} \quad c^m$
  
  with $a,b<c$, divide everything by $c^m$.
  
  Then the comparison becomes
  
  $\qquad \left(\dfrac{a}{c}\right)^m + \left(\dfrac{b}{c}\right)^m \stackrel{?}{<} 1$

  This is usually much easier. Example Compare $\qquad 92^{100}+93^{100} \quad \text{and} \quad 94^{100}$ Divide by $94^{100}$: $\qquad \left(\dfrac{92}{94}\right)^{100} + \left(\dfrac{93}{94}\right)^{100} \stackrel{?}{<} 1$ Now $\qquad \dfrac{92}{94} < \dfrac{93}{94} < 1$ So $\qquad \left(\dfrac{92}{94}\right)^{100} < \left(\dfrac{93}{94}\right)^{100}$ Hence the sum is less than $\qquad 2\left(\dfrac{93}{94}\right)^{100}$ It remains to show $\qquad 2\left(\dfrac{93}{94}\right)^{100} < 1$ That is equivalent to $\qquad 2 < \left(\dfrac{94}{93}\right)^{100}$ which is true because $\left(1+\dfrac{1}{93}\right)^{100} > 2$ is easy to verify using binomial expansion or exponential growth. So $\qquad 92^{100}+93^{100} < 94^{100}$ ---

  Fifth Tool: Replace by Simpler Bounds

  💡 Bounding Strategy

  If exact comparison is hard, replace each expression by an easier upper or lower bound.
  
  Typical examples:
  

  • $\sqrt5 > 2$
    
    • $\log$ is increasing
    
    
    • $1+x < e^x$ for $x>0$
    
    
    • $(1+t)^n > 1+nt$ for $t>-1$ and integer $n\ge1$

  This often reduces a difficult-looking inequality to a short argument. ---

  Standard Growth Hierarchy

  📐 Very Important Order of Growth

  For large $n$, the typical growth order is
  
  $\qquad \log n \ll n^\alpha \ll a^n \ll n! \ll n^n$
  
  for fixed $\alpha>0$ and fixed $a>1$.

  This is useful intuition, but in olympiad-style or exam questions you still need a proof for the exact comparison asked. ---

  Minimal Worked Examples

  Example 1 Compare $\sqrt{17}$ and $4$. Since the square root function is increasing, $\qquad \sqrt{17} > 4 \iff 17 > 16$ which is true. --- Example 2 Compare $9^5+9^5$ and $10^5$. We have $\qquad 9^5+9^5 = 2\cdot 9^5$ So compare $2$ and $\left(\dfrac{10}{9}\right)^5$. Since $\qquad \left(\dfrac{10}{9}\right)^5 > 1+\dfrac{5}{9}$ we get $\left(\dfrac{10}{9}\right)^5 > \dfrac{14}{9} > 1.5$, but this is not enough alone to beat $2$. Compute directly: $\qquad 2\cdot 9^5 = 118098$ and $\qquad 10^5 = 100000$ So here $\qquad 9^5+9^5 > 10^5$ This shows that normalisation is useful, but the estimate must be strong enough. ---

  Common Mistakes

  ⚠️ Avoid These Errors

  • ❌ Comparing large expressions by raw decimal approximation too early

  ✅ First look for structure, monotonicity, or factorisation

  • ❌ Using concavity or convexity in the wrong direction

  ✅ Check whether the function bends upward or downward

  • ❌ Forgetting that logarithms are increasing only on positive inputs

  ✅ Always check domain

  • ❌ Using “growth order” intuition as a proof in a finite problem

  ✅ A finite comparison still needs a clear argument

  • ❌ Applying AM-GM to nonpositive terms

  ✅ AM-GM requires positivity
  ---

  CMI Strategy

  💡 How to Attack Growth Comparison Questions

  • First identify the expression type: radical, logarithm, power, factorial, or sum of powers.

  
  

  • Check whether monotonicity solves it immediately.

  
  

  • If an average appears, test concavity or convexity.

  
  

  • If factorials appear, try pairing factors or AM-GM.

  
  

  • If large powers appear, divide by the largest power.

  
  

  • Use a clean inequality, not a messy numerical approximation.

  ---

  Practice Questions

  :::question type="MCQ" question="Which function is concave on $(0,\infty)$?" options=["$x^2$","$x^3$","$\log x$","$e^x$"] answer="C" hint="Check second derivatives or known standard facts." solution="Among the given functions, $\log x$ is concave on $(0,\infty)$. The functions $x^2$, $x^3$ on $(0,\infty)$, and $e^x$ are convex there. Hence the correct option is $\boxed{C}$." ::: :::question type="NAT" question="Evaluate the sign of $(6!)^2-6^6$ by entering $1$ if positive, $0$ if zero, and $-1$ if negative." answer="1" hint="Use the pairing trick $(n!)^2=\prod_{k=1}^n k(n+1-k)$." solution="We write $\qquad (6!)^2 = \prod_{k=1}^{6} k(7-k)$ Now each factor $k(7-k)$ is at least $6$, and for some values it is strictly bigger than $6$. Hence $\qquad (6!)^2 > 6^6$ So the quantity $(6!)^2-6^6$ is positive. Therefore the required answer is $\boxed{1}$." ::: :::question type="MSQ" question="Which of the following statements are true?" options=["$\sqrt{10}>3$","$\log_2 8<\log_2 9$","If $f$ is concave, then $f\left(\frac{x+y}{2}\right)\ge \frac{f(x)+f(y)}{2}$","For all positive integers $n$, $(n!)^2\ge n^n$"] answer="A,B,C,D" hint="Use monotonicity, concavity, and factorial pairing." solution="1. True, since $10>9$ and square root is increasing.
• True, since $\log_2 x$ is increasing and $8<9$.
• True, this is the midpoint form of concavity.
• True, because $(n!)^2=\prod_{k=1}^n k(n+1-k)$ and each factor is at least $n$.
Hence the correct answer is $\boxed{A,B,C,D}$." ::: :::question type="SUB" question="Explain why $(n!)^2\ge n^n$ for every positive integer $n$." answer="Write $(n!)^2=\prod_{k=1}^n k(n+1-k)$ and note each factor is at least $n$." hint="Pair the $k$th and $(n+1-k)$th factors." solution="We have $\qquad (n!)^2 = \left(\prod_{k=1}^n k\right)\left(\prod_{k=1}^n (n+1-k)\right)=\prod_{k=1}^n k(n+1-k)$ Now for each $k$, $\qquad k(n+1-k)-n=(k-1)(n-k)\ge 0$ Hence $\qquad k(n+1-k)\ge n$ Therefore every factor in the product is at least $n$, so $\qquad (n!)^2 = \prod_{k=1}^n k(n+1-k)\ge \prod_{k=1}^n n = n^n$ Thus $(n!)^2\ge n^n$ for every positive integer $n$." ::: ---

Summary

❗ Key Takeaways for CMI

• Growth comparison depends more on choosing the right tool than on long computation.




• Use monotonicity for radicals, logs, and exponentials.




• Use concavity and convexity when averages or midpoint comparisons appear.




• Use AM-GM or factor pairing for factorial comparisons.




• Use normalisation when comparing sums of large powers with a single larger power.




• A clean structural proof is usually better than a decimal estimate.

---

Chapter Summary

❗ Inequality in sequences and sums — Key Points

Term-wise Comparison and Monotonicity: The fundamental approach for establishing inequalities between sequences and their sums often relies on demonstrating term-by-term dominance or consistent monotonicity.
Bounding Sums and Products: Integral comparison, the AM-GM inequality, and Cauchy-Schwarz are indispensable tools for deriving precise upper and lower bounds for both finite and infinite sums and products.
Growth Rates and Asymptotic Analysis: Understanding the asymptotic behavior of sequence terms is crucial for evaluating limits of sums, determining convergence properties, and making accurate approximations, frequently involving comparisons with known series.
Jensen's Inequality: This powerful principle connects the convexity or concavity of functions to inequalities involving sums and averages, offering a versatile approach for problems with functional transformations.
Classic Inequalities: Proficient application of foundational inequalities such as AM-GM, Cauchy-Schwarz, Bernoulli's, Rearrangement, and Chebyshev's forms the bedrock for solving a wide spectrum of complex problems involving sequences and sums.
Inductive and Constructive Methods: Many sequence-based inequalities can be rigorously proven through mathematical induction or by ingeniously constructing auxiliary sequences or functions that simplify the comparison process.

---

Chapter Review Questions

:::question type="MCQ" question="Consider the sum $S_n = \sum_{k=1}^n \frac{1}{\sqrt{k}}$. Which of the following is true for sufficiently large $n$?" options=["$S_n < \sqrt{n}$", "$S_n \approx 2\sqrt{n}$", "$S_n > n$", "$S_n$ converges to a finite value."] answer="$S_n \approx 2\sqrt{n}$" hint="Compare the sum with an integral of a decreasing function." solution="The function $f(x) = 1/\sqrt{x}$ is decreasing. We can bound the sum using integrals:


$$\int_1^{n+1} \frac{1}{\sqrt{x}} dx < \sum_{k=1}^n \frac{1}{\sqrt{k}} < 1 + \int_1^n \frac{1}{\sqrt{x}} dx$$

Evaluating the integrals:

$$[2\sqrt{x}]_1^{n+1} < S_n < 1 + [2\sqrt{x}]_1^n$$


$$2\sqrt{n+1} - 2 < S_n < 1 + 2\sqrt{n} - 2$$


$$2\sqrt{n+1} - 2 < S_n < 2\sqrt{n} - 1$$

As $n \to \infty$, $S_n$ grows approximately as $2\sqrt{n}$. Therefore, $S_n \approx 2\sqrt{n}$ is the correct statement."
:::

:::question type="NAT" question="Let $a_1, a_2, \dots, a_n$ be positive real numbers such that $\sum_{i=1}^n a_i = 1$. Find the minimum value of $\sum_{i=1}^n \frac{1}{a_i}$." answer="n^2" hint="Consider using the Cauchy-Schwarz inequality or AM-HM inequality." solution="By the Cauchy-Schwarz inequality, for positive real numbers $x_i, y_i$:


$$\left(\sum_{i=1}^n x_i^2\right) \left(\sum_{i=1}^n y_i^2\right) \ge \left(\sum_{i=1}^n x_i y_i\right)^2$$

Let $x_i = \sqrt{a_i}$ and $y_i = \frac{1}{\sqrt{a_i}}$. Then $x_i y_i = 1$.

$$\left(\sum_{i=1}^n (\sqrt{a_i})^2\right) \left(\sum_{i=1}^n \left(\frac{1}{\sqrt{a_i}}\right)^2\right) \ge \left(\sum_{i=1}^n 1\right)^2$$


$$\left(\sum_{i=1}^n a_i\right) \left(\sum_{i=1}^n \frac{1}{a_i}\right) \ge n^2$$

Given $\sum_{i=1}^n a_i = 1$, we substitute this into the inequality:

$$(1) \left(\sum_{i=1}^n \frac{1}{a_i}\right) \ge n^2$$

Thus, $\sum_{i=1}^n \frac{1}{a_i} \ge n^2$. Equality holds when $a_i = 1/n$ for all $i$.
The minimum value is $n^2$."
:::

:::question type="MCQ" question="Let $a_n = \left(1 + \frac{1}{n}\right)^n$ and $b_n = \sum_{k=0}^n \frac{1}{k!}$. Which of the following statements is true?" options=["$a_n > b_n$ for all $n \ge 1$.", "$\lim_{n \to \infty} a_n = \lim_{n \to \infty} b_n$.", "$a_n$ converges but $b_n$ diverges.", "$b_n$ converges but $a_n$ diverges."] answer="$\lim_{n \to \infty} a_n = \lim_{n \to \infty} b_n$." hint="Recall the definitions of the mathematical constant $e$." solution="Both sequences $a_n$ and $b_n$ are fundamental definitions or approximations of the mathematical constant $e$.
The limit of $a_n = \left(1 + \frac{1}{n}\right)^n$ as $n \to \infty$ is $e$.
The limit of $b_n = \sum_{k=0}^n \frac{1}{k!}$ as $n \to \infty$ is also $e$ (this is the Taylor series expansion for $e^x$ evaluated at $x=1$).
Since both sequences converge to the same value $e$, their limits are equal.
(Note: It is also true that $a_n < b_n$ for all $n \ge 1$, but the question asks for a true statement, and the limit equality is unequivocally true.)"
:::

:::question type="NAT" question="Let $S = \sum_{n=1}^{99} \frac{1}{\sqrt{n} + \sqrt{n+1}}$. Find the value of $S$." answer="9" hint="Rationalize the denominator of each term in the sum." solution="Each term of the sum can be simplified by rationalizing the denominator:


$$\frac{1}{\sqrt{n} + \sqrt{n+1}} = \frac{1}{\sqrt{n} + \sqrt{n+1}} \cdot \frac{\sqrt{n+1} - \sqrt{n}}{\sqrt{n+1} - \sqrt{n}}$$


$$= \frac{\sqrt{n+1} - \sqrt{n}}{(\sqrt{n+1})^2 - (\sqrt{n})^2} = \frac{\sqrt{n+1} - \sqrt{n}}{(n+1) - n} = \sqrt{n+1} - \sqrt{n}$$

The sum $S$ is therefore a telescoping sum:

$$S = \sum_{n=1}^{99} (\sqrt{n+1} - \sqrt{n})$$


$$S = (\sqrt{2} - \sqrt{1}) + (\sqrt{3} - \sqrt{2}) + (\sqrt{4} - \sqrt{3}) + \dots + (\sqrt{100} - \sqrt{99})$$

All intermediate terms cancel out, leaving:

$$S = \sqrt{100} - \sqrt{1} = 10 - 1 = 9$$

The value of $S$ is 9."
:::

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What's Next?

💡 Continue Your CMI Journey

Having gained proficiency in comparing sequences and bounding sums, you are now well-prepared to extend these concepts. The principles explored here form a crucial foundation for related chapters on inequalities involving functions, which delve into continuous analogues through tools like convexity, concavity, and integral inequalities. Furthermore, chapters on advanced asymptotic analysis and estimation will build upon the growth comparisons introduced, providing more sophisticated methods for approximating and analyzing the behavior of sequences and functions at large scales, which is vital for advanced CMI problems.