JEE Shortcut Reading Time: 15 min 5 Powerful Applications

The AM-GM Shortcut: Solving Maxima/Minima Without Calculus

Discover how to solve complex optimization problems in seconds using the Arithmetic Mean - Geometric Mean inequality.

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Why AM-GM is a JEE Superpower

The AM-GM inequality is one of the most powerful tools in a JEE aspirant's arsenal. While calculus can solve optimization problems, AM-GM often provides elegant, one-line solutions that save precious minutes in the exam.

🎯 JEE Advantage

Solves problems in 30 seconds instead of 3 minutes
Works where calculus becomes messy or complicated
Provides elegant verification for calculus answers
Especially useful for symmetric expressions

The AM-GM Inequality

$$ \frac{a_1 + a_2 + \cdots + a_n}{n} \geq \sqrt[n]{a_1 a_2 \cdots a_n} $$

Equality occurs when $a_1 = a_2 = \cdots = a_n$

📐 For 2 Variables

$$ \frac{a + b}{2} \geq \sqrt{ab} $$

Most commonly used form in JEE problems

📊 For 3 Variables

$$ \frac{a + b + c}{3} \geq \sqrt[3]{abc} $$

Useful for three-variable optimization

🚀 Application Areas

Two Variables Three Variables Fixed Sum Fixed Product Reciprocals

Application 1: Two Variable Optimization

Classic JEE Problem

If $x + y = 12$, find the maximum value of $xy$.

AM-GM Solution (15 seconds)

Step 1: Apply AM-GM to x and y:

$$ \frac{x + y}{2} \geq \sqrt{xy} $$

Step 2: Substitute $x + y = 12$:

$$ \frac{12}{2} \geq \sqrt{xy} \Rightarrow 6 \geq \sqrt{xy} $$

Step 3: Square both sides:

$$ 36 \geq xy $$

Step 4: Maximum occurs when equality holds: $x = y = 6$

Maximum value: $36$

💡 Key Insight

For fixed sum, product is maximized when numbers are equal. This works for any number of variables!

Application 2: Three Variable Problems

Volume Optimization

If $x + y + z = 18$, find the maximum value of $xyz$.

AM-GM Solution (20 seconds)

Step 1: Apply AM-GM to x, y, z:

$$ \frac{x + y + z}{3} \geq \sqrt[3]{xyz} $$

Step 2: Substitute $x + y + z = 18$:

$$ \frac{18}{3} \geq \sqrt[3]{xyz} \Rightarrow 6 \geq \sqrt[3]{xyz} $$

Step 3: Cube both sides:

$$ 216 \geq xyz $$

Step 4: Maximum when $x = y = z = 6$

Maximum value: $216$

⚠️ Common Mistake

Students often forget that all variables must be positive for AM-GM to apply. Always check this condition!

Application 3: Fixed Sum with Conditions

Advanced JEE Problem

If $x > 0, y > 0$ and $2x + 3y = 60$, find the maximum value of $x^2 y$.

AM-GM Solution (45 seconds)

Step 1: Express in symmetric form. Write $x^2 y$ as:

$$ x^2 y = \frac{4}{3} \cdot \frac{x}{2} \cdot \frac{x}{2} \cdot \frac{3y}{2} $$

Step 2: Notice the sum:

$$ \frac{x}{2} + \frac{x}{2} + \frac{3y}{2} = \frac{2x + 3y}{2} = \frac{60}{2} = 30 $$

Step 3: Apply AM-GM to three terms:

$$ \frac{\frac{x}{2} + \frac{x}{2} + \frac{3y}{2}}{3} \geq \sqrt[3]{\frac{x}{2} \cdot \frac{x}{2} \cdot \frac{3y}{2}} $$

Step 4: Simplify:

$$ 10 \geq \sqrt[3]{\frac{3x^2 y}{8}} $$

Step 5: Cube and solve:

$$ 1000 \geq \frac{3x^2 y}{8} \Rightarrow x^2 y \leq \frac{8000}{3} $$

Maximum value: $\frac{8000}{3}$ when $x = 20, y = \frac{20}{3}$

💡 Pro Tip

The key is to split variables to make the sum symmetric. Here we split x as x/2 + x/2 to match the coefficients.

Application 4: Fixed Product Problems

Minimizing Sum

If $xy = 144$, find the minimum value of $x + y$ for $x > 0, y > 0$.

AM-GM Solution (10 seconds)

Step 1: Direct application of AM-GM:

$$ \frac{x + y}{2} \geq \sqrt{xy} $$

Step 2: Substitute $xy = 144$:

$$ \frac{x + y}{2} \geq \sqrt{144} = 12 $$

Step 3: Multiply by 2:

$$ x + y \geq 24 $$

Minimum value: $24$ when $x = y = 12$

📚 General Rule

For fixed product, sum is minimized when numbers are equal. This is why squares have minimum perimeter for given area!

Application 5: Reciprocal Problems

Harmonic Mean Connection

If $x + y = 10$, find the minimum value of $\frac{1}{x} + \frac{1}{y}$ for $x > 0, y > 0$.

AM-GM Solution (20 seconds)

Step 1: Write the expression:

$$ \frac{1}{x} + \frac{1}{y} = \frac{x + y}{xy} = \frac{10}{xy} $$

Step 2: From AM-GM, maximum of xy is when $x = y = 5$:

$$ xy \leq \left(\frac{x + y}{2}\right)^2 = 25 $$

Step 3: Therefore:

$$ \frac{1}{x} + \frac{1}{y} = \frac{10}{xy} \geq \frac{10}{25} = \frac{2}{5} $$

Minimum value: $\frac{2}{5}$ when $x = y = 5$

🎯 Quick Method

For positive numbers, $\frac{1}{x} + \frac{1}{y}$ is minimized when x and y are equal (for fixed sum).

🎯 When to Use AM-GM vs Calculus

✅ Use AM-GM When:

Problem involves symmetric expressions
Variables are positive real numbers
There's a fixed sum or product constraint
You need a quick verification of calculus result
The answer suggests equal values at optimum

❌ Use Calculus When:

Expressions are highly asymmetric
Constraints are complex inequalities
Variables can be negative
You need to find local extrema
Problem involves trigonometric functions

💪 Practice Your AM-GM Skills

1. If $x + 2y = 24$, find maximum value of $xy^2$

Hint: Write as $x + y + y = 24$

2. If $xyz = 64$, find minimum value of $x + y + z$

Hint: Direct AM-GM application

3. If $x + y + z = 12$, find maximum value of $x^2 y^3 z$

Hint: Split variables to make sum symmetric

4. Find minimum value of $x + \frac{1}{x}$ for $x > 0$

Classic AM-GM problem

5. If $2x + y = 15$, find maximum value of $x^2 y$

JEE Main 2022 type

📋 AM-GM Quick Reference

Situation	AM-GM Application	Result
Fixed sum $S$	$x + y = S$	Max $xy = (S/2)^2$ at $x = y$
Fixed product $P$	$xy = P$	Min $x + y = 2\sqrt{P}$ at $x = y$
Three variables	$x + y + z = S$	Max $xyz = (S/3)^3$ at $x = y = z$
Reciprocal sum	$x + y = S$	Min $1/x + 1/y = 4/S$ at $x = y$

🚀 Exam Strategy

Before Exam:

Memorize the 2-variable and 3-variable forms
Practice the "splitting" technique for asymmetric problems
Learn to recognize AM-GM applicable patterns quickly
Solve at least 10 varied AM-GM problems

During Exam:

Check if variables are positive
Look for fixed sum/product constraints
Try AM-GM first for symmetric-looking problems
Use as quick verification for calculus answers
Remember: Equality at equal values is the key

Mastered AM-GM? Level Up Further!

Learn Cauchy-Schwarz, Jensen's inequality and other powerful tools

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