AM-GM-HM Inequalities

Master AM-GM-HM inequalities - proofs, applications, and optimization problems for JEE

11 min read

Interactive Demo: Visualize Mean Inequalities

Explore how AM, GM, and HM relate to each other for different values.

Real-Life Hook: The Farmer’s Optimization Problem

A farmer has 100 meters of fence and wants to create a rectangular enclosure with maximum area. Should he make it square or rectangular?

Using AM-GM inequality, we can prove that for a fixed perimeter, a square gives maximum area! This powerful inequality helps solve countless optimization problems in physics, economics, and engineering.

From proving $x + \frac{1}{x} \geq 2$ to finding maximum volumes and minimum distances, AM-GM-HM inequalities are the Swiss Army knife of JEE Mathematics!

The Three Means

For positive numbers $a$ and $b$:

1. Arithmetic Mean (AM)

$$\boxed{\text{AM} = \frac{a+b}{2}}$$

For $n$ numbers: $\text{AM} = \frac{a_1 + a_2 + \cdots + a_n}{n}$

Interpretation: Simple average

2. Geometric Mean (GM)

$$\boxed{\text{GM} = \sqrt{ab}}$$

For $n$ numbers: $\text{GM} = \sqrt[n]{a_1 \cdot a_2 \cdot \ldots \cdot a_n}$

Interpretation: $n$th root of product

3. Harmonic Mean (HM)

$$\boxed{\text{HM} = \frac{2ab}{a+b} = \frac{2}{\frac{1}{a} + \frac{1}{b}}}$$

For $n$ numbers: $\text{HM} = \frac{n}{\frac{1}{a_1} + \frac{1}{a_2} + \cdots + \frac{1}{a_n}}$

Interpretation: Reciprocal of average of reciprocals

The Fundamental Inequality

For positive numbers $a$ and $b$:

$$\boxed{\text{AM} \geq \text{GM} \geq \text{HM}}$$ $$\boxed{\frac{a+b}{2} \geq \sqrt{ab} \geq \frac{2ab}{a+b}}$$

Equality Condition: All three are equal if and only if $a = b$

For $n$ positive numbers:

$$\boxed{\frac{a_1 + a_2 + \cdots + a_n}{n} \geq \sqrt[n]{a_1 a_2 \ldots a_n} \geq \frac{n}{\frac{1}{a_1} + \frac{1}{a_2} + \cdots + \frac{1}{a_n}}}$$

Proofs

Proof 1: AM ≥ GM (for two numbers)

Method 1: Algebraic

We need to prove: $\frac{a+b}{2} \geq \sqrt{ab}$

Multiply both sides by 2: $a + b \geq 2\sqrt{ab}$

Square both sides (both positive): $(a+b)^2 \geq 4ab$

$a^2 + 2ab + b^2 \geq 4ab$

$a^2 - 2ab + b^2 \geq 0$

$(a-b)^2 \geq 0$ ✓ (always true!)

Equality when $(a-b)^2 = 0$, i.e., $a = b$

Method 2: Geometric

Consider a semicircle with diameter $a + b$. The chord at distance from diameter that divides it into segments $a$ and $b$ has height $\sqrt{ab}$ (by geometric mean theorem).

The radius (which is $\frac{a+b}{2}$) is always greater than or equal to any chord, so:

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

Proof 2: GM ≥ HM (for two numbers)

We need to prove: $\sqrt{ab} \geq \frac{2ab}{a+b}$

Divide both sides by $\sqrt{ab}$ (positive): $1 \geq \frac{2\sqrt{ab}}{a+b}$

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

This is exactly the AM ≥ GM inequality! ✓

So GM ≥ HM follows from AM ≥ GM.

Alternative: Direct Proof AM ≥ HM

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

$(a+b)^2 \geq 4ab$

$a^2 + 2ab + b^2 \geq 4ab$

$a^2 - 2ab + b^2 \geq 0$

$(a-b)^2 \geq 0$ ✓

Important Relationships

1. Connection Formula

If AM, GM, HM are the three means of $a$ and $b$:

$$\boxed{\text{GM}^2 = \text{AM} \times \text{HM}}$$

Proof:

$\text{GM}^2 = ab$

$\text{AM} \times \text{HM} = \frac{a+b}{2} \times \frac{2ab}{a+b} = ab$ ✓

This is a KEY formula for JEE!

2. For Equal AM and GM

If AM = GM, then:

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

$(a+b)^2 = 4ab$

$a^2 + 2ab + b^2 = 4ab$

$a^2 - 2ab + b^2 = 0$

$(a-b)^2 = 0$

$a = b$ ✓

3. Weighted AM-GM

For positive $a, b$ and positive weights $m, n$:

$$\boxed{\frac{ma + nb}{m+n} \geq \sqrt[m+n]{a^m b^n}}$$

With equality when $a = b$.

Common Mistakes & How to Avoid Them

❌ Mistake 1: Negative Numbers

Wrong: Applying AM-GM to negative numbers ✗

Example: $a = -2, b = -3$

$\text{AM} = \frac{-2-3}{2} = -2.5$

$\text{GM} = \sqrt{(-2)(-3)} = \sqrt{6} \approx 2.45$

Here GM > AM! ✗ (Inequality reverses!)

Correct: AM-GM-HM inequalities apply ONLY to positive numbers ✓

❌ Mistake 2: HM Formula Confusion

Wrong: $\text{HM} = \frac{a+b}{2ab}$ ✗

Correct: $\text{HM} = \frac{2ab}{a+b}$ ✓

Memory Trick: “High Product” → Numerator has product $ab$

❌ Mistake 3: Equality Condition

Wrong: AM = GM when $a \approx b$ ✗

Correct: AM = GM only when $a = b$ (exactly!) ✓

Example: $a = 5, b = 5.001$

$\text{AM} = 5.0005$, $\text{GM} = \sqrt{25.005} \approx 5.0004999$ (not equal!)

❌ Mistake 4: Multi-variable Application

Wrong: For $x + y + z = 12$, minimum of $xyz$ is when $x = y = z = 4$ ✗

Correct: That gives MAXIMUM of $xyz$! ✓

By AM-GM: $\frac{x+y+z}{3} \geq \sqrt[3]{xyz}$

$4 \geq \sqrt[3]{xyz}$ → $xyz \leq 64$ (maximum, not minimum!)

For minimum: No lower bound unless constraints added (can approach 0)

Problem-Solving Strategies

Strategy 1: Basic Optimization

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

Solution:

By AM-GM: $\frac{x + \frac{1}{x}}{2} \geq \sqrt{x \cdot \frac{1}{x}} = \sqrt{1} = 1$

$x + \frac{1}{x} \geq 2$

Minimum value = $2$, achieved when $x = \frac{1}{x}$, i.e., $x = 1$

Answer: Minimum = $2$ at $x = 1$

Strategy 2: Constraint Optimization

Problem: If $xy = 16$ and $x, y > 0$, find minimum value of $x + y$.

Solution:

By AM-GM: $\frac{x+y}{2} \geq \sqrt{xy} = \sqrt{16} = 4$

$x + y \geq 8$

Minimum = $8$, achieved when $x = y = 4$

Answer: Minimum = $8$

Strategy 3: Creative Grouping

Problem: Find minimum value of $\frac{x^2 + 1}{x}$ for $x > 0$.

Solution:

$\frac{x^2 + 1}{x} = x + \frac{1}{x}$

By AM-GM: $x + \frac{1}{x} \geq 2\sqrt{x \cdot \frac{1}{x}} = 2$

Minimum = $2$ at $x = 1$

Answer: Minimum = $2$

Strategy 4: Multi-term Problems

Problem: If $a + b + c = 12$ and $a, b, c > 0$, find maximum of $abc$.

Solution:

By AM-GM: $\frac{a+b+c}{3} \geq \sqrt[3]{abc}$

$\frac{12}{3} \geq \sqrt[3]{abc}$

$4 \geq \sqrt[3]{abc}$

$abc \leq 64$

Maximum = $64$, achieved when $a = b = c = 4$

Answer: Maximum = $64$

Strategy 5: Weighted AM-GM

Problem: Find minimum of $2x + \frac{9}{x}$ for $x > 0$.

Solution:

Using weighted AM-GM with weights 2:1:

$\frac{2x + 2x + \frac{9}{x}}{3} \geq \sqrt[3]{2x \cdot 2x \cdot \frac{9}{x}} = \sqrt[3]{36x}$

This doesn’t simplify nicely. Let me try different grouping:

$2x + \frac{9}{x} = x + x + \frac{9}{x}$

By AM-GM: $\frac{x + x + \frac{9}{x}}{3} \geq \sqrt[3]{x \cdot x \cdot \frac{9}{x}} = \sqrt[3]{9x}$

Still not clean. Better approach:

Let $2x = a$ and $\frac{9}{x} = b$. We want minimum of $a + b$ subject to $a \cdot \frac{b}{2} = 2x \cdot \frac{9}{2x} = 9$.

Actually, simplest: By AM-GM on two groups:

$2x + \frac{9}{x} \geq 2\sqrt{2x \cdot \frac{9}{x}} = 2\sqrt{18} = 6\sqrt{2}$

Minimum when $2x = \frac{9}{x}$ → $x^2 = \frac{9}{2}$ → $x = \frac{3}{\sqrt{2}}$

Answer: Minimum = $6\sqrt{2}$

Practice Problems

Level 1: JEE Main Basics

Problem 1: Find AM, GM, and HM of 4 and 9. Verify AM ≥ GM ≥ HM.

Solution

$\text{AM} = \frac{4+9}{2} = 6.5$

$\text{GM} = \sqrt{4 \times 9} = \sqrt{36} = 6$

$\text{HM} = \frac{2 \times 4 \times 9}{4+9} = \frac{72}{13} \approx 5.54$

Check: $6.5 \geq 6 \geq 5.54$ ✓

Answer: AM = 6.5, GM = 6, HM = 72/13

Problem 2: Find the minimum value of $x^2 + \frac{4}{x^2}$ for $x > 0$.

Solution

By AM-GM: $\frac{x^2 + \frac{4}{x^2}}{2} \geq \sqrt{x^2 \cdot \frac{4}{x^2}} = \sqrt{4} = 2$

$x^2 + \frac{4}{x^2} \geq 4$

Minimum = 4, when $x^2 = \frac{4}{x^2}$ → $x^4 = 4$ → $x = \sqrt[4]{4} = \sqrt{2}$

Answer: Minimum = 4 at $x = \sqrt{2}$

Problem 3: If AM and GM of two numbers are 10 and 8 respectively, find the numbers.

Solution

$\frac{a+b}{2} = 10$ → $a + b = 20$ … (1)

$\sqrt{ab} = 8$ → $ab = 64$ … (2)

From (1) and (2), $a$ and $b$ are roots of: $t^2 - 20t + 64 = 0$

$t = \frac{20 \pm \sqrt{400 - 256}}{2} = \frac{20 \pm \sqrt{144}}{2} = \frac{20 \pm 12}{2}$

$t = 16$ or $t = 4$

Answer: The numbers are 4 and 16

Level 2: JEE Main/Advanced

Problem 4: If $a, b, c > 0$ and $a + b + c = 3$, find the maximum value of $a^2b^3c^4$.

Solution

Using weighted AM-GM with weights 2:3:4:

$\frac{2a + 3b + 4c}{2+3+4} \geq \sqrt[9]{a^2 b^3 c^4}$

Wait, we need the terms to multiply correctly. Let me use:

$\frac{\frac{a}{2} + \frac{a}{2} + \frac{b}{3} + \frac{b}{3} + \frac{b}{3} + \frac{c}{4} + \frac{c}{4} + \frac{c}{4} + \frac{c}{4}}{9} \geq \sqrt[9]{\left(\frac{a}{2}\right)^2 \left(\frac{b}{3}\right)^3 \left(\frac{c}{4}\right)^4}$

Hmm, this gets messy. Let me try a different approach.

By AM-GM on $\underbrace{a, a}_{2}, \underbrace{b, b, b}_{3}, \underbrace{c, c, c, c}_{4}$ (9 terms):

$\frac{2a + 3b + 4c}{9} \geq \sqrt[9]{a^2 b^3 c^4}$

But we know $a + b + c = 3$, not $2a + 3b + 4c = $ constant.

For maximum of $a^2 b^3 c^4$, at equality in AM-GM: $a = b = c$, but this doesn’t match the weights.

Actually, for weighted AM-GM, equality occurs when all terms being averaged are equal: $\frac{a}{2} = \frac{b}{3} = \frac{c}{4} = k$

So $a = 2k, b = 3k, c = 4k$

From $a + b + c = 3$: $2k + 3k + 4k = 3$ → $k = \frac{1}{3}$

$a = \frac{2}{3}, b = 1, c = \frac{4}{3}$

Maximum value: $\left(\frac{2}{3}\right)^2 \cdot 1^3 \cdot \left(\frac{4}{3}\right)^4 = \frac{4}{9} \cdot \frac{256}{81} = \frac{1024}{729}$

Answer: Maximum = $\frac{1024}{729}$ at $a = \frac{2}{3}, b = 1, c = \frac{4}{3}$

Problem 5: Prove that for positive $a, b, c$: $\frac{a}{b} + \frac{b}{c} + \frac{c}{a} \geq 3$

Solution

By AM-GM on the three terms:

$\frac{\frac{a}{b} + \frac{b}{c} + \frac{c}{a}}{3} \geq \sqrt[3]{\frac{a}{b} \cdot \frac{b}{c} \cdot \frac{c}{a}} = \sqrt[3]{1} = 1$

$\frac{a}{b} + \frac{b}{c} + \frac{c}{a} \geq 3$

Equality when $\frac{a}{b} = \frac{b}{c} = \frac{c}{a} = 1$, i.e., $a = b = c$ ✓

Answer: Proven

Problem 6: If $x + y = 8$, find the maximum value of $xy(16-xy)$ where $x, y > 0$.

Solution

Let $xy = t$. Since $x + y = 8$:

By AM-GM: $\frac{x+y}{2} \geq \sqrt{xy}$ → $4 \geq \sqrt{t}$ → $t \leq 16$

So $0 < t \leq 16$.

We want to maximize $f(t) = t(16-t) = 16t - t^2$

This is a parabola opening downward. Maximum at $t = 8$:

$f(8) = 8(16-8) = 64$

Check if $t = 8$ is achievable: $xy = 8$ and $x + y = 8$

This gives $x = y = 4$ (by symmetry or solving $x^2 - 8x + 8 = 0$… wait:

$x^2 - 8x + 8 = 0$ → $x = \frac{8 \pm \sqrt{64-32}}{2} = \frac{8 \pm \sqrt{32}}{2} = 4 \pm 2\sqrt{2}$

So $x = 4 + 2\sqrt{2}, y = 4 - 2\sqrt{2}$ (or vice versa)

And $xy = (4+2\sqrt{2})(4-2\sqrt{2}) = 16 - 8 = 8$ ✓

Answer: Maximum = 64 at $xy = 8$

Level 3: JEE Advanced

Problem 7: Find the minimum value of $\frac{(x+y+z)^2}{xy+yz+zx}$ for positive $x, y, z$.

Solution

Let $S = x + y + z$ and $P = xy + yz + zx$.

We need to minimize $\frac{S^2}{P}$.

Note: $(x+y+z)^2 = x^2 + y^2 + z^2 + 2(xy + yz + zx)$

So $S^2 = x^2 + y^2 + z^2 + 2P$

By Cauchy-Schwarz or AM-GM: $x^2 + y^2 + z^2 \geq xy + yz + zx$ (can be proven using $(x-y)^2 + (y-z)^2 + (z-x)^2 \geq 0$)

So $S^2 = x^2 + y^2 + z^2 + 2P \geq P + 2P = 3P$

$\frac{S^2}{P} \geq 3$

Minimum = 3, achieved when $x = y = z$

Answer: Minimum = 3

Problem 8: Prove that if $a, b, c > 0$ and $abc = 1$, then $a + b + c \geq 3$.

Solution

By AM-GM: $\frac{a + b + c}{3} \geq \sqrt[3]{abc} = \sqrt[3]{1} = 1$

$a + b + c \geq 3$

Equality when $a = b = c = 1$ ✓

Answer: Proven

Cross-Topic Connections

Optimization problems using derivatives often confirm AM-GM results. See Differential Calculus for comparison.

Cauchy-Schwarz inequality and Holder’s inequality are generalizations of AM-GM. Essential for JEE Advanced.

In triangles, inequalities involving sides and angles use AM-GM principles.

$|z_1 + z_2| \leq |z_1| + |z_2|$ (triangle inequality) relates to mean inequalities. See Complex Numbers.

Quick Revision Checklist

  • AM ≥ GM ≥ HM (only for positive numbers!)
  • Equality holds only when all numbers are equal
  • $\text{GM}^2 = \text{AM} \times \text{HM}$ (KEY formula!)
  • HM = $\frac{2ab}{a+b}$ (product on top!)
  • For max/min with product constant → use AM-GM
  • For max/min with sum constant → use AM-GM
  • $x + \frac{k}{x} \geq 2\sqrt{k}$ (standard result)

Memory Palace Technique

Imagine a podium (means hierarchy):

  1. Gold Medal (AM - highest): Stands at $\frac{a+b}{2}$
  2. Silver Medal (GM - middle): Stands at $\sqrt{ab}$
  3. Bronze Medal (HM - lowest): Stands at $\frac{2ab}{a+b}$
  4. Equality Platform: All three stand together when $a = b$
  5. Positive Zone: Podium only works in positive number territory!

Final Tips for JEE

  1. Check positivity first - AM-GM-HM needs all positive numbers!
  2. Equality condition - Always verify when equality holds
  3. Creative grouping - Sometimes need to rewrite expression cleverly
  4. Weighted AM-GM - For expressions with different coefficients
  5. Maximum vs Minimum - AM-GM gives upper/lower bound depending on constraint
  6. Verify answer - Substitute back to check if equality condition is achievable
  7. Common pattern: $x + \frac{a}{x} \geq 2\sqrt{a}$ appears frequently!

Advanced Results (JEE Advanced)

Power Mean Inequality

For $r > s$:

$$\left(\frac{a_1^r + a_2^r + \cdots + a_n^r}{n}\right)^{1/r} \geq \left(\frac{a_1^s + a_2^s + \cdots + a_n^s}{n}\right)^{1/s}$$

Special cases:

  • $r = 1, s = 0$: AM ≥ GM
  • $r = 0, s = -1$: GM ≥ HM

Cauchy-Schwarz Inequality

$(a_1^2 + a_2^2 + \cdots + a_n^2)(b_1^2 + b_2^2 + \cdots + b_n^2) \geq (a_1b_1 + a_2b_2 + \cdots + a_nb_n)^2$

Last Updated: October 22, 2025