AM-GM Inequality and its Applications: Algebra Notes for IMO (Maths Olympiad)

AM‑GM Inequality & Its Applications

Algebra – Beginner Friendly Notes

1. What is AM‑GM?

AM = Arithmetic Mean (average)
GM = Geometric Mean (the “root” way to average)

The rule
For any two non‑negative numbers x and y

(x + y) / 2   ≥   √(x · y)

In plain words
The ordinary average of two numbers is never smaller than the “root” average.
Equality (they are equal) happens only when the numbers are the same.

2. Why is it true? (Very quick proof)

Take the difference of the two sides squared:

((x + y)/2)²  –  (√(x·y))²  =  (x² + 2xy + y²)/4 – xy

Simplify:

= (x² – 2xy + y²)/4
= (x – y)² / 4

A square is always ≥ 0, so the whole expression is ≥ 0.
That means the left side is ≥ the right side.

No fancy tricks needed—just play with squares.

3. Everyday Analogy

Imagine you have two piles of candies, one with x candies and the other with y candies.

Arithmetic mean: If you combine the piles and split them evenly, each pile gets (x+y)/2 candies.
Geometric mean: Think of multiplying the pile sizes and then taking the square root – it’s like asking, “what single pile size would give the same product when squared?”

The rule tells us that the even split (average) will never give you fewer candies than the “root” pile.

4. Tiny Example

Let x = 4, y = 9.

AM = (4 + 9) / 2 = 6.5
GM = √(4 · 9) = √36 = 6

6.5 ≥ 6 → the rule holds.

5. Quick Applications

Goal	How AM‑GM Helps	Small Example
1. Show a product is big when the sum is fixed	Put numbers together: (x+y)/2 ≥ √(xy) → (x+y)² ≥ 4xy	If the perimeter of a rectangle is 20, the area is at most 25.
2. Prove a² + b² ≥ 2ab	Multiply both sides of AM‑GM by 2: x + y ≥ 2√(xy) → square both sides → x² + 2xy + y² ≥ 4xy → x² + y² ≥ 2xy	For 3 and 5: 9 + 25 = 34 ≥ 30.
3. Find the maximum/minimum of a sum or product	Use AM‑GM to set the two numbers equal for the best balance	The product of two numbers with sum 10 is largest when both are 5 (5·5 = 25).

6. Step‑by‑Step Mini‑Problem

Problem
Find the maximum area of a rectangle whose perimeter is 24 cm.

Solution

Let side lengths be x and y.
Perimeter: 2x + 2y = 24 → x + y = 12.

Area = x·y.
Use AM‑GM:

(x + y)/2  ≥  √(x·y)
12/2       ≥  √(x·y)
6          ≥  √(x·y)

Square both sides:
```
36 ≥ x·y
```
The largest area is 36 cm², achieved when x = y = 6.

Answer – 36 cm².

7. Quick “Cheat Sheet”

Situation	AM‑GM Formula	Result
Two numbers, sum fixed	(x + y)/2 ≥ √(xy)	product ≤ (sum/2)²
Three numbers, sum fixed	(x + y + z)/3 ≥ (xyz)^(1/3)	product ≤ (sum/3)³
Equality	x = y (or all equal for >3 numbers)	AM = GM

8. Simple Practice Problems

Show that for any non‑negative a and b:
a² + b² ≥ 2ab
Hint: Start with AM‑GM: (a + b)/2 ≥ √(ab).
If a + b = 10, find the largest possible value of a·b.
For what values of x does x + 1/x reach its minimum?
Hint: Think of AM‑GM with x and 1/x.

9. Final Thoughts

AM‑GM is a simple tool that turns “average” into a solid inequality.
It works for any number of non‑negative items, not just two.
Whenever you want to compare a sum to a product, try AM‑GM first.

Happy practicing! 🎉