AM GM Inequality (original) (raw)

Last Updated : 23 Jul, 2025

**AM-GM Inequality is one of the most famous inequalities in algebra. Before going through AM-GM Inequality, we first need to go through arithmetic and geometric mean concepts.

In this article, we will learn about **AM-GM Inequality, the relationship between AM and GM, and solve examples and problems on it.

Table of Content

What is AM-GM Inequality?

Arithmetic Mean-Geometric Mean (AM-GM) Inequality is a fundamental result in algebra that provides a relationship between the arithmetic mean and the geometric mean of a set of non-negative real numbers. This inequality states that for any list of non-negative real numbers, the arithmetic mean (average) is at least as great as the geometric mean.

For any, two numbers 'a', and 'b' their arithmetic mean is 1/2(a + b) and geometric mean is √(ab). Then AM-GM inequality is,

AM-and-GM-Inequality

AM GM Inequality

AM–GM Inequality Relationship

AM–GM Inequality is discussed below in the article,

For two positive numbers a and b

Arithmetic Mean: A.M = (a+b)/2

Geometric Mean: G.M = √(ab)

A.M ≥ G.M

****(a+b)/2 ≥ √(ab)**

For 'n' positive numbers a1, a2, a3, a4, ... an

Arithmetic Mean: A.M = (a1+ a2+ a3+ a4+.......... +an) / n

Geometric Mean: G.M = (a1 a2 a3 a4 .......... an) 1/n

A.M ≥ G.M

****(a** 1 + a 2 + a 3 + a 4 +.......... +a n ) / n ≥ (a 1 **a 2 a 3 **a 4 .......... a n ) **1/n

AM-GM Inequality Formula

For two positive numbers a and b,

****(a+b)/2 ≥ √(ab)**

For 'n' positive numbers a1, a2, a3, a4, ... an

****(a1+ a2+ a3+ a4+.......... +an) / n ≥ (a1 a2 a3 a4 .......... an) 1/n**

AM–GM Inequality Relationship Proof

Statement: For any n positive numbers a1, a2, ... an Arithmetic Mean is always greater than equal to Geometric Mean. **A.M ≥ G.M

**Proof:

For two numbers,

A.M - G.M = (a+b)/2 - √(ab)

A.M - G.M = ½ (a+b - 2 √(ab) )

A.M - G.M = ½ (√a - √b)2

We know that square of any number is positive,( i.e ≥0) Hence,

A.M - G.M ≥0

A.M ≥ G.M

Hence we conclude by above proof that for all positive Numbers, A.M ≥ G.M

**Article Related to AM GM Inequality:

Example on AM-GM Inequality

**Example 1: Find the arithmetic mean of 3 and 27

**Solution:

Arithmetic Mean: A.M = (a+b)/2

A.M = (3+27)/2

A.M = 15

**Example 2: Find the Geometric Mean of 3 and 27

**Solution:

Geometric Mean: G.M = √(ab)

G.M = √(27 × 3) = √81

G.M = 9

**Example 3: If x>0, then Prove That: x+ (1/x) ≥ 2

**Solution:

Since x>0 we can apply A.M-G.M Inequality here,

A.M ≥ G.M

(x+1/x) / 2 ≥ (x . 1/x) ½

(x+1/x) /2 ≥ 1

x + (1/x) ≥ 2 (Proved)

**Example 4: If x,y>0,then Prove That: x 2 +y 2 ≥ 2xy

**Solution:

Since x,y>0 we can apply A.M-G.M Inequality here,

A.M ≥ G.M

For two variables x and y

(x+y) /2 ≥ (xy)1/2

squaring both sides:

(x+y)2 /4 ≥ xy

x2+ y2 ≥ 2xy

**Example 5: If x,y,z ≤ 0,then can we Prove That: (x+y)(y+z)(z+x) ≥ 8xyz through A.M-G.M Inequality

**Solution:

Since x, y, z ≤ 0

So, we can't apply A.M-G.M Inequality here,

**Example 6: If a,b,c ∈ R+, such that a+b+c = 3, find the maximum value of abc.

**Solution:

Since a,b,c >0 we can apply A.M-G.M Inequality here.

We need to find the value of the product of a,b,c i.e abc.

Applying A.M ≥ G.M,

(a+b+c)/3 ≥ 3√(abc)

3/3 ≥ (abc)1/3

1 ≥ (abc)1/3

cubing both sides, 13 ≥ (abc)

so, abc ≤1

Hence the maximum value of abc = 1

Summary - AM-GM Inequality

AM-GM Inequality is a fundamental mathematical principle stating that for any set of non-negative real numbers, the arithmetic mean (AM) is always greater than or equal to the geometric mean (GM). Specifically, the inequality formula is expressed as

****(a** 1 + a 2 + a 3 + a 4 +.......... +a n ) / n ≥ (a 1 **a 2 a 3 **a 4 .......... a n ) **1/n

where,

The equality holds if and only if all the numbers in the set are equal. This inequality is crucial in various mathematical contexts, especially in proving bounds and optimizing algebraic expressions.

It finds applications across diverse fields such as economics, engineering, and optimization problems, making it a versatile and powerful tool in theoretical and applied mathematics.