In Mathematics, inequality is a relation that makes a non-equal comparison between mathematical expressions or two numbers. The **AMâ€“GM inequality,** or inequality of arithmetic and geometric means, states that the arithmetic means of a list of non-negative real numbers is greater than or equal to the geometric mean of the same list.

If every number in the list is the same, only then there is a possibility that two means are equal. In this section, we will practise problems related to AMâ€“GM inequality.

The simplest non-trivial case, i.e., with more than one variable for two non-negative numbers x and y, is the statement that

with equality if and only if x = y. This case can be seen from the fact that the square of a real number is always non-negative (greater than or equal to zero) and from the elementary case

Hence, (x + y)^{2} â‰¥ 4xy, with equality precisely when (x – y)^{2} = 0, i.e., x = y.

The AMâ€“GM inequality then follows by taking the positive square root of both sides.

For a geometrical interpretation, consider a rectangle with sides of length x and y, hence it has a perimeter 2x + 2y and area xy. Similarly, a square with all sides of length âˆš(xy) has a perimeter of 4âˆš(xy)Â and the same area as the rectangle. The simplest non-trivial case of the AMâ€“GM inequality implies for the perimeters that

### AM-GM Inequality Relationship

Let a_{1}, a_{2},…, a_{n} be n positive real numbers. The Arithmetic Mean and Geometric Mean are defined as:

The AMâ€“GM inequality states that

AM â‰¥ GM

## Solved Examples on AM-GM Inequality

**Example 1:** Prove that

**Solution: **Applying AM > GM

On multiplying equations (1) and (2),

We get

Therefore,

Hence proved.

**Example 2:** Insert 3 GMs between 1 and 16

**Solution:** Let the terms be 1, g_{1}, g_{2}, g_{3}, 16

Three geometric means between 1 and 16 are 2, 4, 8.

**Example 3:** If P, Q, R are positive real numbers such that P + Q + R = g, find the maximum value of P^{2}Q^{3}R^{4}.

**Solution: **P + Q + R = g

Applying AM > GM

3g, 22, 33, 44 > P^{2}Q^{2}R^{4}

**Example 4:** Evaluate 1^{2} + 2^{2} + 3^{2} + 4^{2} + 5^{2}.

**Solution:** Above expression can be written as

We know, sum of squares of first five natural numbers = 1/6 {n(n+1)(2n+1)}

Here, n = 5.

Therefore,

**Example 5: **Prove (a + b + c) (ab + bc + ca) > 9abc.

**Solution: **

Multiplying the above results, we get

= (a^{3}b^{3}c^{3})^{1/3}

= abc

â‡’Â (a + b + c)(ab + bc + ca) > 9abc.

Hence proved.

**Example 5:**Â Consider the functionÂ

**Solution:Â **

Applying the AMâ€“GM inequality forÂ *n*Â = 6, we get

Further, we know that the two sides are equal exactly when all the terms of the mean are equal.

All the points (x, y, z), satisfying these conditions lie on a half-line starting at the origin and are given by,

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