**Proof: Arithmetic Mean is Greater than Geometric Mean**

To prove that the arithmetic mean of two different positive numbers is greater than the geometric mean, let's consider two positive numbers, a and b, where a ≠ b.

**Arithmetic Mean**

The arithmetic mean, denoted by AM, is the sum of two numbers divided by 2. Mathematically, it can be expressed as:

AM = (a + b) / 2

**Geometric Mean**

The geometric mean, denoted by GM, is the square root of the product of two numbers. Mathematically, it can be expressed as:

GM = √(a * b)

**Proof by Contradiction**

To prove the statement, let's assume that the arithmetic mean is not greater than the geometric mean. In other words:

AM ≤ GM

Now, let's substitute the expressions for AM and GM:

(a + b) / 2 ≤ √(a * b)

**Squaring Both Sides**

To simplify the inequality, let's square both sides:

(a + b)^2 / 4 ≤ a * b

Expanding the numerator:

(a^2 + 2ab + b^2) / 4 ≤ a * b

Multiplying both sides by 4:

a^2 + 2ab + b^2 ≤ 4ab

**Simplifying the Inequality**

Rearranging the terms:

a^2 + b^2 - 2ab ≤ 0

Factoring the left side:

(a - b)^2 ≤ 0

Since a and b are positive numbers and a ≠ b, (a - b)^2 will always be greater than zero. However, the inequality states that (a - b)^2 is less than or equal to zero, which is a contradiction.

Therefore, our assumption that AM ≤ GM is incorrect.

**Conclusion**

From the contradiction, we can conclude that the arithmetic mean of two different positive numbers is always greater than the geometric mean.

Hence, the proof is complete.