Prove that the A.M. of two positive real numbers is greater than their G.M.

Explanation

To prove that the Arithmetic Mean (A.M.) of two positive real numbers is greater than their Geometric Mean (G.M.), we will follow these steps:

### Step 1: Define the two positive real numbers
Let the two positive real numbers be \( a \) and \( b \).

### Step 2: Calculate the Arithmetic Mean (A.M.)
The Arithmetic Mean of \( a \) and \( b \) is given by:
\[ A.M. = \frac{a + b}{2} \]

### Step 3: Calculate the Geometric Mean (G.M.)
The Geometric Mean of \( a \) and \( b \) is given by:
\[ G.M. = \sqrt{ab} \]

### Step 4: Set up the inequality
We want to show that:
\[ A.M. > G.M. \]
This translates to:
\[ \frac{a + b}{2} > \sqrt{ab} \]

### Step 5: Rearranging the inequality
To eliminate the fraction, we can multiply both sides by 2 (since both \( a \) and \( b \) are positive, this does not change the direction of the inequality):
\[ a + b > 2\sqrt{ab} \]

### Step 6: Square both sides
Next, we square both sides of the inequality to eliminate the square root:
\[ (a + b)^2 > (2\sqrt{ab})^2 \]
This simplifies to:
\[ a^2 + 2ab + b^2 > 4ab \]

### Step 7: Rearranging the terms
Now, we can rearrange the terms:
\[ a^2 + b^2 + 2ab - 4ab > 0 \]
This simplifies to:
\[ a^2 + b^2 - 2ab > 0 \]

### Step 8: Recognizing a perfect square
Notice that \( a^2 + b^2 - 2ab \) can be rewritten as:
\[ (a - b)^2 > 0 \]

### Step 9: Conclusion
Since \( a \) and \( b \) are positive real numbers, \( (a - b)^2 \) is always non-negative and is equal to zero only when \( a = b \). Therefore, for \( a \neq b \), we have:
\[ (a - b)^2 > 0 \]
This implies that:
\[ A.M. > G.M. \]
Thus, we have proved that the Arithmetic Mean of two positive real numbers is greater than their Geometric Mean.