If a and b are two different positive real numbers then which of the following statement is true?

Options

A. `2sqrt(ab)gta+b`
B. `2sqrt(ab)lta+b`
C. `2sqrt(ab)=a+b`
D. none of these

Answer

To solve the problem, we need to apply the concept of the Arithmetic Mean-Geometric Mean (AM-GM) inequality. Let's go through the steps systematically.

### Step-by-Step Solution:

1. **Understanding AM-GM Inequality**:
The Arithmetic Mean (AM) of two positive real numbers \( a \) and \( b \) is given by:
\[ AM = \frac{a + b}{2} \]
The Geometric Mean (GM) of the same numbers is given by:
\[ GM = \sqrt{ab} \]
According to the AM-GM inequality, for any two positive real numbers \( a \) and \( b \):
\[ AM \geq GM \]

2. **Applying AM-GM to Our Numbers**:
We apply the AM-GM inequality to our numbers \( a \) and \( b \):
\[ \frac{a + b}{2} \geq \sqrt{ab} \]

3. **Rearranging the Inequality**:
To manipulate the inequality, we can multiply both sides by 2 (which is positive, so the direction of the inequality remains unchanged):
\[ a + b \geq 2\sqrt{ab} \]

4. **Considering Different Cases**:
Since \( a \) and \( b \) are different positive real numbers, we can also note that:
\[ \sqrt{ab} < \frac{a + b}{2} \]
This implies that:
\[ 2\sqrt{ab} < a + b \]

5. **Conclusion**:
Therefore, the correct statement that holds true for two different positive real numbers \( a \) and \( b \) is:
\[ 2\sqrt{ab} < a + b \]

### Final Answer:
The correct option is that \( 2\sqrt{ab} < a + b \).