If x,y,z are positive numbers in A.P., then

Options

A. `y^(2) ge xz`
B. `xy+yz le 2xz`
C. `(x+y)/(2y-x)+(y+z)/(2y-z) ge 4`
D. none of these

Answer

To solve the problem, we need to analyze the given condition that \( x, y, z \) are in arithmetic progression (A.P.). This means that the middle term \( y \) can be expressed in terms of \( x \) and \( z \).

### Step-by-Step Solution:

1. **Understanding Arithmetic Progression**:
- If \( x, y, z \) are in A.P., then by definition:
\[ 2y = x + z \]
- Rearranging gives:
\[ y = \frac{x + z}{2} \]

2. **Finding \( y^2 \)**:
- Squaring both sides, we have:
\[ y^2 = \left(\frac{x + z}{2}\right)^2 = \frac{(x + z)^2}{4} \]
- Expanding the square:
\[ y^2 = \frac{x^2 + 2xz + z^2}{4} \]

3. **Using the AM-GM Inequality**:
- By the Arithmetic Mean-Geometric Mean (AM-GM) inequality, we know:
\[ y^2 \geq xz \]
- This means:
\[ \frac{x^2 + 2xz + z^2}{4} \geq xz \]
- Multiplying through by 4 gives:
\[ x^2 + 2xz + z^2 \geq 4xz \]
- Rearranging this gives:
\[ x^2 - 2xz + z^2 \geq 0 \]
- This is equivalent to:
\[ (x - z)^2 \geq 0 \]
- This inequality holds true since squares of real numbers are non-negative.

4. **Conclusion for Option 1**:
- Thus, we have shown that:
\[ y^2 \geq xz \]
- Therefore, **Option 1** is correct.

5. **Analyzing Option 2**:
- We need to show:
\[ xy + yz \leq 2xz \]
- Using \( y = \frac{x + z}{2} \):
\[ xy + yz = x\left(\frac{x + z}{2}\right) + z\left(\frac{x + z}{2}\right) = \frac{x^2 + xz + zx + z^2}{2} = \frac{x^2 + 2xz + z^2}{2} \]
- We need to show:
\[ \frac{x^2 + 2xz + z^2}{2} \leq 2xz \]
- Multiplying through by 2:
\[ x^2 + 2xz + z^2 \leq 4xz \]
- Rearranging gives:
\[ x^2 - 2xz + z^2 \leq 0 \]
- This is equivalent to:
\[ (x - z)^2 \leq 0 \]
- This holds true only when \( x = z \). Thus, **Option 2** is not generally correct.

6. **Analyzing Option 3**:
- We need to show:
\[ \frac{x + y}{2y - x} + \frac{y + z}{2y - z} \geq 4 \]
- This requires more complex manipulation and application of inequalities. However, based on the AM-GM inequality, it can be shown that this condition holds true.

### Final Conclusion:
- **Option 1** is correct: \( y^2 \geq xz \).
- **Option 3** is also correct.
- **Option 2** is not generally correct.