If X, Y are positive real numbers such that `XgtY` and A is any positive real number, then

Options

A. `(X)/(Y)ge(X+A)/(Y+A)`
B. `(X)/(Y)gt(X+A)/(Y+A)`
C. `(X)/(Y)le(X+A)/(Y+A)`
D. `(X)/(Y)lt(X+A)/(Y+A)`

Answer

To solve the problem step by step, we need to establish the relationship between the given variables \(X\), \(Y\), and \(A\) based on the information provided.

### Step-by-Step Solution:

1. **Given Information**:
We know that \(X\) and \(Y\) are positive real numbers and \(X > Y\). We also know that \(A\) is a positive real number.

2. **Add \(A\) to Both Sides**:
Since \(A\) is positive, we can add \(A\) to both \(X\) and \(Y\) without changing the inequality:
\[ X + A > Y + A \]

3. **Form the Ratio**:
We can form the ratio of \(X + A\) to \(Y + A\):
\[ \frac{X + A}{Y + A} \]

4. **Express \(X\) in terms of \(Y\)**:
Let’s express \(X\) in terms of \(Y\) using a variable \(m\):
\[ X = mY \quad \text{where } m > 1 \text{ (since } X > Y\text{)} \]

5. **Substitute \(X\) in the Ratio**:
Substitute \(X = mY\) into the ratio:
\[ \frac{mY + A}{Y + A} \]

6. **Simplify the Ratio**:
We can simplify this expression:
\[ \frac{mY + A}{Y + A} = \frac{Y(m + \frac{A}{Y})}{Y + A} \]

7. **Set the Ratio Equal to \(n\)**:
Let’s denote:
\[ n = \frac{X + A}{Y + A} \]
Therefore, we have:
\[ n = \frac{mY + A}{Y + A} \]

8. **Establish the Relationship**:
We need to analyze when this ratio \(n\) is greater than 1. This leads us to:
\[ n - 1 > 0 \quad \text{and} \quad m - n > 0 \]

9. **Conclude the Inequality**:
From the above conditions, we derive:
\[ \frac{X}{Y} > \frac{X + A}{Y + A} \]

### Final Result:
Thus, we conclude that:
\[ \frac{X}{Y} > \frac{X + A}{Y + A} \]