The inequality `x(lambda - x)lt y(lambda - y)AA x, y` with `x lt y lt 1` is always true if

Options

A. `lambda le 2`
B. `lambda ge 2`
C. `lambda lt -1`
D. for no value of `lambda`

Explanation

To solve the inequality \( x(\lambda - x) < y(\lambda - y) \) given the conditions \( x < y < 1 \), we will follow these steps:

### Step 1: Rewrite the Inequality
Start with the original inequality:
\[ x(\lambda - x) < y(\lambda - y) \]
This can be rewritten as:
\[ x\lambda - x^2 < y\lambda - y^2 \]

### Step 2: Rearrange the Terms
Rearranging gives us:
\[ x\lambda - y\lambda < x^2 - y^2 \]
Factoring both sides, we have:
\[ \lambda(x - y) < x^2 - y^2 \]

### Step 3: Factor the Right Side
The right side can be factored using the difference of squares:
\[ x^2 - y^2 = (x - y)(x + y) \]
Substituting this back into the inequality gives:
\[ \lambda(x - y) < (x - y)(x + y) \]

### Step 4: Divide by \( (x - y) \)
Since \( x < y \), we know \( x - y < 0 \). Thus, we can divide both sides by \( (x - y) \) and reverse the inequality:
\[ \lambda > x + y \]

### Step 5: Consider the Range of \( x \) and \( y \)
Given that \( x < y < 1 \), we know that both \( x \) and \( y \) are less than 1. Therefore, the maximum value of \( x + y \) occurs when both \( x \) and \( y \) approach 1. Hence, we can conclude:
\[ x + y < 2 \]
This implies:
\[ \lambda > x + y \quad \text{and} \quad x + y < 2 \implies \lambda < 2 \]

### Step 6: Final Condition
Combining the results, we find that for the inequality to hold true under the given conditions, we need:
\[ \lambda < 2 \]

### Conclusion
Thus, the inequality \( x(\lambda - x) < y(\lambda - y) \) is always true if:
\[ \lambda < 2 \]