Let $x, y$ be positive real numbers with $x + y + xy= 3$. Prove that$$x + y\ge 2.$$ When does equality holds?

(K. Czakler, GRG 21, Vienna)
Given the problem, we need to prove that \( x + y \ge 2 \) given the condition \( x + y + xy = 3 \). Let's proceed step-by-step.

1. **Given Condition:**
   \[
   x + y + xy = 3
   \]

2. **Substitute \( x = a - 1 \) and \( y = b - 1 \):**
   Let \( x = a - 1 \) and \( y = b - 1 \). Then the given condition becomes:
   \[
   (a - 1) + (b - 1) + (a - 1)(b - 1) = 3
   \]
   Simplifying this, we get:
   \[
   a + b - 2 + (a - 1)(b - 1) = 3
   \]
   \[
   a + b - 2 + ab - a - b + 1 = 3
   \]
   \[
   ab - 1 = 3
   \]
   \[
   ab = 4
   \]

3. **Using AM-GM Inequality:**
   By the Arithmetic Mean-Geometric Mean (AM-GM) inequality, we know:
   \[
   \frac{a + b}{2} \ge \sqrt{ab}
   \]
   Substituting \( ab = 4 \):
   \[
   \frac{a + b}{2} \ge \sqrt{4}
   \]
   \[
   \frac{a + b}{2} \ge 2
   \]
   \[
   a + b \ge 4
   \]

4. **Relating \( a + b \) to \( x + y \):**
   Recall that \( x = a - 1 \) and \( y = b - 1 \). Therefore:
   \[
   x + y = (a - 1) + (b - 1) = a + b - 2
   \]
   Since we have \( a + b \ge 4 \):
   \[
   x + y = a + b - 2 \ge 4 - 2
   \]
   \[
   x + y \ge 2
   \]

5. **Equality Condition:**
   Equality holds in the AM-GM inequality when \( a = b \). If \( a = b \), then:
   \[
   ab = 4 \implies a^2 = 4 \implies a = 2
   \]
   Thus, \( x = a - 1 = 2 - 1 = 1 \) and \( y = b - 1 = 2 - 1 = 1 \).

Therefore, equality holds when \( x = 1 \) and \( y = 1 \).

\(\blacksquare\)

The final answer is \( \boxed{ x + y \ge 2 } \). Equality holds when \( x = 1 \) and \( y = 1 \).