Let $x,y$ be positive real numbers such that $x+y+xy=3$. Prove that $x+y\geq 2$. For what values of $x$ and $y$ do we have $x+y=2$?
1. We start with the given equation:
   \[
   x + y + xy = 3
   \]
2. We rewrite the equation by adding 1 to both sides:
   \[
   x + y + xy + 1 = 4
   \]
3. Notice that the left-hand side can be factored:
   \[
   (x + 1)(y + 1) = 4
   \]
4. We apply the Arithmetic Mean-Geometric Mean (AM-GM) inequality to the terms \(x+1\) and \(y+1\):
   \[
   \frac{(x+1) + (y+1)}{2} \geq \sqrt{(x+1)(y+1)}
   \]
5. Substituting the known value of \((x+1)(y+1)\):
   \[
   \frac{(x+1) + (y+1)}{2} \geq \sqrt{4}
   \]
6. Simplifying the right-hand side:
   \[
   \frac{(x+1) + (y+1)}{2} \geq 2
   \]
7. Multiplying both sides by 2:
   \[
   (x+1) + (y+1) \geq 4
   \]
8. Simplifying the left-hand side:
   \[
   x + y + 2 \geq 4
   \]
9. Subtracting 2 from both sides:
   \[
   x + y \geq 2
   \]
10. To find the values of \(x\) and \(y\) for which \(x + y = 2\), we substitute \(x + y = 2\) into the original equation:
    \[
    x + y + xy = 3
    \]
    \[
    2 + xy = 3
    \]
    \[
    xy = 1
    \]
11. We solve the system of equations:
    \[
    x + y = 2
    \]
    \[
    xy = 1
    \]
12. These are the roots of the quadratic equation:
    \[
    t^2 - (x+y)t + xy = 0
    \]
    \[
    t^2 - 2t + 1 = 0
    \]
    \[
    (t-1)^2 = 0
    \]
    \[
    t = 1
    \]
    Therefore, \(x = 1\) and \(y = 1\).

Conclusion:
\[
\boxed{x + y \geq 2}
\]
\[
\text{For } x + y = 2, \text{ we have } x = 1 \text{ and } y = 1.
\]