Prove for all real numbers $x, y$, $$(x^2 + 1)(y^2 + 1) + 4(x - 1)(y - 1) \geq 0.$$ Determine when equality holds.
1. We start with the given inequality:
   \[
   (x^2 + 1)(y^2 + 1) + 4(x - 1)(y - 1) \geq 0
   \]

2. First, we expand \((x^2 + 1)(y^2 + 1)\):
   \[
   (x^2 + 1)(y^2 + 1) = x^2 y^2 + x^2 + y^2 + 1
   \]

3. Next, we expand \(4(x - 1)(y - 1)\):
   \[
   4(x - 1)(y - 1) = 4(xy - x - y + 1)
   \]

4. Combining these, we get:
   \[
   (x^2 + 1)(y^2 + 1) + 4(x - 1)(y - 1) = x^2 y^2 + x^2 + y^2 + 1 + 4xy - 4x - 4y + 4
   \]

5. Simplifying the expression:
   \[
   x^2 y^2 + x^2 + y^2 + 1 + 4xy - 4x - 4y + 4 = x^2 y^2 + x^2 + y^2 + 4xy - 4x - 4y + 5
   \]

6. We now rewrite the expression in a different form. Notice that:
   \[
   (x^2 + 1)(y^2 + 1) = (x + yi)(x - yi)(y + i)(y - i)
   \]

7. This can be simplified using complex numbers:
   \[
   (x + yi)(x - yi)(y + i)(y - i) = \{(xy - 1) + (x + y)i\}\{(xy - 1) - (x + y)i\}
   \]

8. This results in:
   \[
   (xy - 1)^2 + (x + y)^2
   \]

9. Therefore, we have:
   \[
   (x^2 + 1)(y^2 + 1) = (xy - 1)^2 + (x + y)^2
   \]

10. Substituting back into the original inequality:
    \[
    (xy - 1)^2 + (x + y)^2 + 4(x - 1)(y - 1) \geq 0
    \]

11. We already expanded \(4(x - 1)(y - 1)\) as:
    \[
    4(xy - x - y + 1)
    \]

12. Combining these, we get:
    \[
    (xy - 1)^2 + (x + y)^2 + 4(xy - x - y + 1) = (xy - 1)^2 + (x + y)^2 + 4xy - 4x - 4y + 4
    \]

13. Simplifying further:
    \[
    (xy - 1)^2 + (x + y)^2 + 4xy - 4x - 4y + 4 = (x + y - 2)^2 + (xy + 1)^2
    \]

14. Since both \((x + y - 2)^2\) and \((xy + 1)^2\) are squares, they are always non-negative:
    \[
    (x + y - 2)^2 + (xy + 1)^2 \geq 0
    \]

15. Equality holds if and only if both terms are zero:
    \[
    (x + y - 2)^2 = 0 \quad \text{and} \quad (xy + 1)^2 = 0
    \]

16. Solving these, we get:
    \[
    x + y = 2 \quad \text{and} \quad xy = -1
    \]

17. These are the roots of the quadratic equation:
    \[
    t^2 - 2t - 1 = 0
    \]

18. Solving this quadratic equation, we find:
    \[
    t = 1 \pm \sqrt{2}
    \]

19. Therefore, the pairs \((x, y)\) that satisfy the equality are:
    \[
    (x, y) = (1 - \sqrt{2}, 1 + \sqrt{2}) \quad \text{or} \quad (1 + \sqrt{2}, 1 - \sqrt{2})
    \]

The final answer is \( \boxed{ (x, y) = (1 - \sqrt{2}, 1 + \sqrt{2}) \quad \text{or} \quad (1 + \sqrt{2}, 1 - \sqrt{2}) } \)