Let $x$, $y$, and $z$ be positive real numbers such that $x + y + z = 1$. Prove that

$\frac{(x+y)^3}{z} + \frac{(y+z)^3}{x} + \frac{(z+x)^3}{y} + 9xyz \ge 9(xy + yz + zx)$.

When does equality hold?
1. Given the problem, we need to prove the inequality:
   \[
   \frac{(x+y)^3}{z} + \frac{(y+z)^3}{x} + \frac{(z+x)^3}{y} + 9xyz \ge 9(xy + yz + zx)
   \]
   where \(x + y + z = 1\) and \(x, y, z\) are positive real numbers.

2. We start by applying the AM-GM inequality to each term. For the term \(\frac{(x+y)^3}{z}\), we can write:
   \[
   \frac{(x+y)^3}{z} + 4z^2 = \frac{(x+y)^3}{2z} + \frac{(x+y)^3}{2z} + 4z^2 \ge 3 \sqrt[3]{\left(\frac{(x+y)^3}{2z}\right)^2 \cdot 4z^2}
   \]
   Simplifying the right-hand side:
   \[
   3 \sqrt[3]{\frac{(x+y)^6}{4z^2} \cdot 4z^2} = 3(x+y)^2
   \]

3. Using similar inequalities for the other terms, we get:
   \[
   \frac{(x+y)^3}{z} + \frac{(y+z)^3}{x} + \frac{(z+x)^3}{y} \ge 3(x+y)^2 + 3(y+z)^2 + 3(z+x)^2
   \]

4. We need to show that:
   \[
   3(x+y)^2 + 3(y+z)^2 + 3(z+x)^2 + 9xyz \ge 9(xy + yz + zx)
   \]

5. Simplifying the left-hand side:
   \[
   3(x+y)^2 + 3(y+z)^2 + 3(z+x)^2 = 3(x^2 + y^2 + 2xy + y^2 + z^2 + 2yz + z^2 + x^2 + 2zx) = 6(x^2 + y^2 + z^2) + 12(xy + yz + zx)
   \]

6. Therefore, we need to prove:
   \[
   6(x^2 + y^2 + z^2) + 12(xy + yz + zx) + 9xyz \ge 9(xy + yz + zx)
   \]

7. Simplifying further:
   \[
   6(x^2 + y^2 + z^2) + 3(xy + yz + zx) + 9xyz \ge 0
   \]

8. Since \(x + y + z = 1\), we can substitute \(x + y + z\) into the inequality:
   \[
   2(x^2 + y^2 + z^2) + 9xyz \ge 3(xy + yz + zx)
   \]

9. Expanding and simplifying:
   \[
   2(x^3 + y^3 + z^3 + xy(x+y) + yz(y+z) + zx(z+x)) + 9xyz \ge 3(xy(x+y) + yz(y+z) + zx(z+x) + 3xyz)
   \]

10. This reduces to:
    \[
    2(x^3 + y^3 + z^3) \ge xy(x+y) + yz(y+z) + zx(z+x)
    \]

11. This is true by Muirhead's inequality, rearrangement inequality, or AM-GM inequality. For example, using AM-GM:
    \[
    x^3 + y^3 \ge xy(x+y)
    \]
    and similarly for the other terms.

12. Therefore, the inequality holds, and equality occurs when \(x = y = z = \frac{1}{3}\).

The final answer is \( \boxed{ x = y = z = \frac{1}{3} } \)