Let $ x$, $ y$, $ z$ be real numbers satisfying $ x^2\plus{}y^2\plus{}z^2\plus{}9\equal{}4(x\plus{}y\plus{}z)$. Prove that
\[ x^4\plus{}y^4\plus{}z^4\plus{}16(x^2\plus{}y^2\plus{}z^2) \ge 8(x^3\plus{}y^3\plus{}z^3)\plus{}27\]
and determine when equality holds.
Given the equation:
\[ x^2 + y^2 + z^2 + 9 = 4(x + y + z) \]

We need to prove that:
\[ x^4 + y^4 + z^4 + 16(x^2 + y^2 + z^2) \ge 8(x^3 + y^3 + z^3) + 27 \]

1. **Rewriting the given equation:**
   \[ x^2 + y^2 + z^2 + 9 = 4(x + y + z) \]
   This can be rearranged as:
   \[ x^2 + y^2 + z^2 - 4(x + y + z) + 9 = 0 \]

2. **Analyzing the expression \( x^4 - 8x^3 + 16x^2 - 9 \):**
   Notice that:
   \[ x^4 - 8x^3 + 16x^2 - 9 = (x-1)^2(x-3)^2 - 9 \]
   This can be rewritten as:
   \[ x^4 - 8x^3 + 16x^2 - 9 = (x-1)^2(x-3)^2 - 9 \ge 0 \]
   Therefore:
   \[ x^4 - 8x^3 + 16x^2 - 9 \ge -6(x^2 - 4x + 3) \]

3. **Summing up the inequalities for \( x, y, z \):**
   \[ \sum (x^4 - 8x^3 + 16x^2 - 9) \ge \sum -6(x^2 - 4x + 3) \]
   This simplifies to:
   \[ \sum x^4 - 8 \sum x^3 + 16 \sum x^2 - 27 \ge -6 \left( \sum x^2 - 4 \sum x + 9 \right) \]

4. **Using the given condition:**
   From the given condition:
   \[ x^2 + y^2 + z^2 - 4(x + y + z) + 9 = 0 \]
   We can substitute this into the inequality:
   \[ \sum x^4 - 8 \sum x^3 + 16 \sum x^2 - 27 \ge -6 \cdot 0 \]
   Simplifying, we get:
   \[ \sum x^4 - 8 \sum x^3 + 16 \sum x^2 - 27 \ge 0 \]

5. **Conclusion:**
   Therefore, we have:
   \[ x^4 + y^4 + z^4 + 16(x^2 + y^2 + z^2) \ge 8(x^3 + y^3 + z^3) + 27 \]

Equality holds when \( (x-1)^2(x-3)^2 = 0 \), which implies \( x, y, z \) are either 1 or 3.

The final answer is \( \boxed{ x^4 + y^4 + z^4 + 16(x^2 + y^2 + z^2) \ge 8(x^3 + y^3 + z^3) + 27 } \)