We call a function $f: \mathbb{Q}^+ \to \mathbb{Q}^+$ [i]good[/i] if for all $x,y \in \mathbb{Q}^+$ we have: $$f(x)+f(y)\geq 4f(x+y).$$
a) Prove that for all good functions  $f: \mathbb{Q}^+ \to \mathbb{Q}^+$ and $x,y,z \in \mathbb{Q}^+$  $$f(x)+f(y)+f(z) \geq 8f(x+y+z)$$
b) Does there exists a good functions  $f: \mathbb{Q}^+ \to \mathbb{Q}^+$ and $x,y,z \in \mathbb{Q}^+$ such that $$f(x)+f(y)+f(z) < 9f(x+y+z) ?$$
1. We start with the given inequality for a good function \( f \):
   \[
   f(x) + f(y) \geq 4f(x+y)
   \]
   for all \( x, y \in \mathbb{Q}^+ \).

2. To prove that \( f(x) + f(y) + f(z) \geq 8f(x+y+z) \) for all \( x, y, z \in \mathbb{Q}^+ \), we will use the given inequality multiple times.

3. Consider the inequality:
   \[
   f(x) + f(y) \geq 4f(x+y)
   \]
   and apply it to \( f(y) \) and \( f(z) \):
   \[
   f(y) + f(z) \geq 4f(y+z)
   \]

4. Now, we use the inequality again for \( f(x) \) and \( f(y+z) \):
   \[
   f(x) + f(y+z) \geq 4f(x + (y+z)) = 4f(x+y+z)
   \]

5. To combine these results, we need to manipulate the inequalities. First, we rewrite the second inequality:
   \[
   f(y) + f(z) \geq 4f(y+z) \implies \frac{1}{4}(f(y) + f(z)) \geq f(y+z)
   \]

6. Substitute this into the first inequality:
   \[
   f(x) + \frac{1}{4}(f(y) + f(z)) \geq f(x) + f(y+z) \geq 4f(x+y+z)
   \]

7. Similarly, we can write:
   \[
   f(y) + \frac{1}{4}(f(x) + f(z)) \geq f(y) + f(x+z) \geq 4f(x+y+z)
   \]
   and
   \[
   f(z) + \frac{1}{4}(f(x) + f(y)) \geq f(z) + f(x+y) \geq 4f(x+y+z)
   \]

8. Adding these three inequalities together:
   \[
   \left( f(x) + \frac{1}{4}(f(y) + f(z)) \right) + \left( f(y) + \frac{1}{4}(f(x) + f(z)) \right) + \left( f(z) + \frac{1}{4}(f(x) + f(y)) \right) \geq 12f(x+y+z)
   \]

9. Simplifying the left-hand side:
   \[
   f(x) + f(y) + f(z) + \frac{1}{4}(f(y) + f(z)) + \frac{1}{4}(f(x) + f(z)) + \frac{1}{4}(f(x) + f(y)) = f(x) + f(y) + f(z) + \frac{1}{4}(2f(x) + 2f(y) + 2f(z)) = f(x) + f(y) + f(z) + \frac{1}{2}(f(x) + f(y) + f(z)) = \frac{3}{2}(f(x) + f(y) + f(z))
   \]

10. Therefore, we have:
    \[
    \frac{3}{2}(f(x) + f(y) + f(z)) \geq 12f(x+y+z)
    \]

11. Dividing both sides by \( \frac{3}{2} \):
    \[
    f(x) + f(y) + f(z) \geq 8f(x+y+z)
    \]

Thus, we have proven the desired inequality.

The final answer is \( \boxed{ f(x) + f(y) + f(z) \geq 8f(x+y+z) } \).