Find  all functions  $f:Q\rightarrow Q$ such that  \[ f(x+y)+f(y+z)+f(z+t)+f(t+x)+f(x+z)+f(y+t)\ge 6f(x-3y+5z+7t) \]  for all  $x,y,z,t\in Q.$
1. Let \( P(x, y, z, t) \) be the assertion of the given functional inequality:
   \[
   f(x+y) + f(y+z) + f(z+t) + f(t+x) + f(x+z) + f(y+t) \ge 6f(x - 3y + 5z + 7t)
   \]
   for all \( x, y, z, t \in \mathbb{Q} \).

2. Consider the assertion \( P(x, 0, 0, 0) \):
   \[
   f(x+0) + f(0+0) + f(0+0) + f(0+x) + f(x+0) + f(0+0) \ge 6f(x - 3 \cdot 0 + 5 \cdot 0 + 7 \cdot 0)
   \]
   Simplifying, we get:
   \[
   4f(x) + 2f(0) \ge 6f(x)
   \]
   Rearranging terms, we obtain:
   \[
   2f(0) \ge 2f(x) \implies f(0) \ge f(x)
   \]
   This implies that \( f(x) \le f(0) \) for all \( x \in \mathbb{Q} \).

3. Next, consider the assertion \( P\left(\frac{x}{2}, \frac{x}{2}, -\frac{x}{2}, \frac{x}{2}\right) \):
   \[
   f\left(\frac{x}{2} + \frac{x}{2}\right) + f\left(\frac{x}{2} - \frac{x}{2}\right) + f\left(-\frac{x}{2} + \frac{x}{2}\right) + f\left(\frac{x}{2} + \frac{x}{2}\right) + f\left(\frac{x}{2} - \frac{x}{2}\right) + f\left(\frac{x}{2} + \frac{x}{2}\right) \ge 6f\left(\frac{x}{2} - 3\left(\frac{x}{2}\right) + 5\left(-\frac{x}{2}\right) + 7\left(\frac{x}{2}\right)\right)
   \]
   Simplifying, we get:
   \[
   3f(x) + 3f(0) \ge 6f(x)
   \]
   Rearranging terms, we obtain:
   \[
   3f(0) \ge 3f(x) \implies f(0) \ge f(x)
   \]
   This implies that \( f(x) \ge f(0) \) for all \( x \in \mathbb{Q} \).

4. Combining the results from steps 2 and 3, we have:
   \[
   f(0) \ge f(x) \quad \text{and} \quad f(x) \ge f(0)
   \]
   Therefore, \( f(x) = f(0) \) for all \( x \in \mathbb{Q} \).

5. Since \( f(x) = f(0) \) for all \( x \in \mathbb{Q} \), the function \( f \) is a constant function. Let \( c = f(0) \). Then \( f(x) = c \) for all \( x \in \mathbb{Q} \).

6. To verify, substitute \( f(x) = c \) into the original inequality:
   \[
   c + c + c + c + c + c \ge 6c
   \]
   Simplifying, we get:
   \[
   6c \ge 6c
   \]
   which is always true.

Therefore, the only solution is \( f(x) = c \) for some constant \( c \in \mathbb{Q} \).

The final answer is \( \boxed{ f(x) = c } \) for some constant \( c \in \mathbb{Q} \).