Let $x\geq 5, y\geq 6, z\geq 7$ such that $x^2+y^2+z^2\geq 125$. Find the minimum value of $x+y+z$.
1. Given the conditions \( x \geq 5 \), \( y \geq 6 \), \( z \geq 7 \), and \( x^2 + y^2 + z^2 \geq 125 \), we need to find the minimum value of \( x + y + z \).

2. Let's introduce new variables to simplify the problem:
   \[
   a = x - 5, \quad b = y - 6, \quad c = z - 7
   \]
   This transformation ensures that \( a \geq 0 \), \( b \geq 0 \), and \( c \geq 0 \).

3. Substituting these into the given inequality:
   \[
   (a + 5)^2 + (b + 6)^2 + (c + 7)^2 \geq 125
   \]

4. Expanding the squares:
   \[
   a^2 + 10a + 25 + b^2 + 12b + 36 + c^2 + 14c + 49 \geq 125
   \]

5. Simplifying the inequality:
   \[
   a^2 + b^2 + c^2 + 10a + 12b + 14c + 110 \geq 125
   \]

6. Subtracting 110 from both sides:
   \[
   a^2 + b^2 + c^2 + 10a + 12b + 14c \geq 15
   \]

7. To minimize \( x + y + z \), we need to minimize \( a + b + c \). Notice that:
   \[
   x + y + z = (a + 5) + (b + 6) + (c + 7) = a + b + c + 18
   \]

8. We need to find the minimum value of \( a + b + c \) such that:
   \[
   a^2 + b^2 + c^2 + 10a + 12b + 14c \geq 15
   \]

9. Consider the case where \( a = 0 \), \( b = 0 \), and \( c = 1 \):
   \[
   0^2 + 0^2 + 1^2 + 10 \cdot 0 + 12 \cdot 0 + 14 \cdot 1 = 1 + 14 = 15
   \]
   This satisfies the inequality.

10. Therefore, the minimum value of \( a + b + c \) is:
    \[
    0 + 0 + 1 = 1
    \]

11. Substituting back, the minimum value of \( x + y + z \) is:
    \[
    x + y + z = a + b + c + 18 = 1 + 18 = 19
    \]

The final answer is \(\boxed{19}\)