Find minimum of $x+y+z$ where $x$, $y$ and $z$ are real numbers such that $x \geq 4$, $y \geq 5$, $z \geq 6$ and $x^2+y^2+z^2 \geq 90$
1. We are given the constraints:
   \[
   x \geq 4, \quad y \geq 5, \quad z \geq 6, \quad \text{and} \quad x^2 + y^2 + z^2 \geq 90
   \]
   We need to find the minimum value of \(x + y + z\).

2. First, let's consider the sum of squares constraint:
   \[
   x^2 + y^2 + z^2 \geq 90
   \]
   Since \(x \geq 4\), \(y \geq 5\), and \(z \geq 6\), we can substitute these values to check if they satisfy the inequality:
   \[
   4^2 + 5^2 + 6^2 = 16 + 25 + 36 = 77
   \]
   Clearly, \(77 < 90\), so the minimum values of \(x\), \(y\), and \(z\) do not satisfy the constraint. We need to find values that do.

3. To find the minimum \(x + y + z\), we need to increase \(x\), \(y\), or \(z\) such that the sum of their squares is at least 90. Let's try increasing \(z\) first, since it has the smallest initial increase effect on the sum \(x + y + z\).

4. Let \(z = 7\):
   \[
   x^2 + y^2 + 7^2 = x^2 + y^2 + 49
   \]
   We need:
   \[
   x^2 + y^2 + 49 \geq 90 \implies x^2 + y^2 \geq 41
   \]
   Now, we need to find the minimum \(x + y\) such that \(x^2 + y^2 \geq 41\) and \(x \geq 4\), \(y \geq 5\).

5. Let's try \(x = 4\) and solve for \(y\):
   \[
   4^2 + y^2 \geq 41 \implies 16 + y^2 \geq 41 \implies y^2 \geq 25 \implies y \geq 5
   \]
   Since \(y = 5\) satisfies the inequality, we have:
   \[
   x = 4, \quad y = 5, \quad z = 7
   \]
   Then:
   \[
   x + y + z = 4 + 5 + 7 = 16
   \]

6. We need to verify if this is indeed the minimum. Suppose we increase \(x\) or \(y\) instead of \(z\). Any increase in \(x\) or \(y\) will result in a higher sum \(x + y + z\) due to the quadratic nature of the sum of squares constraint.

Therefore, the minimum value of \(x + y + z\) is indeed 16.

The final answer is \(\boxed{16}\)