Find the minimum and the maximum value of the expression $\sqrt{4 -a^2} +\sqrt{4 -b^2} +\sqrt{4 -c^2}$
where $a,b, c$ are positive real numbers satisfying the condition $a^2 + b^2 + c^2=6$
To find the minimum and maximum values of the expression \(\sqrt{4 - a^2} + \sqrt{4 - b^2} + \sqrt{4 - c^2}\) where \(a, b, c\) are positive real numbers satisfying \(a^2 + b^2 + c^2 = 6\), we will proceed as follows:

1. **Transform the variables:**
   Let \(x = \sqrt{4 - a^2}\), \(y = \sqrt{4 - b^2}\), and \(z = \sqrt{4 - c^2}\). Then, the problem transforms into finding the minimum and maximum of \(x + y + z\) subject to \(x^2 + y^2 + z^2 = 6\) and \(0 \leq x, y, z \leq 2\).

2. **Establish the range for \(x, y, z\):**
   Since \(a, b, c\) are positive real numbers and \(a^2 + b^2 + c^2 = 6\), it follows that \(0 \leq a, b, c \leq \sqrt{6}\). Consequently, \(0 \leq x, y, z \leq 2\).

3. **Find the minimum value:**
   Without loss of generality, assume \(x = \max \{x, y, z\}\). Therefore, \(x \in [\sqrt{2}, 2]\). Since \(y, z\) are non-negative, we have:
   \[
   6 = x^2 + y^2 + z^2 \leq x^2 + (y + z)^2
   \]
   This implies:
   \[
   x + y + z \geq x + \sqrt{6 - x^2}
   \]
   To find the minimum value, consider the function \(f(x) = x + \sqrt{6 - x^2}\) for \(x \in [\sqrt{2}, 2]\). We need to find the minimum value of \(f(x)\).

4. **Analyze the function \(f(x)\):**
   \[
   f(x) = x + \sqrt{6 - x^2}
   \]
   To find the minimum, we take the derivative and set it to zero:
   \[
   f'(x) = 1 - \frac{x}{\sqrt{6 - x^2}}
   \]
   Setting \(f'(x) = 0\):
   \[
   1 = \frac{x}{\sqrt{6 - x^2}} \implies \sqrt{6 - x^2} = x \implies 6 - x^2 = x^2 \implies 6 = 2x^2 \implies x^2 = 3 \implies x = \sqrt{3}
   \]
   Evaluating \(f(x)\) at \(x = \sqrt{3}\):
   \[
   f(\sqrt{3}) = \sqrt{3} + \sqrt{6 - 3} = \sqrt{3} + \sqrt{3} = 2\sqrt{3}
   \]
   However, we need to check the boundary values \(x = \sqrt{2}\) and \(x = 2\):
   \[
   f(\sqrt{2}) = \sqrt{2} + \sqrt{6 - 2} = \sqrt{2} + 2
   \]
   \[
   f(2) = 2 + \sqrt{6 - 4} = 2 + \sqrt{2}
   \]
   Comparing these values, the minimum value is \(2 + \sqrt{2}\).

5. **Find the maximum value:**
   To find the maximum value, consider the sum of the maximum possible values of \(x, y, z\):
   \[
   x^2 + y^2 + z^2 = 6 \implies x, y, z \leq \sqrt{6}
   \]
   However, since \(x, y, z \leq 2\), the maximum value occurs when \(x = y = z = \sqrt{2}\):
   \[
   x + y + z = \sqrt{2} + \sqrt{2} + \sqrt{2} = 3\sqrt{2}
   \]

Therefore, the minimum value is \(2 + \sqrt{2}\) and the maximum value is \(3\sqrt{2}\).

The final answer is \( \boxed{ 2 + \sqrt{2} \leq \sqrt{4 - a^2} + \sqrt{4 - b^2} + \sqrt{4 - c^2} \leq 3\sqrt{2} } \)