Knowing that $ b$ is a real constant such that $b\ge 1$, determine the sum of the real solutions of the equation $$x =\sqrt{b-\sqrt{b+x}}$$
To determine the sum of the real solutions of the equation 
\[ x = \sqrt{b - \sqrt{b + x}}, \]
we will first analyze the given equation and then solve for \( x \).

1. **Square both sides of the equation:**
   \[ x^2 = b - \sqrt{b + x}. \]

2. **Isolate the nested square root:**
   \[ \sqrt{b + x} = b - x^2. \]

3. **Square both sides again to eliminate the square root:**
   \[ b + x = (b - x^2)^2. \]

4. **Expand the right-hand side:**
   \[ b + x = b^2 - 2bx^2 + x^4. \]

5. **Rearrange the equation to form a polynomial:**
   \[ x^4 - 2bx^2 + x^2 + b^2 - b = 0. \]

6. **Combine like terms:**
   \[ x^4 - 2bx^2 + x + b^2 - b = 0. \]

Now, we need to find the real solutions of this polynomial equation. To do this, we will analyze the possible values of \( x \).

7. **Check for possible solutions:**
   Let's test the solutions provided in the initial solution:
   \[ x = \frac{1}{2} (\sqrt{4b + 1} + 1), \quad x = \frac{1}{2} (1 - \sqrt{4b + 1}), \quad x = \frac{1}{2} (\sqrt{4b - 3} - 1), \quad x = \frac{1}{2} (-\sqrt{4b - 3} - 1). \]

8. **Verify each solution:**
   - For \( x = \frac{1}{2} (\sqrt{4b + 1} + 1) \):
     \[ x = \frac{1}{2} (\sqrt{4b + 1} + 1). \]
     Substitute \( x \) back into the original equation to verify if it satisfies the equation.

   - For \( x = \frac{1}{2} (1 - \sqrt{4b + 1}) \):
     \[ x = \frac{1}{2} (1 - \sqrt{4b + 1}). \]
     Substitute \( x \) back into the original equation to verify if it satisfies the equation.

   - For \( x = \frac{1}{2} (\sqrt{4b - 3} - 1) \):
     \[ x = \frac{1}{2} (\sqrt{4b - 3} - 1). \]
     Substitute \( x \) back into the original equation to verify if it satisfies the equation.

   - For \( x = \frac{1}{2} (-\sqrt{4b - 3} - 1) \):
     \[ x = \frac{1}{2} (-\sqrt{4b - 3} - 1). \]
     Substitute \( x \) back into the original equation to verify if it satisfies the equation.

9. **Sum the real solutions:**
   After verifying which of the solutions are real and satisfy the original equation, sum the valid solutions.

Conclusion:
The sum of the real solutions of the equation \( x = \sqrt{b - \sqrt{b + x}} \) is given by the sum of the valid solutions found in step 8.

The final answer is \(\boxed{\frac{1}{2} (\sqrt{4b + 1} + 1) + \frac{1}{2} (1 - \sqrt{4b + 1}) + \frac{1}{2} (\sqrt{4b - 3} - 1) + \frac{1}{2} (-\sqrt{4b - 3} - 1)}\).