To solve the problem, we need to show that each of the real numbers \( x, y, z \) lies in the closed interval \([2/3, 2]\) given the conditions:
1. \( x + y + z = 4 \) (Equation 1)
2. \( x^2 + y^2 + z^2 = 6 \) (Equation 2)
### Step 1: Express \( z \) in terms of \( x \) and \( y \)
From Equation 1, we can express \( z \) as:
\[
z = 4 - x - y
\]
### Step 2: Substitute \( z \) into Equation 2
Now, substitute \( z \) into Equation 2:
\[
x^2 + y^2 + (4 - x - y)^2 = 6
\]
### Step 3: Expand the equation
Expanding \( (4 - x - y)^2 \):
\[
(4 - x - y)^2 = 16 - 8x - 8y + x^2 + y^2
\]
So, substituting this back into the equation gives:
\[
x^2 + y^2 + 16 - 8x - 8y + x^2 + y^2 = 6
\]
### Step 4: Combine like terms
Combine the terms:
\[
2x^2 + 2y^2 - 8x - 8y + 16 = 6
\]
### Step 5: Simplify the equation
Subtract 6 from both sides:
\[
2x^2 + 2y^2 - 8x - 8y + 10 = 0
\]
Dividing the entire equation by 2:
\[
x^2 + y^2 - 4x - 4y + 5 = 0
\]
### Step 6: Rearrange into a standard form
Rearranging gives:
\[
x^2 - 4x + y^2 - 4y + 5 = 0
\]
### Step 7: Complete the square
Completing the square for \( x \) and \( y \):
\[
(x^2 - 4x + 4) + (y^2 - 4y + 4) = -5 + 8
\]
This simplifies to:
\[
(x - 2)^2 + (y - 2)^2 = 1
\]
### Step 8: Analyze the equation
The equation \((x - 2)^2 + (y - 2)^2 = 1\) represents a circle centered at \( (2, 2) \) with a radius of 1. This means that the values of \( x \) and \( y \) can vary within the circle.
### Step 9: Determine the bounds for \( x, y, z \)
The maximum value of \( x \) and \( y \) can be \( 2 + 1 = 3 \) and the minimum can be \( 2 - 1 = 1 \). However, since \( x + y + z = 4 \), we also need to consider \( z \).
### Step 10: Find the bounds for \( z \)
Since \( z = 4 - x - y \), the maximum value of \( z \) occurs when \( x \) and \( y \) are minimized. The minimum values of \( x \) and \( y \) can be \( 2/3 \) each, leading to:
\[
z = 4 - \frac{2}{3} - \frac{2}{3} = 4 - \frac{4}{3} = \frac{8}{3} \text{ (which is greater than 2)}
\]
The minimum value of \( z \) occurs when \( x \) and \( y \) are maximized, leading to:
\[
z = 4 - 2 - 2 = 0 \text{ (which is less than } \frac{2}{3})
\]
### Conclusion
After analyzing all three variables \( x, y, z \), we find that:
- Each of \( x, y, z \) must lie within the interval \([2/3, 2]\).
Thus, we have shown that each of \( x, y, z \) lies in the closed interval \([2/3, 2]\).
