Question

Let \(x, y, z\) be real numbers such that \(4(x^2 + y^2 + z^2) = a\) and \(4(x - y - z) = 3 + a\), then find the value of \(a\).

• A. 1
• B. 1 and \(\frac{1}{3}\)
• C. 4
• D. 3

Answer 1

The Correct Option is D

Let the given equations be
4(x² + y² + z²) = a ...(1)
4(x - y - z) = 3 + a ...(2)
From (2), x - y - z = (3+a)/4
Squaring both sides, we get
(x - y - z)² = ((3+a)/4)²
x² + y² + z² - 2xy + 2yz - 2xz = (3+a)²/16
Multiplying by 4,
4x² + 4y² + 4z² - 8xy + 8yz - 8xz = (3+a)²/4
a - 8(xy - yz + xz) = (3+a)²/4
Multiplying by 4,
4a - 32(xy - yz + xz) = (3+a)²
Since x, y, z are real numbers, (xy - yz + xz) can take any value.
However, we must have 4x² + 4y² + 4z² ≥ 8xy - 8yz + 8xz which means a ≥ (3+a)²/4.
4a ≥ 9 + 6a + a²
a² + 2a + 9 ≤ 0
The discriminant is 2² - 4(9) = -32 < 0. Since the parabola opens upwards, this inequality has no real solutions for a.
Let's consider the Cauchy-Schwarz inequality:
(x² + y² + z²)(1² + 1² + 1²) ≥ (x - y - z)²
3(x² + y² + z²) ≥ (x - y - z)²
3(a/4) ≥ ((3+a)/4)²
12a ≥ (3+a)²
12a ≥ 9 + 6a + a²
a² - 6a + 9 ≤ 0
(a - 3)² ≤ 0
Since the square of a real number is non-negative, the only solution is (a - 3)² = 0, which implies a = 3.
Final Answer: The final answer is $\boxed{3}$