Let x, y, and z be real numbers satisfying 4(x²+y²+z²)=a and 4(x-y-z)=3+a. Then a equals
Question:
Let x, y, and z be real numbers satisfying 4(x²+y²+z²)=a and 4(x-y-z)=3+a. Then a equals
Let x, y, and z be real numbers satisfying 4(x²+y²+z²)=a and 4(x-y-z)=3+a. Then a equals
Options
Answer: 3
Explanation:
From the second equation: x - y - z = (3+a)/4. Let x - y - z = t, so t = (3+a)/4. From first equation: x²+y²+z² = a/4. By Cauchy-Schwarz: (x-y-z)² ≤ 3(x²+y²+z²). So t² ≤ 3(a/4). Substituting t = (3+a)/4: (3+a)²/16 ≤ 3a/4. This gives: (3+a)² ≤ 12a, so 9 + 6a + a² ≤ 12a. Rearranging: a² - 6a + 9 ≤ 0, which means (a-3)² ≤ 0. Since a square is always non-negative, (a-3)² = 0, so a = 3.
Explanation:
From the second equation: x - y - z = (3+a)/4. Let x - y - z = t, so t = (3+a)/4. From first equation: x²+y²+z² = a/4. By Cauchy-Schwarz: (x-y-z)² ≤ 3(x²+y²+z²). So t² ≤ 3(a/4). Substituting t = (3+a)/4: (3+a)²/16 ≤ 3a/4. This gives: (3+a)² ≤ 12a, so 9 + 6a + a² ≤ 12a. Rearranging: a² - 6a + 9 ≤ 0, which means (a-3)² ≤ 0. Since a square is always non-negative, (a-3)² = 0, so a = 3.
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