Question

If $x,y,z$ are distinct real numbers then the range of $f(x,y,z)=x^2+4y^2+9z^2-6yz+3zx-2xy$ is

• A. $[0,\infty )$
• B. $(-\infty ,0)$
• C. $(1,\infty )$
• D. $(2,\infty )$

Answer 1

Answer: (A)

$[0,\infty )$

The given expression $x^2+4y^2+9z^2-6yz+3zx-2xy$ can be written as $\frac{(x-2y{)}^2+(2y-3z{)}^2+(3z+x{)}^2}{2}$, which is the sum of $3$ square terms. Hence, always positive.

So, the range is $[0,\infty )$.