Question

If $x,y$, and $z$ are real and different and $u=x^2+4y^2+9z^2-6yz-3zx-2xy$, then $u$ is always

• A. non-negative
• B. zero
• C. non-positive
• D. cannot be determined

Answer 1

The correct option is A non-negative

$x,y$, and $z$ are real and different and $u=x^2+4y^2+9z^2-6yz-3zx-2xy$
$\Rightarrow 2u=2x^2+8y^2+9z^2-12yz-6zx-4xy$
$\Rightarrow 2u=(x-2y{)}^2+(x-3z{)}^2+(2y-3z{)}^2$
$\Rightarrow u\ge 0$
$\therefore u$ is always non-negative.
Hence, option A.