If x, y, z are real and distinct, then $x^{2} + 4 y^{2} + 9 z^{2} - 6 y z - 3 z x - 2 x y$ =

Non-negative

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Non-positive

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Zero

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positive only

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Solution

The correct option is A
Non-negative

x, y, z, ∈ R and distinct.

Now, u = $x^{2} + 4 y^{2} + 9 z^{2} - 6 y z - 3 z x - 2 x y$

$= \frac{1}{2} (2 x^{2} + 8 y^{2} + 18 z^{2} - 12 y z - 6 z x - 4 x y)$

$= \frac{1}{2} {(x^{2} - 4 x y + 4 y^{2}) + (x^{2} - 6 z x + 9 z^{2}) + (4 y^{2} - 12 y z + 9 z^{2})}$

$= \frac{1}{2} {(x - 2 y)^{2} + (x - 3 z)^{2} + (2 y - 3 z)^{2}}$

Since it is sum of squares. So U is always non-negative.

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