If x,;y,;z are real and distinct, then u = {x | vedclass

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Question

(a) $x,y,z \in R$ and distinct.

Now, $u = {x^2} + 4{y^2} + 9{z^2} - 6yz - 3zx - 2xy$

$ = \frac{1}{2}(2{x^2} + 8{y^2} + 18{z^2} - 12yz - 6zx - 4x

Vedclass · Accepted Answer

Non-negative

Vedclass · Answer

(a) $x,y,z \in R$ and distinct.

Now, $u = {x^2} + 4{y^2} + 9{z^2} - 6yz - 3zx - 2xy$

$ = \frac{1}{2}(2{x^2} + 8{y^2} + 18{z^2} - 12yz - 6zx - 4xy)$

$ = \frac{1}{2}\left\{ {{x^2} - 4xy + 4{y^2}) + ({x^2} - 6zx + 9{z^2}) + (4{y^2} - 12yz + 9{z^2})} \right\}$

$ = \frac{1}{2}\left\{ {{{(x - 2y)}^2} + {{(x - 3z)}^2} + {{(2y - 3z)}^2}} \right\}$

Since it is sum of squares. So $u$ is always non- negative.

If $x,\;y,\;z$ are real and distinct, then $u = {x^2} + 4{y^2} + 9{z^2} - 6yz - 3zx - zxy$ is always

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