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If $x, y$ and $z$ are real numbers, the $x^{2}+4 y^{2}+9 z^{2}-6 y z-3 z x-2 x y$ is always
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non-negative
$x^{2}+4 y^{2}+9 z^{2}-6 y z-3 z x-2 x y$
$=x^{2}+(2 y)^{2}+(3 z)^{2}-(2 y)(3 z)-(3 z)(x)-$ $x(2 y) \geq 0 \forall x, y, z$
$\left[\because \mathrm{a}^{2}+\mathrm{b}^{2}+\mathrm{c}^{2}-\mathrm{ab}-\mathrm{bc}-\mathrm{ca} \geq 0\right]$
$=x^{2}+(2 y)^{2}+(3 z)^{2}-(2 y)(3 z)-(3 z)(x)-$ $x(2 y) \geq 0 \forall x, y, z$
$\left[\because \mathrm{a}^{2}+\mathrm{b}^{2}+\mathrm{c}^{2}-\mathrm{ab}-\mathrm{bc}-\mathrm{ca} \geq 0\right]$
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