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Question: Answered & Verified by Expert
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
MathematicsBasic of MathematicsBITSATBITSAT 2020
Options:
  • A positive
  • B non-positive
  • C zero
  • D non-negative
Solution:
2891 Upvotes Verified Answer
The correct answer is: non-negative
$x^{2}+4 y^{2}+9 z^{2}-6 y z-3 z x-2 x y$

$=x^{2}+(2 y)^{2}+(3 z)^{2}-(2 y)(3 z)-(3 z)(x)-$ $x(2 y) \geq 0 \forall x, y, z$

$\left[\because \mathrm{a}^{2}+\mathrm{b}^{2}+\mathrm{c}^{2}-\mathrm{ab}-\mathrm{bc}-\mathrm{ca} \geq 0\right]$

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