First slide

Theory of expressions

Question

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

Moderate

A

Non-negative
B

Non-positive
C

Zero
D

None of these

Solution

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

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