If $x^{2} + y^{2} + z^{2} = 1$ . What is the range of xy+yz+zx?

[-1/2, 3]

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[-1,2]

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[-1/2, 1]

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[-1,1/2]

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Solution

The correct option is C [-1/2, 1]
Option (c)

$x^{2} + y^{2} + z^{2} = 1$ ...............................(1)

We know that

$x^{2} + y^{2} + z^{2} + 2 (x y + y z + x z) = (x + y + z)^{2} \geq 0$

1+ 2(xy + yz +xz) $\geq$ 0

$x y + y z + z x \geq \frac{- 1}{2}$ .............................(2)

Since AM $\geq$ GM

$x^{2} + y^{2} \geq 2 x y$
$x^{2} + z^{2} \geq 2 x z$
$y^{2} + z^{2} \geq 2 z y$
$x^{2} + y^{2} + z^{2} \geq (x y + y z + z x)$
$(x y + y z + z x) \leq 1$

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