If $x y + y z + z x = 12$ , where $x, y, z$ are positive values, then the greatest value of $x y z$ is

8

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18

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6

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None of the above.

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Solution

The correct option is A $8$
we know that, $A . M . \geq G . M .$

$⟹ \frac{a + b + c}{3} \geq \sqrt[3]{a b c}$

let $a = x y, b = y z, c = z x$

$⟹ \frac{x y + y z + z x}{3} \geq \sqrt[3]{x y \times y z \times z x}$

$⟹ \frac{12}{3} \geq \sqrt[3]{(x y z)^{2}}$ (since, given that $x y + y z + z x = 12$ )

$⟹ 4 \geq \sqrt[3]{(x y z)^{2}}$

$⟹ \sqrt[3]{(x y z)^{2}} \leq 4$

cubing on both sides

$⟹ (x y z)^{2} \leq 4^{3}$

$⟹ (x y z)^{2} \leq 64$

applying square root on both sides

$x y z \leq 8$

therefore, the maximum value of $x y z = 8$

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