If $x y z = a b c$ , then the least value of $b c x + c a y + a b z$ is

3 a b c

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6 a b c

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a b c

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4 a b c

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Solution

The correct option is A $3 a b c$
We know that $A . M . \geq G . M .$

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

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

Therefore,

$\frac{b c x + a c y + a b z}{3} \geq \sqrt[3]{b c x \times a c y \times a b z}$

$⟹ b c x + a c y + a b z \geq 3 \times \sqrt[3]{x y z \times a^{2} b^{2} c^{2}}$

$⟹ b c x + a c y + a b z \geq 3 \times \sqrt[3]{(a b c)^{3}}$ Since, $x y z = a b c$

$⟹ b c z + a c y + a b z \geq 3 a b c$

Therefore the least value of $b c x + a c y + a b z$ is $3 a b c$

Suggest Corrections

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△ A B C

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