Minimum value of $b c x + c a y + a b z when x y z = a b c$ is

$a b c$

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$2 a b c$

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$3 a b c$

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none of these

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Solution

The correct option is C
$3 a b c$

Explanation for the correct option:
Using AM>GM in-equality:
We know, A.M $\geq$ G.M
$\Rightarrow \frac{b c x + c a y + a b z}{3} \geq {(a^{2} b^{2} c^{2} x y z)}^{\frac{1}{3}} \Rightarrow b c x + c a y + a b z \geq 3 x y z \Rightarrow b c x + a c y + a b c \geq 3 a b c$
The minimum value is $3 a b c$
Hence, the correct option is (C)

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Similar questions

\frac{a x + b y}{a^{2} + b^{2}} = \frac{b y + c z}{b^{2} + c^{2}} = \frac{c z + a x}{c^{2} + a^{2}}

x y z = a b c

b c x + c a y + a b z

a x + b y + c z = 0, b c x + c a y + a b z = 0,

x y z + a b c (a^{3} x + b^{3} y + c^{3} z) = 0.

a x + b y + c z = 0 . . . (1),

x + y + z = 0 . . . (2),

b c x + c a y + a b z = (b - c) (c - a) (a - b) . . . (3)

△ A B C

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S A S

△ A B C

A B = 12, B C = 8, ∠ B = 60

△ X Y Z

∠ Y

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