If x, y, z are positive then minimum value of
$x^{log y - log z} + y^{log z - log x} + z^{log x - log y}$ is

3

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1

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9

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16

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Solution

The correct option is A $3$
Let $a = x^{l o g y - l o g z}, b = y^{l o g z - l o g x}, c = z^{l o g x - l o g y}$
Now, $l o g (a b c) = l o g a + l o g b + l o g c$
$\Rightarrow l o g (a b c) = (l o g y - l o g z) l o g x + (l o g z - l o g x) l o g y + (l o g x - l o g y) l o g z = 0$
$\Rightarrow a b c = 1$
As $a, b, c$ are positive $A M \geq G M$
$∴ \frac{a + b + c}{3} \geq (a b c)^{\frac{1}{3}} = 1$
$∴ a + b + c \geq 3$

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Similar questions

x, y, z

x^{l o g y - l o g z} + y^{l o g z - l o g x} + z^{l o g x - l o g y}

x^{l o g y - l o g z} \times y^{l o g z - l o g x} \times z^{l o g x - l o g y} = 1

x

y

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> 0

1

log x + log y + log z = 0

x^{\frac{1}{log y} + \frac{1}{log z}} \times y^{\frac{1}{log z} + \frac{1}{log x}} \times z^{\frac{1}{log x} + \frac{1}{log y}}

10

\frac{l o g x}{y - z} = \frac{l o g y}{z - x} = \frac{l o g z}{x - y}

x y z = ?

If $\frac{x (y + z - x)}{l o g x} = \frac{y (z + x - y)}{l o g y} = \frac{z (x + y - z)}{l o g z}$ , then $x^{y} y^{x} = z^{y} y^{z} = x^{z} z^{x}$

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