If $a, b, c, d > 1$ , then the minimum value of ${log}_{b} a + {log}_{a} b + {log}_{c} d + {log}_{d} c$ is

1

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2

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3

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4

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Solution

The correct option is D $4$
we know that $A . M . \geq G . M .$

$⟹ \frac{l o g_{b} a + l o g_{a} b + l o g_{c} d + l o g_{d} c}{4} \geq \sqrt[4]{l o g_{b} a * l o g_{a} b * l o g_{c} d * l o g_{d} c}$

$⟹ \frac{l o g_{b} a + l o g_{a} b + l o g_{c} d + l o g_{d} c}{4} \geq \sqrt[4]{\frac{l o g a}{l o g b} * \frac{l o g b}{l o g a} * \frac{l o g d}{l o g c} * \frac{l o g c}{l o g d}}$

$⟹ l o g_{b} a + l o g_{a} b + l o g_{c} d + l o g_{d} c \geq 4 \sqrt[4]{1}$

$⟹ l o g_{b} a + l o g_{a} b + l o g_{c} d + l o g_{d} c \geq 4 * 1$

$⟹ l o g_{b} a + l o g_{a} b + l o g_{c} d + l o g_{d} c \geq 4$

therefore the minimum value of $l o g_{b} a + l o g_{a} b + l o g_{c} d + l o g_{d} c$ is $4$

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