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If a, b, c are positive then the minimum value of  alogblogc+blogcloga+clogalogb is 

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a
3
b
1
c
9
d
16

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detailed solution

Correct option is A

Since A.M. ≥ G.M.∴ alog⁡b−log⁡c+blog⁡c−log⁡a+clog⁡a−log⁡b3≥alog⁡b−log⁡c⋅blog⁡c−log⁡a⋅clog⁡a−log⁡b3           (1) Let  x=alog⁡b−log⁡c⋅blog⁡c−log⁡a⋅clog⁡a−log⁡b⇒log⁡x=(log⁡b−log⁡c)log⁡a+(log⁡c−log⁡a)log⁡b+(log⁡a−log⁡b)log⁡c⇒  logx=0  ⇒x=1 ∴ From (1), alog⁡b−log⁡c+blog⁡c−log⁡a+clog⁡a−log⁡b≥3.


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