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TopNewestFirst off, notice that \( b^{-1} = ac , a^{-1} = bc, c^{-1} = ab \).

Then,

\( \dfrac{1}{a + b^{-1} + 1} = \dfrac{1}{a + ac + 1} \)

\( \dfrac{1}{b + c^{-1} + 1} = \dfrac{1}{b + ab + 1} = \dfrac{abc}{b + ab + abc} = \dfrac{ac}{1+a+ac}\)

\( \dfrac{1}{c + a^{-1} + 1} = \dfrac{1}{c + bc + 1} = \dfrac{abc}{c+cb + abc} = \dfrac{ab}{1+b+ab} = \dfrac{ab}{abc + b + ab} = \dfrac{a}{1+ a + ac} \)

Add them all up and you'll get the answer.

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thank's very much for your answer. I didn't think like that before. See you at other question.

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Please mention the source of this problem. @Raju Pratama

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it is my homework from my math course.

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O! See my inductive process!

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If U give me sufficient time, I can provide you the complete process by trying myself.

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Well by what time do you want the answer?

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at 20.45

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What's time now? ( Your profile suggests that U live in Indonesia)

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Please replace my induction with inductive reasoning, I was mistakened as my teacher gave me a wrong explanation. Thanks!

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Did U get it?

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no, I don't, but would you like to explain the complete process to me?

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Think of some numbers a,b,c such that abc=1, I thought and found out! Then put them in the expression!

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@Raju Pratama It's one! I did it by induction! take a =2, b=1/2 and c = 1 !

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Lol how can it be called induction Swapnil Das ? I think you are confused between Inductive reasoning (guessing) and Mathematical Induction ! See this.

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please comment.

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I think the answer is one.

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how the way?

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What's your age bro?

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17 years old, how about you?

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