Recently, while reading a book, I read that \(xyz = x+y+z+2\) is equivalent to \(\dfrac{1}{1+x} +\dfrac{1}{1+y} + \dfrac{1}{1+z}=1 \). The author implied that this was intuitive, but I cannot see where it comes from. Anyone like to clarify?

Just in case anyone is curious as to what this was getting at, the author then went on and showed how the equation \(xyz=x+y+z+2\) implies the existence of reals \(a,b,c\) such that \(x=\dfrac{b+c}{a}, y=\dfrac{a+c}{b}, z=\dfrac{a+b}{c}\). All of which I did, in fact, understand. :)

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TopNewestLet \(a = 1 + x, b = 1 + y\) and \(c = 1 + z\). Then the first equation can be written as

\((a - 1)(b - 1)(c - 1) = a + b + c - 1 \Longrightarrow abc + (a + b + c) - (ab + ac + bc) - 1 = a + b + c - 1\)

\(\Longrightarrow abc = ab + ac + ab \Longrightarrow 1 = \dfrac{1}{c} + \dfrac{1}{b} + \dfrac{1}{a}.\)

Now rearrange and re-substitute to end up with \(\dfrac{1}{1 + x} + \dfrac{1}{1 + y} + \dfrac{1}{1 + z} = 1\) as desired.

This doesn't seem that "intuitive" to me. :)

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Awesome, thanks!

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You're welcome. :)

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Just in case anyone is curious as to what this was getting at, the author then went on and showed how the equation \(xyz=x+y+z+2\) implies the existence of reals \(a,b,c\) such that \(x=\dfrac{b+c}{a}, y=\dfrac{a+c}{b}, z=\dfrac{a+b}{c}\). All of which I did, in fact, understand. :)

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So will you post a question using the above mentioned Identity ?

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Sure, I'll get on that soon!

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