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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?

Note by Ryan Tamburrino
2 years, 8 months ago

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Let \(a = 1 + x, b = 1 + y\) 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. :)

Brian Charlesworth - 2 years, 8 months ago

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Awesome, thanks!

Ryan Tamburrino - 2 years, 8 months ago

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You're welcome. :)

Brian Charlesworth - 2 years, 8 months ago

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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. :)

Ryan Tamburrino - 2 years, 8 months ago

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

Azhaghu Roopesh M - 2 years, 8 months ago

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Sure, I'll get on that soon!

Ryan Tamburrino - 2 years, 8 months ago

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