Help Wanted!

Recently, while reading a book, I read that xyz=x+y+z+2xyz = x+y+z+2 is equivalent to 11+x+11+y+11+z=1\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
6 years, 2 months ago

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Comments

Let a=1+x,b=1+ya = 1 + x, b = 1 + y and c=1+zc = 1 + z. Then the first equation can be written as

(a1)(b1)(c1)=a+b+c1abc+(a+b+c)(ab+ac+bc)1=a+b+c1(a - 1)(b - 1)(c - 1) = a + b + c - 1 \Longrightarrow abc + (a + b + c) - (ab + ac + bc) - 1 = a + b + c - 1

abc=ab+ac+ab1=1c+1b+1a.\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 11+x+11+y+11+z=1\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 - 6 years, 2 months ago

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Awesome, thanks!

Ryan Tamburrino - 6 years, 2 months ago

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You're welcome. :)

Brian Charlesworth - 6 years, 2 months ago

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+2xyz=x+y+z+2 implies the existence of reals a,b,ca,b,c such that x=b+ca,y=a+cb,z=a+bcx=\dfrac{b+c}{a}, y=\dfrac{a+c}{b}, z=\dfrac{a+b}{c}. All of which I did, in fact, understand. :)

Ryan Tamburrino - 6 years, 2 months ago

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

A Former Brilliant Member - 6 years, 2 months ago

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Sure, I'll get on that soon!

Ryan Tamburrino - 6 years, 2 months ago
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