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?

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

## Comments

Sort by:

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. :) – Brian Charlesworth · 1 year, 7 months ago

Log in to reply

– Ryan Tamburrino · 1 year, 7 months ago

Awesome, thanks!Log in to reply

– Brian Charlesworth · 1 year, 7 months ago

You're welcome. :)Log in to reply

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 · 1 year, 7 months ago

Log in to reply

– Azhaghu Roopesh M · 1 year, 7 months ago

So will you post a question using the above mentioned Identity ?Log in to reply

– Ryan Tamburrino · 1 year, 7 months ago

Sure, I'll get on that soon!Log in to reply