# Help Wanted!

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
3 years, 3 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. :)

- 3 years, 3 months ago

Awesome, thanks!

- 3 years, 3 months ago

You're welcome. :)

- 3 years, 3 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 $$xyz=x+y+z+2$$ 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. :)

- 3 years, 3 months ago

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

- 3 years, 3 months ago

Sure, I'll get on that soon!

- 3 years, 3 months ago