Lost on "Simple Computations" I have come across the follow assertion: $$\text{for } x,y,z >0, xyz = x+y+z+2$$ may be rewritten as $$\frac{1}{1+x}+\frac{1}{1+y}+\frac{1}{1+z}=1$$
and that proving this is a matter of 'simple computations'. However, I can't seem to see the way about this.
I would say what I've tried so far, but every approach I've tried so far as petered out after a step or two.

I would go reverse engineering for that one $$\frac{1}{x+1} + \frac{1}{1+y} + \frac{1}{1+z} = 1$$
First you add $\frac{1}{x+1} + \frac{1}{1+y}$ to get $$\frac{1+y+1+x}{(x+1)(y+1)}= \frac{2+x+y}{(x+1)(y+1)}$$ now we add this result to $\frac{1}{z+1}$ to get  $$\frac{2+x+y}{(x+1)(y+1)} +  \frac{1}{z+1} = \frac{(2+x+y)(1+z) + (x+1)(y+1)}{(x+1)(y+1)(z+1)} =1$$
now you have $(2+x+y)(1+z) + (x+1)(y+1) = (x+1)(y+1)(z+1) $ Now Subtract $(x+1)(y+1)$ from each side , you will end up with $$(2+x+y)(1+z) = (x+1)(y+1)(z+1)-(x+1)(y+1)$$ now we have $$(2+x+y)(1+z) = (x+1)(y+1)(z+1-1)$$ and so we have 
$$(2+x+y)(1+z) = (x+1)(y+1)z$$ Now expand the LHS and the RHS to get $$2 + 2z + x + xz + y + yz = xyz + zx + zy + z$$. Now $xz$ cancels with $xz$ and $yz$ cancels with $yz$ and you end up with $$ 2 + 2z + x + y = xyz + z$$ Now final step , you subtract $z$ from each side to end up with $$2 + z + x + y = xyz$$ and you are done !!!!
If you don't like reverse engineering , you can reverse what I did, and you would be able to go from that to the fraction form. reverse reverse = direct :)
