I have the following functional equation for injective $ f : \mathbb R ^ * \to \mathbb R ^ * $ $$ x + y \ne 0 \implies f ( x + y ) \big( f ( x ) + f ( y ) \big) = f ( x y ) $$
Found $ f ( 2 ) = \frac 1 2 $ and I am trying to find $ f ( 1 ) $, so I let $$ x + y = x y \implies y = \frac x { x - 1 } $$ $$ f \left( \frac x { x - 1 } \right) = 1 - f ( x ) $$ then declared $ g : \mathbb R ^ * \setminus \{ 1 \} \to \mathbb R ^ * \setminus \{ 1 \} $ $$ g(x) = \frac{x}{x-1} $$ and was able to prove $ g $'s bijectivity. So, if I can prove $ f $ is surjetive (or alternatively that it is multiplicative or that, at least $ \exists x \, f ( x ) = 1 $) it follows that $ f ( 1 ) = 1 $, $ \forall q \in \mathbb Q \, f ( q ) = q ^ { - 1 } $ and from this point it should be easy to finish the problem.
Am I going the wrong path? Is there an easier way to deal with this problem? Can someone give me hints?
Update:
I was able to make a system of equations with $ f ( - 1 ) $, $ f ( 1 ) $, $ f ( 2 ) $ and $ f ( - 2 ) $, and found $ f ( - 1 ) = - 1 $, $ f ( - 2 ) = \frac { - 1 } 2 $, $ f ( 1 ) = 1$ and $f \left( \frac 3 2 \right) = \frac 2 3 $. Not as elegant as the method I wanted to use but it works.
Using the identity with $ f \circ g ( x ) $ got the values for $ f \left( \frac k { k - 1 } \right)$ plugging positive integers and $ f \left( \frac { k - 1 } k \right) $ with negative ones. Now I am doing some manipulation so I can get from this to $ \frac 1 k $ and from there it should be easy to induce for the rationals. I will post my calculations soon.
The problem is, having no info about continuity, how could we expand this to the reals? That's harder than I thought it would be. We do know, however, the function is decreasing for positive rational numbers. This may be useful.
Second Update:
I got it, the secret is to look at $ \frac 1 { f ( x ) } $, will probably answer this question myself next week.