How do I find the inverse of $f(x_{1}, x_{2}, x_{3}) = ( \frac{ x_{1} }{1 + x_{1} + x_{2} + x_{3}} , \frac{ x_{2} }{1 + x_{1} + x_{2} + x_{3}} , \frac{ x_{3} }{1 + x_{1} + x_{2} + x_{3}})$
I already managed to prove that the function is inyective but I have no clue on how to find the inverse.
Any help is very appreciated.
Given $(y_1,y_2,y_3)$, you need to find $(x_1,x_2,x_3)$ such that $f(x_1,x_2,x_3)=(y_1,y_2,y_3)$. That is,
$$\left(\frac{x_1}{1+x_1+x_2+x_3},\frac{x_2}{1+x_1+x_2+x_3},\frac{x_3}{1+x_1+x_2+x_3}\right)=(y_1,y_2,y_3).$$
So each coordinate is equal. Then we have $$1=\frac{1+x_1+x_2+x_3}{1+x_1+x_2+x_3}=\frac{1}{1+x_1+x_2+x_3}+y_1+y_2+y_3$$ so $$1+x_1+x_2+x_3=\frac{1}{1-y_1-y_2-y_3}.$$ Then for $1 \leq i \leq 3$, $\frac{x_i}{1+x_1+x_2+x_3}=y_i$, so $x_i=y_i(1+x_1+x_2+x_3)=\frac{y_i}{1-y_1-y_2-y_3}$. That is, $$(x_1,x_2,x_3)=\left(\frac{y_1}{1-y_1-y_2-y_3},\frac{y_2}{1-y_1-y_2-y_3},\frac{y_3}{1-y_1-y_2-y_3}\right).$$ and you should be able to check this works. You have to be careful about the domain of $f$ though - it is not defined if $x_1+x_2+x_3=1$. Additionally, showing $f$ is injective doesn't mean $f^{-1}$ exists on all of $\mathbb{R}^3$, only on $\text{Im}(f)$.
Let $f(x_1,x_2,x_3) = (y_1,y_2,y_3)$. Let us list all that we can find about $y_i$:
1) Add all the $y_i$: $$ \sum y_i = \frac{x_1+x_2+x_3}{1 + x_1+x_2+x_3} = 1-\frac 1{1+\sum x_i} \implies \sum x_i = \frac{1}{1-\sum y_i} - 1 $$ Keep this in mind.
2) Note that $\frac {y_1}{x_1} = \frac {y_2}{x_2} = \frac{y_3}{x_3} = \frac{1}{1+\sum x_i}$.
From $2$, we can say that $x_2 = \frac{x_1y_2}{y_1} ,x_3 =\frac{x_1y_3}{y_1}$.
Substitute this in $1$, and factorize out $x_1$: $$ x_1\left( 1 + \frac{y_2}{y_1} + \frac{y_3}{y_1}\right) = \frac 1{1-\sum y_i} - 1 $$
Hence, $$x_1 = \frac{\frac 1{1-\sum y_i} -1}{\left( 1 + \frac{y_2}{y_1} + \frac{y_3}{y_1}\right)}$$
I leave you to simplify this. Now, from here, $x_2$ and $x_3$ can be found via $2$. Thus, $(x_1,x_2,x_3)$ can be written in terms of $y_1,y_2,y_3$, which entirely describes the inverse function (wherever it is defined i.e. the range of $f(x)$)
