Finding Global Inverse of $ f(x,y,z)= \left( \frac{x}{a+x+y+z}, \frac{y}{a+x+y+z}, \frac{z}{a+x+y+z} \right).$

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Let $a\in \mathbb{R}$, $a\neq 0$, and $E=\{(x,y,z)\in \mathbb{R}^3:a+x+y+z\neq 0\}$ and $f:E\rightarrow \mathbb{R}^3$ defined by

$$ f(x,y,z)= \left( \frac{x}{a+x+y+z}, \frac{y}{a+x+y+z}, \frac{z}{a+x+y+z} \right).$$

Compute the inverse of $f$.


I know I can use the Inverse Function Theorem to show local invertibility, however, that does not necessarily imply global invertibility. Thus, even though I can invert the Jacobian of the function, it does not follow that I can recover the inverse function from its Jacobian. Therefore, I have no idea what to try for this problem. Could you please provided a small hint on something to try for this problem? Thank you in advance.

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In order to solve the equation $f(x, y, z) = (u, v, w)$ for $(x, y, z)$, add the three components and conclude that $$ \frac{x+y+z}{a+x+y+z} = u+v+w \implies a+x+y+z = \frac{a}{1-(u+v+w)} \, . $$ Then view at $f(x, y, z) = (u, v, w)$ again.