Inverse of power-2 rational function

Question

I have a function $f(a,b) = \frac{ab}{(\frac{a+b}{2})^2}$, and (to me) it has some cool properties (e.g $f(a,b) = f(b,a)$, $f(x,0) = 0$, $f(x, x) = 1$, $0 \leq f \leq 1$, etc.). Now I wanted to know the inverse function for this. It's a rational function, but I couldn't figure out how to do this. $$ y=\frac{xb}{(\frac{x+b}{2})^2}=\frac{4xb}{(x+b)^2} \\ x=\frac{4yb}{(y+b)^2} \implies x(y+b)^2=4yb \implies \frac{(y+b)^2}{y}=\frac{4b}{x} \\ \frac{y^2 + b^2}{y}=\frac{4b}{x}-2b \implies y+\frac{b^2}{y}=\frac{4b}{x}-2b $$ At this point, I have no idea where to go. I think that I'm missing out on some fundamental rule of solving this, but I don't have a clue about where to go. How do I simplify the $y+\frac{b^2}{y}$ ?
NOTE: The final function $f^{-1}(y,b)$ is such that $f^{-1}(f(x,b),b) = x$ for any $x$ and $b$ such that $x$ and $b$ are not both 0. One can be zero, but not both (since $f(0, 0)$ is undefined).

Answer 1

You can write $$y=\frac {4xb}{(x+b)^2}\\ x^2y+2bxy+-4bx+b^2y=0\\ x=\frac{4b-2by\pm\sqrt{(4b-2by)^2-4b^2y^2}}{2y}\\ x=\frac{2b-by\pm\sqrt{4b^2-4by}}{y}$$