# Inverse of power-2 rational function

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).

• Can you write the condition for an inverse of $f$? Like, it is a function $g(a,b)$ such that ...? – Dietrich Burde Jan 25 at 19:45
• @DietrichBurde Updated. – ARaspiK Jan 26 at 10:01

You can write $$y=\frac {4xb}{(x+b)^2}\\ x^2y+2bxy+-4bx+b^2y=0\\ x=\frac{4b-2by\pm\sqrt{(4b-2by)^2-4b^2y^2}}{2y}\\ x=\frac{2b-by\pm\sqrt{4b^2-4by}}{y}$$
• Can you please detail how you went from $x^2y+2bxy+-4bx+b^2y=0$ to the next step? – ARaspiK Jan 26 at 7:31
• I just plugged into the quadratic formula. We have a quadratic in $x$. – Ross Millikan Jan 26 at 15:16