How to solve $\frac{f^{-1}(x)f(x)}{x}=\frac{f^{-1}(x)+f(x)}{2}$? I've come across the following functional equation:
Determine all surjective functions $f:\mathbb{R_{>0}}\to\mathbb{R_{>0}}$ which satisfy for all $x\in\mathbb{R_{>0}}$:
$$
2xf(f(x))=f(x)\left(x+f(f(x))\right)
$$
My approach so far:
The equation is equivalent to:
$$
x=\frac{f(x)f(f(x))}{2f(f(x))-f(x)}
$$
Which implies injectivity. Thus, $f$ is bijective and therefore, there exists a bijective function $f^{-1}:\mathbb{R_{>0}}\to\mathbb{R_{>0}}$ such that $f^{-1}(f(x))=x$. This allows the, in my opinion, very beautiful reformulation:
Find all (bijective) functions such that for all $x\in\mathbb{R_{>0}}$:
$$
\frac{f^{-1}(x)f(x)}{x}=\frac{f^{-1}(x)+f(x)}{2}
$$
But now I don't know how to proceed further. So any help will be appreciated.
 A: I will write a sketch of a solution:
As suggested in my comment above, we fix some $x_0>0$. Define $x_{n+1}=f(x_n)$ recursively.
Note that your equation implies $x_k>0$ and $2x_kx_{k+2}=x_kx_{k+1}+x_{k+1}x_{k+2}$ for all $k$.
Let $y_k=\frac{1}{x_k}$.
Then this equation is equivalent to $y_{n+2}+y_n=2y_{n+1}$ i.e. $y_n=an+b$ for some $a,b$. Since $y_n>0$ for all $n$ we conclude $a \ge 0$.
Also, by your ideas $f$ must be bijective i.e. $f^{-1}$ exists and so we can define $y_{-1},y_{-2},\dotsc$ which satisfy the same equation and are also positive.
Now, similar to above we can conclude $a \le 0$ because otherwise $y_k$ would be negative for $k \to -\infty$. So $a=0$ and hence $y_n$ is constant.
In particular, we find $f(x_0)=x_1=\frac{1}{y_1}=\frac{1}{y_0}=x_0$.
Since $x_0$ was chosen arbitrarily, we find $f(x)=x$ for all $x$ which is indeed a solution.
Remark: This really should be the standard approach when you see problems that only involve $x,f(x),f(f(x)),\dotsc$ in some combination. But often for these methods to work, you need some additional constraint such as continuity or, in this case, that $f(x)>0$ for all $x$.
