(Belarus 1998) functional equation with weird question: $f[f(x)+1/f(x)]=x+a$

Question

If $a \leq 1$, then there is no function $f : \mathbb{R}^+ → \mathbb{R}^+$ such that $$f\left [f(x)+\frac 1{f(x)}\right]=x+a$$ I showed that f is injective and for a=0 there is no such function because if we substitute x by f(x)+1/f(x) and taking f from both sides of the equation we get f(f(x+1/x))=x and by substituting x by 1/x we get x=1/x which is a contradiction

Answer 1

The equality has two implications: $f(x) + 1/f(x)$ is injective, and the range of $f$ contains $(a, \infty)$. Given $a < 1$ (stronger than in question), there is a positive number $y$ in $(a, 1)$ that is in the range of $f$. The number $z = 1/y > 1$ is also in the range of $f$ and it satisfies that $y+1/y=z+1/z$. This contradicts the injectivity of $f(x) + 1/f(x)$.

I think we can then argue $a=1$ as a special case, but I have yet to come up with a solution. I'll update if I have an idea, or maybe someone else can come up with something better.