Functional Equation $f(x)f(f(x)+\frac{1}{x})=1$

Question

I'd like to ask how to find all solutions to the functional equation $f(x)\cdot f(f(x)+\frac{1}{x})=1$, where $f: (0, +\infty)\to\mathbb{R}$ is strictly increasing?

Answer 1

I'll answer the problem by myself. For any $x$, let $f(x)=a$. Then, $f(a+\frac{1}{x})=\frac{1}{a}$. Furthermore, we have $f(a+\frac{1}{x})f(f(a+\frac{1}{x})+\frac{1}{a+\frac{1}{x}})=1$. Therefore, $f(\frac{1}{a}+\frac{1}{a+\frac{1}{x}})=a$. Since $f$ is strictly increasing, we have $\frac{1}{a}+\frac{1}{a+\frac{1}{x}}=x$. Thus, the value of $a$ can be obtained by solving the quadratic equation.

Answer 2

Hint:

Let $g(x)=f(x)+\dfrac{1}{x}$ ,

Then $f(x)=g(x)-\dfrac{1}{x}$

$f\left(f(x)+\dfrac{1}{x}\right)=g\left(f(x)+\dfrac{1}{x}\right)-\dfrac{1}{f(x)+\dfrac{1}{x}}=g(g(x))-\dfrac{1}{g(x)}$

$\therefore\left(g(x)-\dfrac{1}{x}\right)\left(g(g(x))-\dfrac{1}{g(x)}\right)=1$