if $\frac{1}{1+x+f(y)}+\frac{1}{1+y+f(z)}+\frac{1}{1+z+f(x)}=1$ find the function $f(x)$ Find all functions $f(x):(0,\infty)\to(0,\infty) $satisfying
$$\dfrac{1}{1+x+f(y)}+\dfrac{1}{1+y+f(z)}+\dfrac{1}{1+z+f(x)}=1$$
whenever $x,y,z$ are positive numbers and $xyz=1$
I know this if
$$xyz=1\Longrightarrow \dfrac{1}{1+x+xy}+\dfrac{1}{1+y+yz}+\dfrac{1}{1+z+zx}=1$$
because
\begin{align*}\dfrac{1}{1+x+xy}+\dfrac{1}{1+y+yz}+\dfrac{1}{1+z+zx}&=\dfrac{1}{1+x+xy}+\dfrac{x}{x+xy+xyz}+\dfrac{xy}{xy+xyz+x^2yz}\\
&=\dfrac{1+x+xy}{1+x+xy}\\
&=1
\end{align*}
so I guess
$$f(x)=\dfrac{1}{x}$$
But I can't prove it.Thank you 

Consider a function $f$ that satisfies the proposed condition.


*

*Setting $(x,y,z)=(1,1,1)$ we see that $f(1)=1$.

*Now, consider $t>0$ and let $a=f(t)$, $b=f(1/t)$. Setting $(x,y,z)=(t,1/t,1)$ we get
$$
\frac{1}{1+t+b}+\frac{1}{2+1/t}+\frac{1}{2+a}=1\tag{1}
$$
and setting $(x,y,z)=(1/t,t,1)$ we get
$$
\frac{1}{1+1/t+a}+\frac{1}{2+t}+\frac{1}{2+b}=1\tag{2}
$$
Now it is easy to check that this system of two equations with unknowns $a$ and $b$ has a unique solution $(a,b)=(1/t,t)$. In particular, $f(t)=1/t$.
Indeed, from the $(1)$ we get
$$
b=\frac{1+3t-a t^2}{1+a+a t}
$$
and from the $(2)$ we get
$$
b=\frac{3t+(1-a)t^2}{1+at+a t^2}
$$
Equating these two expressions yields a simple equation equivalent to $(at-1)^2=0$,  (for $t\ne1$.) So, $a=1/t$.

*Conversely, it is easy to check that $t\mapsto 1/t$ is a solution to the proposed problem. So, it is the only one.$\qquad\square$
