Substituting x=(b+c)/2a, ... gives g(x)+g(y)+g(z)=1 <= x+y+z=1, 0<x y z<1 where g(x)=f((1-x)/2x). Now I'm stuck..

@above you can do something like the caucy function.The domain are rationals.

@above I tried it in the test and got lost...

since it is defined over $\mathbb{Q}$, we can do the same as this problem without needingthe function to be bounded (after doing what @3above did) https://artofproblemsolving.com/community/c5h1629603p10226145

nice problem! proof sketch $\frac{1}{2x+1}+\frac{1}{2y+1}+\frac{1}{2z+1}=\frac{4xy+4yz+4zx+4x+4y+4z+3}{8xyz+4xy+4yz+4zx+2x+2y+2z+1}=1$ then Let function $g$ such that $f(x)=g(\frac{1}{2x+1})$ then $x+y+z=1 \Leftrightarrow g(x)+g(y)+g(z)=1 $ by Cauchy, $g(x)=kx+\frac{1-k}{3}$ then we can find f(x)

Is the answer f(x)=k(2-2x/3+6x)+1/3?

deleted;;;

Such a bad problem for #3. I will assume every variables are rational numbers. [Main observations] First change the given equation $x+y+z+1=4xyz$ into $$\frac{1}{2x+1}+ \frac{1}{2y+1}+ \frac{1}{2z+1}=1$$After this you can easily see that both $f(x)=1/3$ and $f(x)= 1/(2x+1)$ are the possible answers. Moreover, any linear combination of those are possible answer. Note that $f(1)=1/3$ for any cases. [Step1] Change the equation into usual Cauchy equation Define bijective function $g : \mathbb{Q}^{+} \mapsto \mathbb{Q} \cap (0,1)$ by $g(x)= \frac{1}{2x+1}$. Our condition is just “For $x, y, z \in \mathbb{Q}^{+}$ with $g(x)+g(y)+g(z)=1$, $f(x)+f(y)+f(z)=1$. Let $h(x)+1/3=f(g^{-1}(x+1/3))$. Then the condition is equivalent to “For $x,y,z \in (-1/3, 2/3)$ with $x+y+z=0$, $h(x)+h(y)+h(z)=0$.

the problem is equivalent to “For $0 \le x,y <1/3$, $h(x+y)=h(x)+h(y)$.” [Step2] Solving Cauchy equation for restricted domain Let $h(1/2)=a/2$ for some $a \in \mathbb{R}$. Starting from $h(1/2n)+h(1/2n)=h(2/2n)$, we can get $h(m/2n)=m\times h(1/2n)$ for natural number $m \le n$. Since $h(n/2n)=a/2$, $h(m/2n)=a\times m/2n$ for $m \le n$. Finally, for any $0<x=p/q<2/3$, $h(p/q)=2\times h(p/2q)$ $=2a \times p/2q = a\times p/q.$ From $h(-x)=-h(x)$ and $h(0)=0$, we get $h(x)=ax$ for $-1/3<x<2/3$. Returning back to $f$, we get $f(x)=\frac{1}{3}+a \times (\frac{1}{2x+1}-\frac{1}{3}) = a\times \frac{1}{2x+1} + (1-a) \frac{1}{3}\blacksquare$.

Wow..

Basically the same problem as USAMO 2018/2…