System of equations with $3^x+4^x+5^x$ Solve over reals:

\begin{align}
3^x + 4^x + 5^x &amp;= 2^{x} 3^{x-1} y \\
3^y + 4^y + 5^y &amp;= 2^{y} 3^{y-1} z \\
3^z + 4^z + 5^z &amp;= 2^{z} 3^{z-1} x
\end{align}

I searched for it, but can't find anything similar. I don't really know how to begin. Maybe divide by $2^{x}3^{x-1}$ and then what? Can somebody give me a starting idea? Thank you.
Define
$$
f(x) = \frac{3^x+4^x+5^x}{2^x3^{x-1}}
$$
The system is: $f(x) = y, f(y) = z, f(z)= x$.

The function $f(x)$ decreasing, so the function $f(f(f(x)))$ is decreasing.  Thus, there is at most one solution for $f(f(f(x))) = x$.  As Ilya noted, $3$ is that solution.
One of the answers is $x = 3 , y = 3 , z = 3$