System of equations: $3^x + 4^x + 5^x = 2^x \cdot 3^{x -1} \cdot y$ This was taken from a local mathematical olympiad in Romania. It is from the year 2003.
Solve for $x, y, z \in \mathbb{R}$:
$$
\left\{ 
\begin{array}{c}
3^x + 4^x + 5^x = 2^x \cdot 3^{x -1} \cdot y \\ 
3^y + 4^y + 5^y = 2^y \cdot 3^{y -1} \cdot z \\ 
3^z + 4^z + 5^z = 2^z \cdot 3^{z -1} \cdot x
\end{array}
\right. 
$$
So, the first thing I notice was $4^x = 2^{2x} = a$ so we get a system of quadratic equations. This proved to be quite hard to deal with, so I gave it up. I also tried dividing by $2^x \cdot 3^{x - 1} \cdot y$ and the others, but it was weird to deal with. Do you have any ideas?
 A: Observe that your system is equivalent to $$\begin{cases}f(x)=y\\ f(y)=z\\f(z)=x\end{cases}$$ Where $$f(x):=3\cdot \left[\left(\frac{3}{6}\right)^x+\left(\frac{4}{6}\right)^x+\left(\frac{5}{6}\right)^x\right]$$ Let $g(x)=f(x)-x$. Adding up the three equations yields $\sum g(x)=0$, so at least one of $g(x), g(y), g(z)$ has to be $\leqslant 0$. It's easy to see that $g(x)$ is strictly decreasing and has its only root at $x=3$.
Case 1. $g(x)\leqslant 0\iff x\geqslant 3$. Since $f(x)$ is also strictly decreasing, this implies that $y=f(x)\leqslant f(3)=3$. Thus $y\leqslant 3\implies z=f(y)\geqslant f(3)=3$. Hence, $z\geqslant 3\implies x=f(z)\leqslant f(3)=3$. So $x\leqslant 3$. But this is absurd unless $x=y=z=3$.
Cases 2 and 3 are analogous.
A: We have:
$$\left( \frac{3}{6} \right)^x+\left( \frac{4}{6} \right)^x+\left( \frac{5}{6} \right)^x =\frac{y}{3} \tag{1}$$
$$\left( \frac{3}{6} \right)^y+\left( \frac{4}{6} \right)^y+\left( \frac{5}{6} \right)^y =\frac{z}{3}\tag{2}$$
$$\left( \frac{3}{6} \right)^z+\left( \frac{4}{6} \right)^z+\left( \frac{5}{6} \right)^z =\frac{x}{3}\tag{3}$$
Let's denote $f(t) =\left( \frac{3}{6} \right)^t+\left( \frac{4}{6} \right)^t+\left( \frac{5}{6} \right)^t  $, this function is decreasing in $t$ (you can prove it easily).
WLOG, suppose $x=\min{\{x,y,z\}}$. This implies that $x \le y$.
From (1) and (3), we have: $x \le y \implies f(x) \ge f(z) \implies x \le z$.
From (2) and (3), we have: $ x \le z \implies  f(z) \le f(y) \implies z \ge y$.
From (1) and (2), we have: $ y \le z \implies  f(x) \le f(y) \implies x \ge y$.
But we already suppose $x=\min{\{x,y,z\}}$ , hence, $x=y$. And from (1) and (2), we have $f(x) = f(y) \implies y = z$. So, $x=y=z$.
It suffices now to solve the equation
$$f(x)=\left( \frac{3}{6} \right)^x+\left( \frac{4}{6} \right)^x+\left( \frac{5}{6} \right)^x =\frac{x}{3}$$
We notice that $f(x)$ is decreasing function while $=\frac{x}{3}$ is an increasing function. Thus, there is at most 1 solution. And because $x=3$ is the solution, then $x=y=z=3$ is the unique solution of the system equation.
A: This is mostly a variant of NN2's answer, deriving the unique solution from an argument about the middle of the three variables rather than the minimum. I'm posting it mainly because it occurs to me that proving that each variable must be "between" the other two might be easier to follow than eyeballing the directions that a bunch of inequalities point.
Let
$$f(u)=3\left(\left(1\over2\right)^u+\left(2\over3\right)^u+\left(5\over6\right)^u\right)$$
and rewrite the system as
$$\begin{align}
f(x)&=y\\
f(y)&=z\\
f(z)&=x
\end{align}$$
Now since $1\over2$, $2\over3$, and $5\over6$ are all less than $1$, $f$ is a decreasing function of $u$. Hence if $x$ is between $y$ and $z$ (i.e., greater than or equal to one and less than or equal to the other), then $f(x)=y$ is between $f(y)=z$ and $f(z)=x$, which in turn implies $f(y)=z$ is between $f(z)=x$ and $f(x)=y$ (which in turn implies $f(z)=x$ is between $f(x)=y$ and $f(y)=z$). Since at least one variable is between the other two, we conclude that each is between the others, i.e., $x=y=z$.
Finally, since $f$ is decreasing, the equation $f(u)=u$ can have at most one solution. By inspection, we see that the unique solution is $u=3$. So the OP's system of equations has $(x,y,z)=(3,3,3)$ as its unique solution.
