Determine the solutions $(x,y,z)$ for the following system: $$\begin{cases}3^x + 4^y = 5^x \\\ 3^y + 4^z = 5^y \\\ 3^z + 4^x = 5^z \end{cases}$$
My initial guess was that $(2,2,2)$ is the only possible solution, and after checking on Desmos, that is correct as far as I can tell. However, I'm not quite sure how to prove that is the only solution. What I tried is the following: $$4^y = 5^x-3^x \iff y\ln(4) = \ln(5^x-3^x) \iff y = \frac{\ln(5^x-3^x)}{\ln(4)} \\\ z = \frac{\ln(5^y-3^y)}{\ln(4)} \\\ x = \frac{\ln(5^z-3^z)}{\ln(4)} \ \text{and by substituting } z \text{ with } y \text{ and } y \text{ with } x \text{ I reach an equation only in } x.$$
From where what I would normally do is creating a helping function by subtracting one of the sides of the equation and then trying to find the roots (although in this case it would be the numbers of roots) of that function by differentiating, based on monotony. However, differentiating such a function is probably not the point of this problem, so I'm quite stuck with how one should start (I'm also thinking that I shouldn't prove that $2$ is the only solution by guessing, but to prove that $2$ is the solution algebraically). I'd appreciate any hints towards solving this exercise!
First note that $f(t) = (\frac35)^t+(\frac45)^t$ is strictly decreasing and hence has only one solution for $f(t)=1$, viz. $t=2$.
Let $x\geqslant y\geqslant z$ first. Then if $x>2$, as $f(x)<1$, we have $5^x>3^x+4^x\geqslant 3^x+4^y$ so this cannot be. Similarly if $z<2$, then $f(z)>1 \implies 5^z< 3^z+4^z\leqslant 3^z+4^x$ so we cannot have this either.
Combining $2\geqslant x$ and $z\geqslant 2$ together gives $2\geqslant x\geqslant y \geqslant z \geqslant 2 \implies x=y=z=2$.
Due to cyclic symmetry in the system, we need to check only one more case, viz. $x\geqslant z\geqslant y$ which can be tackled similarly.