Cycling powers

We'll prove there are no such triples.

Let $y=ux$ . Then from $x^y=z^x$ , we find $\begin{aligned} x^{ux} &= z^x \\ x^u &= z \end{aligned}$

Now from $x^y=y^z$ , $\begin{aligned} x^{ux} &= (ux)^{x^u} \\ ux \log x &= x^u \log (ux) \\ u \log x &= x^{u-1} \log (ux) \\ u \log x &= x^{u-1} \log u + x^{u-1} \log x \\ \log x &= \frac{x^{u-1} \log u}{u-x^{u-1}} \end{aligned}$

Let $g(x)=\frac{x^{u-1} \log u}{u-x^{u-1}}$ . Differentiating, we have $g'(x)=\frac{u(u-1)x^u \log u }{\left(x^u-ux\right)^2}$

There are two cases to consider: $u>1$ or $u<1$ . Let's start with...

---

Case $u>1$ :

In this case, each of the quantities that multiply to make the numerator of $g'(x)$ is positive: we have $u>0,u-1>0,\log u>0,x^u>0$ . The denominator is a square, so also positive; hence $g'(x)$ is positive everywhere.

$g(x)$ has a vertical asymptote when $u -x^{u-1}=0$ - that is at $x=\sqrt[u-1]{u}$ . Call this critical value $c$ . Note that $c>1$ .

When $x<c$ , $g(x)>0$ . When $x>c$ , $g(x)<0$ .

Now let's look at the equation. When $0<x<1$ , $\log x<0$ and $g(x)>0$ . So there are no solutions in this region.

When $c<x$ , $\log x>0$ and $g(x)<0$ . So there are no solutions here either.

The only place we might find a solution then is when $1<x<c$ . Taking logs of this, we have $0<\log x<\log c = \frac{\log u}{u-1}$ in this interval.

But $g(1)=\frac{\log u}{u-1}$ , and $g(x)$ is increasing; so there are no solutions for $1<x<c$ either.

Therefore there are no solutions when $u>1$ .

---

Case $u<1$ :

If $(x,y,z)=(a,b,c)$ is a solution, then so are both $(b,c,a)$ and $(c,a,b)$ .

In this case, we have $a>b$ . Now, either $b<c$ , or $b>c$ and $c<a$ .

If $b<c$ , then the triple $(x,y,z)=(b,c,a)$ satisfies $x<y$ , and we have a $u>1$ case.

If $b>c$ , then $c<a$ and the triple $(x,y,z)=(c,a,b)$ satisfies $x<y$ , and again we have a $u>1$ case.

---

In conclusion, there are no solutions in positive reals to the original equation.

Let x^y=y^z=z^x=e^k(sup) =>ln(x)=k/y, ln(y)=k/z, ln(z)=k/x. On eliminating y and z we get, ln(ln(x))=ln(k)-ke^(-k/x). Now observe that this equation must have 3 real roots(for correspondence to 3 variables x,y,z in the question). However both the functions are monotonic in nature for x>0, thus only 1 solution is possible which does not serve our purpose for 3 solution