Let $x$, $y$, $z$ be three positive reals such that $x+y+z=\sqrt{10+\sqrt{19}}$ and $\frac1x+\frac1y+\frac1z=\sqrt{10-\sqrt{19}}$ ...

Solution 1:

Let's write $\sum_{cyc} \frac{x}{y}+\frac{y}{x}$ explicitly as $\frac{x}{y}+\frac{y}{x}+\frac{y}{z}+\frac{z}{y}+\frac{z}{x}+\frac{x}{z}$ and denote this by $A$. Then the arithmetic mean of these $6$ numbers is $\frac{A}{6}$ and the geometric mean of these $6$ numbers is $\sqrt[6]{1}=1$. By AM-GM we have $\frac{A}{6}\geq 1$ with equality if and only if all $6$ numbers are equal, that is, $A\geq 6$ with equality if and only if $\frac{x}{y}=\frac{y}{x}=\frac{y}{z}=\ldots$. Since it was shown before that indeed $A=6$ then all $6$ numbers must be equal and in particular $\frac{x}{y}=\frac{y}{x}$ therefore $x=y$, and similarly for $y=z$.