Cyclic inequality, need help

Question

$x+\frac{1}{y}=10$; $y+\frac{1}{z}=10$; $z+\frac{1}{x}=10$;

What is the highest possible value of z?

Answer 1

$x=\frac{1}{10-z}$ and $1=y\left(10-\frac{1}{10-z}\right)$,

which gives $y=\frac{10-z}{99-10z}$.

Hence, $$\frac{(10-z)z}{99-10z}+1=10z$$ or $$z^2-10z+1=0,$$ which gives the answer: $5+\sqrt{24}$.

Answer 2

As you put it, it's an equation with two solutions. The one with the highest z has $z=5+2\sqrt 6$