Solve $x+\frac{2}{y}=3,y+\frac{2}{z}=3,z+\frac{2}{x}=3 $ in reals 
Find answers of this system of equations in real numbers$$
\left\{ 
\begin{array}{c}
x+\frac{2}{y}=3 \\ 
y+\frac{2}{z}=3 \\ 
z+\frac{2}{x}=3 
\end{array}
\right. 
$$

Things I have done: first I show that all of these numbers should be bigger than $0$. If we take $x<0$ then this would led us to $z>3$ and $\frac{2}{y}>3$. By this we got $y+\frac{2}{z} < \frac{4}{3}$ which is false.summing these three equation and using AM-GM gives: $$9=x+\frac{1}{y}+\frac{1}{y}+y+\frac{1}{z}+\frac{1}{z}+z+\frac{1}{x}+\frac{1}{x}\ge9\sqrt[9]{\frac{1}{xyz}} \rightarrow xyz\ge1$$
so as $xyz\ge1$ we can conclude that $x\ge1,y\ge1,z\ge1$ because $x<1$ would led us to $y<1 $ and $z<1$ which makes $xyz<1$. I tried to show that $x>1,y>1,z>1$ leads to contradiction but I was not successful(well I was wrong as $x=y=z=2$ is an answer). Any other approaches on solving this or help on continuing my solution is appreciated. 

We have $z=3-\frac{2}{x}$, so
$$
y=3-\frac{2}{z}=3-\frac{2}{3-\frac{2}{x}}=
3-\frac{2x}{3x-2}=\frac{7x-6}{3x-2} \tag{1}
$$
Then
$$
0=x+\frac{2}{y}-3=x+\frac{2(3x-2)}{7x-6}-3=
\frac{7x^2 - 21x + 14}{7x-6}
$$
We deduce $7x^2 - 21x + 14=0$, i.e. $x^2-3x+2=0$, or $(x-1)(x-2)=0$, so that
$x=1$ or $x=2$.
In the end we have two solutions : $x=y=z=1$ or $x=y=z=2$.
