Find all the solutions $\in \Bbb R^+$ that satisfy $x(6-y)=9$, $y(6-z)=9$, $z(6-x)=9$ Find all the solutions $\in \Bbb R^+$  that satisfy $x(6-y)=9$, $y(6-z)=9$, $z(6-x)=9$
My try
I found that $0 \lt x,y,z \lt 6$
Multiplying the equations we got $x(6-y)y(6-z)z(6-x)=9^3$
$x(6-x)y(6-y)z(6-x)=9^3$
And here is the problem, i applied AM-GM inequality for $(x \;, \;6-x)$
$$\Biggl (\frac{(x+(6-x)}{2}\Biggr) \ge \sqrt {x(x-6)}$$
Expanding out we get $$(x-3)^2\ge0$$
Holding the equality when $x=3$
We can do the same with $(y \;, \; 6-y)$ and $(z \;, \; 6-z)$ getting $(y-3)^2\ge0$ and $(z-3)^2\ge0$ holding when $y,z =3$ and getting that one solution for the system is $x=y=z=3$ but i don't know if this is enough for proving that those are the only solutions.

You got till $x(6-x)y(6-y)z(6-x)=9^3$.  
Now it just remains to note by AM-GM, $x(6-x)y(6-y)z(6-x) \leqslant  \left(\frac16(x+6-x+y+6-y+z+6-z) \right)^6=3^6=9^3$ with equality possible iff $x=6-x=y=6-y=z=6-z$.  So $x=y=z=3$.
