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.
 A: 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 $x=y=z=3$.
A: we have $$x=\frac{9}{6-y}$$ and $$z=\frac{9}{6-x}$$ putting things together we have
$$6-\frac{9}{y}=\frac{9}{6-\frac{9}{6-y}}$$
simplifying we get
$$9\,{\frac { \left( y-3 \right) ^{2}}{y \left( -9+2\,y \right) }}=0$$
A: If two of $x,y,z$ are equal, wlog, $x=y$ then $x(6-x)=9 \to x=3$ and $z(6-3)=9 \to z=3$.
If all are $\neq$, wlog $0<x<y<z$, then $9=x(6-y)<z(6-x)=9$, abs.
Finally, the solution is $(3,3,3)$
A: Repeated substitution gives
\begin{eqnarray*}
x \left( 6- \frac{9}{6- \frac{9}{6-x}} \right) =9
\end{eqnarray*}
and this simplifies to $(x-3)^2=0$.
A: Replace $x$ with $3+e$
If $x = 3 + e$ then $z(3-e) = 9; z=\frac 9{3-e}$ and $(3+e)(6-y)=9$ while $y(6- \frac 9{3-e}) = 9$ or $y(2-\frac 3{3-e}) = 3$.
So $y = \frac 3{2-\frac 3{3-e}} = 6-\frac 9{3+e}$
Let solve for $e$
$ \frac 3{2-\frac 3{3-e}}= \frac {3(3-e)}{2(3-e) - 3}=3\frac {3-e}{3-2e}=6-\frac9{3+e}=3(2 - \frac 3{3+e})$
$\frac{(3-e)(3+e)}{3-2e}=(2(3+e) - 3)=3+2e$
$(3-e)(3+e) = (3+2e)(3- 2e)$
$9 - e^2 = 9 - 4e^2$ so $e = 0$
So $x = 3+0 = 3$ and $z=\frac 9{3-0} = 3$ and $y = \frac 3{2-\frac 3{3-0}} = 6-\frac 9{3+0} = 3$
So, unless I made an error, $x = y = z = 3$ is the only solution.
(Actually, I believe this shows that $x=y=z=3$ are the only solutions in $\mathbb R$ whether positive or negative.)
======
If you want to use A.M-G.M (and be perverse)
Note $a + b \ge 2\sqrt ab$ with equality holding if and only if $a=b$. ($a- 2\sqrt ab + b = (\sqrt a - \sqrt b)^2 \ge 0$ with equality holdig if and only if $a = b$.)
So suppose  $x = 3+e; e > 0$
then $x(3-e) = (3+e)(3-e) = 9-e^2 < 9=x(6-y)$ so $3-e < 6-y; y< 3+e$.  But if $x > 3$ then $6-y < 3$ so $y > 3$.
So $y = 3 + d; d < e$.
But the same argument $y(3-d) = (3+d)(3-d) = 9 -d^2 < 9 = y(6-z)$ so $3-d < 6-z; z< 3+d$ but $z > 3$ so
$z = 3+ c; c < d$.
And by the same argument as above $z(3-c) = (3+c)(3-c) = 9 - c^2 < 9 z(6-x)$ so $3-c < 6-x so $x < 3 + c < 3 + d < 3+ e = x$ which is a contradiction.
THe same argument would apply if we had assume $x < 3$ so $x = 3-e$ or if we had begun with any variable.
So $x = y = z = 3$ is the only possible solution.  (If there are any solution; which obviously $x=y=z=3$ is one.)
