Solve the non-linear system of equations For real $x,y,z>0$ solve the system of equation
\begin{cases}
 \dfrac{1}{x}-3 y+4 z=5,\\
   \dfrac{1}{y}-4 z+5 x=3,\\
      \dfrac{1}{z}-5 x+3 y=4,
\end{cases}
It is easy to check out that
$$
x =\frac 1 5, y= \frac 1 3,  z=\frac 1 4
$$ 
is a solution. 
How I could  prove that there is no any other real solutions?
Edit.  By adding all equations we get 
$$
\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=12.
$$
 A: One possibility is to compute $S$-polynomials from the given polynomials, namely
$$
25x + 9y + 16z - 12=0, \\
24yz + 16z^2 - 32z + 5 =0.
$$
Then we can eliminate $x$ and $y$ and substitute into the three original equations. We obtain
$$
(64z^3 - 80z^2 - 52z + 5)(4z - 1)=0.
$$
Although we have all roots real here, we see that the only positive real solution is the one given above.
A: eliminating $y,z$ we get for $x$:
$$125 x^4-60 x^3-58 x^2+18 x=1$$
A: Multiply the first equation by $5x$, the second by $3y$, the third by $4z$. Then add all three up. This gives you $25x + 9y + 16z = 12$. Now this is clearly a useful intermediate result. For example, you can substitute it into the original equations to eliminate one variable. This yields six new relations:
[1] $4/x -25x - 21y = 8$; 
[2] $3/x + 25x + 28z = 27$
[3] $5/y - 9y - 36z = 3$
[4] $4/y + 9y + 45x = 24$
[5] $5/z + 16z + 24y = 32$
[6] $3/z - 16z - 40x = 0$
From eqs. $1$, $3$ and $6$ we obtain these bounds: $0 < x < \frac {2}{5}$; $0 < y < \frac {1}{3} \sqrt{5}$; $0 < z < \frac {1}{4} \sqrt{3}$. 
Eliminating $y$ from eqs. $1$, $4$ [or alternatively $z$ from eqs. $2$, $6$] yields a fourth order equation for $x$. In the same way fourth order expressions for $y$ and $z$ can be derived. 
