Maximum possible value with 3 variables and fractional equations A cool problem I was trying to solve today but I got stuck on:
Find the maximum possible value of $x + y + z$ in the following system of equations:
$$\begin{align}
x^2 – (y– z)x – yz &= 0 \tag1 \\[4pt]
y^2 – \left(\frac8{z^2}– x\right)y – \frac{8x}{z^2}&= 0 \tag2\\[4pt]
z^2 – (x – y)z – xy &= 0 \tag3
\end{align}$$
I tried extending the first equation to
$$x^2 - xy + xz - yz = 0 \tag4$$
I then did the same thing for the second equation and got
$$y^2 - \frac{8y}{z^2} + xy - \frac{8x}{z^2} = 0 \tag5$$
For the 3rd equation:
$$z^2 - xz + yz - xy = 0 \tag6$$
I realized that adding the first and third equations got
$$x^2 + z^2 - 2xy = 0 \tag7$$
I also realized that the 2nd equation could be written as
$$y^2 - \frac{8}{z^2}(x + y) + xy = 0 \tag8$$
but I couldn't get much more. I was thinking that graphs could help me here but I don't have much of an idea.
It would be really helpful if someone could explain to me what I could do further to solve the problem.
 A: From (1):
\begin{align*}
 0 &= x^2 -(y-z)x-yz  \\
    &= (x+y+z)(x-y) -y(x-y)  \text{,}  
\end{align*}
so either
$$ x+y+z = y  \qquad \text{ or } \qquad x-y = 0  \text{.}  $$
We conclude either $x = -z$ or $x = y$.
From (2):
\begin{align*}
0 &= z^2 y^2 - (8-xz^2)y-8x  \\
    &= (x+y+z)(yz^2 - 8) - z(yz^2-8)  \text{,}  
\end{align*}
so either
$$  x+y+z = z  \qquad \text{ or } \qquad yz^2 - 8 = 0  \text{.}  $$
We conclude either $x = -y$ or $yz^2 = 8$.
From (3):
\begin{align*}
0 &= z^2 -(x-y)z-xy  \\
    &= (x+y+z)(z-x) - x(z-x)  \text{,}  \end{align*}
so either
$$  x+y+z = x  \qquad \text{ or } \qquad z-x = 0  \text{.}  $$
We conclude either $y = -z$ or $x = z$.
By cases:

*

*$x = -z$, $x = -y$, $y = -z$: The first two give $y = z$.  With the third, this forces $y = z = 0$ and then $x = 0$, giving $x+y+z = 0$.

*$x = -z$, $x = -y$, $x = z$: The first and third require $x = z = 0$.  With the second, $y = 0$, giving $x+y+z = 0$.

*$x = -z$, $yz^2 = 8$, $y = -z$: Using the third in the second, $-z^3 = 8$, which has no solutions in real numbers.

*$x = -z$, $yz^2 = 8$, $x = z$: The first and third require $x = z = 0$, making the second unsatisfiable.

*$x = y$, $x = -y$, $y = -z$: The first and second require $x = y = 0$.  With the third, $z = 0$, giving $x+y+z = 0$.

*$x = y$, $x = -y$, $x = z$: Again, $x+y+z = 0$ in this case.

*$x = y$, $yz^2 = 8$, $y = -z$:  Again, the second and third are unsatisfiable.

*$x = y$, $yz^2 = 8$, $x = z$: Eliminating $y$ between the first and second and $z$ between the result and third, $x^3 = 8$, so $x = 2$, so $x = y = z = 2$, giving $x+y+z = 6$.

The maximum of $x+y+z$ is $6$ occurring when $x = y = z = 2$.
A: It will be edited, but let's begin with attempt..
From (4): $y=\frac{x(x+z)}{x+z}=x$ (if $x \neq -z$) because otherwise $x + z$ would be $0$.
From (7): $x^2 + z^2 - 2x^2 => x^2 = z^2$.
So if $x \neq -z$, then $x = y = z$
