I am trying to calculate the maximum of $x +y +z$ and $(x_0 ,y_0,z_0)$ is a real solution of the system: $$2x=y+ \frac{2}{y}$$ $$2y=z+ \frac{2}{z}$$ $$2z=x+ \frac{2}{x}$$
I tried to sum them and got that
$$x+y+z=2\left( \frac{1}{y} + \frac{1}{z} +\frac{1}{x} \right)$$
and with $$(x+y+z)\left(\frac{1}{y} + \frac{1}{z} +\frac{1}{x} \right) ≥9,$$ I got
$$x+y+z≥3\sqrt{2}.$$
I am stuck here because they want the maximum of the sum. Any help will be appreciated.
It's easy to notice that $(x,y,z)$ must have the same sign. In other to maximize $x+y+z$, we must have $$\color{red}{x,y,z >0}\tag{1}$$
We have also that $$2x =y+\frac{1}{y} \ge2\sqrt{y\cdot \frac{2}{y}} = 2\sqrt{2}$$ So, we can deduce easily $$\color{red}{x,y,z \ge \sqrt{2}}\tag{2}$$
WOLG, suppose $x\le y$, then
subtract the first and second equations: $$2(x-y) = (y-z)\left(1-\frac{2}{yz} \right) \implies y\le z$$
subtract the second and third equations : $$2(y-z) = (z-x)\left(1-\frac{2}{zx} \right) \implies z\le x$$ we deduce that $\color{red}{x\le y\le z\le x}$. So, $$\color{red}{x=y=z}\tag{3}$$
From $(1),(2)(3)$ we deduce that in order to reach the maximum, we must have $$\color{red}{x=y=z=\sqrt{2}}$$ and the maximum is $\color{red}{x+y+z = 3\sqrt{2}}$.
Remark: there are only 2 real solutions $(x,y,z= = (\sqrt{2},\sqrt{2},\sqrt{2})$ or $(-\sqrt{2},-\sqrt{2},-\sqrt{2})$