Given positives $x, y , z$ such that $x + y + z = xyz$. Calculate the minimum value of $\frac{x - 1}{y^2} + \frac{y - 1}{z^2} + \frac{z - 1}{x^2}$. 
Given positives $x, y , z > 1$ such that $x + y + z = xyz$. Calculate the minimum value of $$\large \frac{x - 1}{y^2} + \frac{y - 1}{z^2} + \frac{z - 1}{x^2}$$

We have that $x + y + z = xyz \implies \dfrac{1}{yz} + \dfrac{1}{zx} + \dfrac{1}{xy} = 1$ and $$\left(\frac{x - 1}{y^2} + \frac{y - 1}{z^2} + \frac{z - 1}{x^2}\right) \cdot \left(\frac{1}{x - 1} + \frac{1}{y - 1} + \frac{1}{z - 1}\right)$$
$$ \ge \left(\frac{1}{x} + \frac{1}{y} + \frac{1}{z}\right)^2 \ge 3 \cdot \left(\frac{1}{yz} + \frac{1}{zx} + \frac{1}{xy}\right) = 3$$
Now we need to find the maximum value of $\dfrac{1}{x - 1} + \dfrac{1}{y - 1} + \dfrac{1}{z - 1}$.
Let $\dfrac{1}{x} = a, \dfrac{1}{y} = b, \dfrac{1}{z} = c$ which implies that $a, b, c \in (0, 1)$.
It could be observed that $ab + bc + ca = 1$ and $$\dfrac{1}{x - 1} + \dfrac{1}{y - 1} + \dfrac{1}{z - 1} = \frac{a}{1 - a} + \frac{b}{1 - b} + \frac{c}{1 - c}$$
$$ = 3 - \left(\frac{1}{1 - a} + \frac{1}{1 - b} + \frac{1}{1 - c}\right) \le 3 - \frac{9}{3 - (a + b + c)} = \frac{-3(a + b + c)}{3 - (a + b + c)}$$
Then I'm stuck.

For $x=y=z=\sqrt3$ we'll get a value $\sqrt3-1$.
But,  $$\sum_{cyc}\frac{x - 1}{y^2}-\sqrt3+1=\sum_{cyc}\left(\frac{x}{y^2}-\frac{2}{y}+\frac{1}{x}\right)-\sum_{cyc}\left(\frac{1}{x^2}-\frac{1}{xy}\right)+\frac{xy+xz+yz}{xyz}-\sqrt3=$$
$$=\sum_{cyc}\frac{(x-y)^2}{y^2x}-\sum_{cyc}\frac{(x-y)^2}{2x^2y^2}+\frac{xy+xz+yz}{xyz}-\sqrt{\frac{3(x+y+z)}{xyz}}=$$
$$=\sum_{cyc}\frac{(x-y)^2(2x-1)}{2x^2y^2}+\frac{\sum\limits_{cyc}z^2(x-y)^2}{2xyz\left(xy+xz+yz+\sqrt{3xyz(x+y+z)}\right)}\geq0.$$
Now, we see that $\sqrt3-1$ is a minimal value even for $\{x,y,z\}\subset\left[\dfrac{1}{2},+\infty\right)$ such that $x+y+z=xyz.$
