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.
 A: First of all, as you observed, the condition can be written as
$$\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}=1$$
Let's call the expression $S$. Write $S$ as:
$$\sum_{cyc}(x-1)\left(\frac{1}{y^2}+\frac{1}{x^2}\right) - \frac{1}{x}-\frac{1}{y}-\frac{1}{z}+\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}$$
Since $x,y,z > 1$, we can apply AM-GM as:
$$\frac{1}{y^2}+\frac{1}{x^2} \geq \frac{2}{xy}$$
to get that:
$$S \geq (x-1)\cdot \frac{2}{xy}+(y-1)\cdot \frac{2}{yz}+(z-1)\cdot \frac{2}{zx} - \frac{1}{x}-\frac{1}{y}-\frac{1}{z}+\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}$$
$$=2\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)-2\left(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}\right) - \frac{1}{x}-\frac{1}{y}-\frac{1}{z}+\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}=$$
$$=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}-2+\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}$$
Now, just use the trivial inequalities $a^2+b^2+c^2 \geq ab+bc+ca$ and $(a+b+c)^2\geq 3(ab+bc+ca)$ to get
$$\frac{1}{x}+\frac{1}{y}+\frac{1}{z} \geq \sqrt{3\left(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}\right)} = \sqrt{3}$$
and
$$\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2} \geq \frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}=1$$
This gives the final result $S\geq \sqrt{3}-1$, attained when $x=y=z=\sqrt{3}$.
A: 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 $x+y+z=xyz.$
A: Two solution, one is mine, one is other people.
Solution 1:
Let $x=\frac{1}{a},y=\frac{1}{b},z=\frac{1}{c}$ then $0<a,b,c<1$ and $ab+bc+ca=1$
Need to prove:$$\sum \frac{b^2-ab^2}{a} \geqq \sqrt{3}-1$$
Or $$(\sum \frac{b^2}{a})^2 (ab+bc+ca) \geqq \Big[(\sqrt{3}-1)(ab+bc+ca) +(a^2+b^2+c^2)\Big]^2 $$
Assume $c=\min\{a,b,c\}$ then let $a=c+u,b=c+v (u,v\geqq 0)$ you get something is obvious, but ugly ;)
Equality holds when $x=y=z=\sqrt{3}$
Solution 2:(of other people)
Let $$P= \frac{x - 1}{y^2} + \frac{y - 1}{z^2} + \frac{z - 1}{x^2}$$

This solution has been solved by me and Nguyễn Việt Lâm, see here: https://hoc24.vn/hoi-dap/question/972598.html
