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.
Best Answer
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 $x=y=z=\sqrt{3}$.