Show that $\frac{x^2}{(x-1)^2} + \frac{y^2}{(y-1)^2} + \frac{z^2}{(z-1)^2}\ge 1$ for real numbers $x$, $y$, and $z$ with $xyz=1$. I've been thinking to prove each one individually then sum them up which seems too complex to because I haven't made any notable success proving $\frac{x^2}{(x-1)^2}> 1$.
I know that $(x-1)^2\ge 0$ , but I'm not able to transfer it to what I'm looking to prove.
Is there is any techniques to look for when solving these kind of inequalities because I'm totally stuck here.

If
$x=1+a, y=1+b, z=1+c$
then
$a, b, c \ge 0$
since all are positive integers
and
$\begin{array}\\
\dfrac{x^2}{(x-1)^2} + \dfrac{y^2}{(y-1)^2} + \dfrac{z^2}{(z-1)^2}
&=\dfrac{(1+a)^2}{a^2} + \dfrac{(1+b)^2}{b^2} + \dfrac{(1+c)^2}{c^2}\\
&=1+\dfrac{1+2a}{a^2} +1+ \dfrac{1+2b}{b^2} +1+ \dfrac{1+2c}{c^2}\\
&=3+\dfrac{1}{a^2}+ \dfrac{1}{b^2}+ \dfrac{1}{c^2}+2(\dfrac{1}{a}+ \dfrac{1}{b} + \dfrac{1}{c})\\
&\ge 3\\
\end{array}
$
