How is this property of real numbers proven? 
Prove that $\dfrac{x^2}{(x − 1)^2} + \dfrac{y^2}{(y − 1)^2} + \dfrac{z^2}{(z − 1)^2} ≥ 1$ for all real numbers $x, y, z$, each different from $1$ and satisfying $xyz = 1$. 

How do I prove this?

it should be $$\frac{(xy+yz+zx-3)^2}{(x-1)^2(y-1)^2(z-1)^2}\geq 0$$
Hint: set $$x=a/b,y=b/c,z=c/a$$ in the given term
after the substituion we obtain
$${\frac {{a}^{2}}{{b}^{2}} \left( {\frac {a}{b}}-1 \right) ^{-2}}+{
\frac {{b}^{2}}{{c}^{2}} \left( {\frac {b}{c}}-1 \right) ^{-2}}+{
\frac {{c}^{2}}{{a}^{2}} \left( {\frac {c}{a}}-1 \right) ^{-2}}-1
$$
factoring all we get 
$${\frac { \left( b{a}^{2}-3\,bca+{c}^{2}a+{b}^{2}c \right) ^{2}}{
 \left( a-b \right) ^{2} \left( b-c \right) ^{2} \left( -c+a \right) ^
{2}}}
\geq 0$$
