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?

Since,$xyz=1$,we have any two of x,y,z is negative or all must be positive and in both case all three are non zero.
CASE 1:
If all x,y,z are positive I hope that  you may easily prove it.
CASE 2:
 if x,y is negative then we have $$x>x-1$$
But $$ x^2≤(x-1)^2.$$since if we assume $x=\frac{1}{2} $ then$x^2=(x-1)^2$
Therefore $$ 0<\frac{x^2}{(x-1)^2}<1$$
Similarly it follows for y,
But for z we have $$\frac{z^2}{(z-1)^2}≤1$$if $z≤\frac{1}{2}$,else z>1
therfore totally we have $$ \frac{x^2}{(x-1)^2}+
\frac{y^2}{(x-1)^2}
+
\frac{z^2}{(z-1)^2}
≥1 $$
hence proved.
