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?
 A: 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.
A: 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 
$${\frac { \left( b{a}^{2}-3\,bca+{c}^{2}a+{b}^{2}c \right) ^{2}}{
 \left( a-b \right) ^{2} \left( b-c \right) ^{2} \left( -c+a \right) ^
{2}}}
\geq 0$$
A: Substitute $a=\frac{x}{x-1}, b=\frac{y}{y-1}, c=\frac{z}{z-1}$.
Then we have $x=\frac{a}{a-1}$ and the similar identities so that the condition implies $abc=(a-1)(b-1)(c-1)$ and hence $ab+ac+bc=a+b+c-1$.
We want to prove $a^2+b^2+c^2 \ge 1$ which is equivalent to $(a+b+c)^2 - 2(ab+ac+bc)-1 \ge 0$ or, using the condition, $(a+b+c)^2 -2(a+b+c)+1 \ge 0$.
But the LHS is $(a+b+c-1)^2$ which is clearly non-negative. Hence the result.
