<s>[INST] 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?
 [/INST] 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.
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