International Mathematical Olympiad

(1) Let $$\frac{x}{x-1}=a,\frac{y}{y-1}=b,\frac{z}{z-1}=c,$$ then $$x=\frac{a}{a-1},y=\frac{b}{b-1},z=\frac{c}{c-1}.$$ Since $xyz=1$, we have $$abc=(a-1)(b-1)(c-1),$$ that is $$a+b+c-1=ab+bc+ca.$$ Therefore $$\begin{array}{rl}{a}^2+b^2+c^2& =(a+b+c{)}^2-2(ab+bc+ca)\\ & =(a+b+c{)}^2-2(a+b+c-1)\\ & =(a+b+c-1{)}^2+1\\ & \ge 1.\end{array}$$ So $$\frac{x^2}{(x-1{)}^2}+\frac{y^2}{(y-1{)}^2}+\frac{z^2}{(z-1{)}^2}\ge 1.$$

(2) Take $(x,y,z)=\left(-\frac{k}{(k-1{)}^2},k-k^2,\frac{k-1}{k^2}\right)$, $k$ is an integer, then $(x,y,z)$ is a triple of rational numbers, with $x$, $y$, $z$ each different from $1$. What is more, a different integer $k$ gives a different triple of rational numbers. $$\begin{array}{rl}& \frac{x^2}{(x-1{)}^2}+\frac{y^2}{(y-1{)}^2}+\frac{z^2}{(z-1{)}^2}\\ & =\frac{k^2}{(k^2-k+1{)}^2}+\frac{(k-k^2{)}^2}{(k^2-k+1{)}^2}+\frac{(k-1{)}^2}{(k^2-k+1{)}^2}\\ & =\frac{k^4-2k^3+3k^2-2k+1}{(k^2-k+1{)}^2}=1.\end{array}$$ So the problem is proved.

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Alternative solution.

(1) By $xyz=1$, let $p=x$, $q=1$, $r=\frac{1}{y}$, then $x=\frac{p}{q}$, $y=\frac{q}{r}$, $z=\frac{1}{xy}=\frac{r}{p}$, where $p$, $q$, $r$ are different from each other. We have $$\begin{array}{rl}\frac{x^2}{(x-1{)}^2}+\frac{y^2}{(y-1{)}^2}+\frac{z^2}{(z-1{)}^2}& \ge 1\\ \iff \frac{p^2}{(p-q{)}^2}+\frac{q^2}{(q-r{)}^2}+\frac{r^2}{(r-p{)}^2}& \ge 1.\end{array}①$$ Let $$a=\frac{p}{p-q},b=\frac{q}{q-r},c=\frac{r}{r-p},$$ after the substitution, ① reduces to $\sum a^2\ge 1$. Since $$\frac{-1+a}{a}=\frac{q}{p},\frac{-1+b}{b}=\frac{r}{q},\frac{-1+c}{c}=\frac{p}{r},$$ therefore $$\begin{array}{rl}& \frac{-1+a}{a}\cdot \frac{-1+b}{b}\cdot \frac{-1+c}{c}=1,\\ & 1-\sum a+\sum ab=0.\end{array}②$$ By ②, we get $$1-\sum a^2=-(a+b+c-1{)}^2\le 0,$$ so $\sum a^2\ge 1$. Hence ① holds.

(2) Let $b=\frac{t^2+t}{t^2+t+1}$, $c=\frac{t+1}{t^2+t+1}$, $a=-\frac{bc}{b+c}$, here $t$ can be any rational number except $0$ and $-1$. While $t$ varies, only finitely many of $t$ values can make $b$, $c$, $a$ be $1$. That is, there are infinitely many triples of rational numbers $a$, $b$, $c$ each different from $1$, satisfying $\sum a=\sum a^2=1$. By $$(x,y,z)=\left(\frac{a}{a-1},\frac{b}{b-1},\frac{c}{c-1}\right),$$ (2) holds.