Problem: Let $x$, $y$, $z$ be real numbers with $xyz = 1$, all different from $1$.
Prove that
\[ \frac{x^2}{(x-1)^2} + \frac{y^2}{(y-1)^2} + \frac{z^2}{(z-1)^2} \ge 1 \]
and show that equality holds for infinitely many choices
of rational numbers $x$, $y$, $z$.

Solution: Let $x=a/b$, $y=b/c$, $z=c/a$, so we want to show
\[ \left(\frac{a}{a-b}\right)^2+\left(\frac{b}{b-c}\right)^2
  +\left(\frac{c}{c-a}\right)^2\ge 1.\]
A boring computation shows this is equivalent to
\[ \frac{(a^2b+b^2c+c^2a-3abc)^2}{(a-b)^2(b-c)^2(c-a)^2} \ge 0 \]
which proves the inequality
(and it is unsurprising we are in such a situation,
given that there is an infinite curve of rationals).

For equality, it suffices to show there are infinitely many integer solutions to
\[ a^2b+b^2c+c^2a=3abc \iff \frac ac + \frac ba + \frac ca = 3 \]
or equivalently that there are infinitely many rational solutions to
\[ u + v + \frac{1}{uv} = 3. \]
For any $0 \neq u \in \QQ$ the real solution for $u$ is
\[ v = \frac{-u + (u-1)\sqrt{1-4/u} + 3}{2} \]
and there are certainly infinitely many rational numbers $u$
for which $1-4/u$ is a rational square
(say, $u = \frac{-4}{q^2-1}$ for $q \neq \pm 1$ a rational number).