Problem: Let $a$, $b$, $c$ be positive real numbers with $abc = 1$.
Show that
\[
  \left( a - 1 + \frac 1b \right)
  \left( b - 1 + \frac 1c \right)
  \left( c - 1 + \frac 1a \right)
  \le 1.
\]

Solution: Let $a = x/y$, $b = y/z$, $c = z/x$ for $x,y,z > 0$.
Then the inequality rewrites as
\[ (-x+y+z)(x-y+z)(x+y-z) \le xyz \]
which when expanded is equivalent to Schur's inequality.
Alternatively, if one wants to avoid appealing to Schur,
then the following argument works:
\begin{itemize}
  \ii At most one term on the left-hand side is negative;
  if that occurs we are done from $xyz > 0 > (-x+y+z)(x-y+z)(x+y-z)$.
  \ii If all terms in the left-hand side are nonnegative,
  let us denote $m = -x+y+z \ge 0$, $n = x-y+z \ge 0$, $p = x+y-z \ge 0$.
  Then it becomes
  \[ mnp \le \frac{(m+n)(n+p)(p+m)}{8} \]
  which follows by AM-GM.
\end{itemize}