2000 IMO #2

Let $a, b, c$ be positive real numbers so that $abc = 1$. Prove that $\left( a - 1 + \frac 1b \right) \left( b - 1 + \frac 1c \right) \left( c - 1 + \frac 1a \right) \leq 1.$

Because $abc = 1$, we know that there exist some positive real numbers $x,y,z$ such that $$a = \frac{x}{y},b=\frac{y}{z},c=\frac{z}{x}.$$ Then it suffices to prove $\left( \frac{x-y+z}y \right) \left( \frac{y-z+x}z \right) \left( \frac{z-x+y}x \right) \leq 1.$ Substitute $p= b + c - a$, $q = c + a - b$ and $r = a + b - c$. Then, it suffices to prove $8pqr \le (p+q)(q+r)(r+p).$ At this point, there are two cases. Either one term on the RHS is negative, or none of them are.

If one is negative, then we have $xyz>0>pqr$, so then the inequality is clearly true.

If all are nonnegative, then we can say $$pqr \le \frac{(p+q)(q+r)(r+p)}{8},$$ which holds true because of the AM-GM inequality.