**Problem**: (2000 IMO - #2) Let $a, b, c$ be positive real numbers so that $abc=1$. Prove that

$\left( a-1+\frac 1b \right) \left( b-1+\frac 1c \right) \left( c-1+\frac 1a \right) \leq 1$.

**Solution**: The standard trick to solve the condition $ abc = 1 $ is to substitute

$ a = \frac{x}{y}$, $ b = \frac{y}{z}$, $ c = \frac{z}{x} $

for positive reals $ x, y, z $ (their existence is guaranteed since it's a three-variable system with three equations).

Our inequality reduces to

$ \left(\frac{x+z-y}{y}\right)\left(\frac{y+x-z}{z}\right)\left(\frac{z+y-x}{x}\right) \le 1 $.

Multiplying through by $ xyz $, expanding, and rearranging everything to the RHS, we have

$ 0 \le x^3+y^3+z^3-x^2y-x^2z-y^2z-y^2x-z^2x-z^2y+3xyz $

or

$ 0 \le x(x-y)(x-z)+y(y-z)(y-x)+z(z-x)(z-y) $,

which is just Schur's Inequality, so the result is proved. QED.

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Comment: The condition $ abc = 1 $ almost always calls for the substitution above. It also works with $ abc \ge 1 $ in which case you simply have another variable multiplied by each of the terms.

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Practice Problem: Go take the 2006 Mock AIME 5 below.

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