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    Serves a Daily Special and an All-You-Can-Eat Course in Problem Solving. Courtesy of me, Jeffrey Wang.
  
Mass Production. Topic: Inequalities. Level: Olympiad. February 28th, 2006

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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