## Monday, February 27, 2006

### Mass Production. Topic: Inequalities. Level: Olympiad.

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.

--------------------

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.

--------------------

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