## Friday, February 10, 2006

### Not Everyone Can Win. Topic: Algebra/Inequalities. Level: AIME/Olympiad.

Problem: (360 Problems for Mathematical Contests - 1.1.23) Let $a,b,c$ be real numbers such that $abc = 1$. Prove that at most two of the numbers

$2a-\frac{1}{b}$, $2b-\frac{1}{c}$, $2c-\frac{1}{a}$

are greater than $1$.

Solution: We proceed by contradiction, that is, assuming the opposite of what we wish to prove and showing that it is impossible. So assume all three are greater than $1$.

Then their product is also greater than $1$, giving

$\left(2a-\frac{1}{b}\right)\left(2b-\frac{1}{c}\right)\left(2c-\frac{1}{a}\right) > 1$

Expanding and substituting $abc = 1$, we have

$2(a+b+c)-\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right) < 3$.

But we also have the sum of the three greater than $3$, which means

$2(a+b+c)-\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right) > 3$,

giving a contraction (since it cannot be both greater than and less than $3$). Hence our assumption must be false and at most two of the numbers can be greater than $1$. QED.

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Comment: Contradiction is a powerful technique for solving problems - oftentimes the words "at most" or "at least" lead to a possible contradiction solution.

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Practice Problem: (Olympiad Problem Solving MidTerm) Prove there are infinitely many primes of the form $6n+5$ by contradiction.