**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.

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