How Does Proof by Contradiction Validate a/b + b/a >= 2?

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SUMMARY

The proof by contradiction for the inequality \( \frac{a}{b} + \frac{b}{a} \geq 2 \) is established by assuming the opposite, leading to the conclusion that \( (a-b)^2 < 0 \). This contradiction arises from the assumption that \( a \) and \( b \) are positive numbers, which ensures that \( (a-b)^2 \) cannot be negative. The key steps involve manipulating the inequality to show that if \( \frac{a^2 + b^2}{ab} < 2 \), it results in an impossible scenario.

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  • Understanding of basic algebraic manipulation
  • Familiarity with inequalities and their properties
  • Knowledge of proof techniques, specifically proof by contradiction
  • Concept of positive numbers and their implications in inequalities
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  • Study the principles of proof by contradiction in mathematical proofs
  • Explore the Cauchy-Schwarz inequality and its applications
  • Learn about the properties of positive numbers in algebra
  • Investigate other mathematical inequalities and their proofs
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[SOLVED] basic math proof by contradiction

Homework Statement



prove: If a and b are positive numbers, a/b +b/a>=2

Homework Equations





The Attempt at a Solution



by contradiction (a^2+b^2)/ab<2 and got lost
 
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If (a^2+b^2)/ab<2 then a^2+b^2<2ab. So a^2+b^2-2ab<0. But a^2+b^2-2ab=(a-b)^2<0. What could be wrong with that?
 
thanks it was that easy
 

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