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where to watch freakier friday - IQnection

📅 May 4, 2026 👤 scraface

May 04, 2026

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📌 subject to $ x, y > 0 $.

📌 Solution: Let $ A = rac{x^2 + y^2}{xy} $ and $ B = rac{xy}{x^2 + y^2} $. Then the expression becomes

📌 A + B = rac{x^2 + y^2}{xy} + rac{xy}{x^2 + y^2}.

📌 Let $ t = rac{x^2 + y^2}{xy} $. By the AM-GM inequality,

📌 x^2 + y^2 \geq 2xy \Rightarrow rac{x^2 + y^2}{xy} \geq 2,

📌 so $ t \geq 2 $. The expression becomes $ t + rac{1}{t} $. Define $ f(t) = t + rac{1}{t} $ for $ t \geq 2 $. The derivative $ f'(t) = 1 - rac{1}{t^2} > 0 $ for $ t > 1 $, so $ f(t) $ is increasing. Therefore, the minimum occurs at $ t = 2 $:

📌 f(2) = 2 + rac{1}{2} = rac{5}{2}.

📌 Equality holds when $ x = y $, which satisfies $ x, y > 0 $. Thus, the minimum value is

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