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Home » Use Pythagorean theorem: \( a^2 + b^2 = c^2 \) - Belip

Use Pythagorean theorem: \( a^2 + b^2 = c^2 \) - Belip

📅 May 2, 2026 👤 scraface

May 02, 2026

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📌 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

📌 oxed{rac{5}{2}}.

📌 Question: A hydrologist models the flow resistance $ R $ through a porous aquifer as

📌 R = rac{p + q}{p - q} + rac{p - q}{p + q} - 2,

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