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📅 May 1, 2026 👤 scraface

May 01, 2026

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📌 since $ rac{x}{z} = rac{x}{y} \cdot rac{y}{z} = ac = rac{1}{b} $, and similarly $ rac{z}{x} = rac{1}{c} $, but $ rac{x}{z} + rac{z}{x} = b + rac{1}{b} $.

📌 a + b + c - \log\left(b + rac{1}{b}

📌 Since $ abc = 1 $, we can assume $ a = rac{x}{y} $, $ b = rac{y}{z} $, $ c = rac{z}{x} $, and by AM-GM,

📌 a + b + c \geq 3\sqrt[3]{abc} = 3.

📌 Equality occurs when $ a = b = c = 1 $, i.e., when $ x = y = z $.

📌 Now consider the second term:

📌 \log\left(b + rac{1}{b}

📌 ight) \geq \log(2),

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