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📅 May 4, 2026 👤 scraface

May 04, 2026

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📌 with equality only if $ rac{p+q}{p-q} = rac{p-q}{p+q} $, which implies $ (p+q)^2 = (p-q)^2 \Rightarrow 4pq = 0 $, impossible. So $ R > 0 $, but infimum is 0. But the problem asks for minimum. However, since $ p > q > 0 $, $ R > 0 $, but can be arbitrarily small. Thus, no minimum? But wait â perhaps we misread. Letâs compute minimum numerically. Let $ x = rac{p}{q} > 1 $, so

📌 R = rac{x + 1}{x - 1} + rac{x - 1}{x + 1} - 2.

📌 Let $ f(x) = rac{x+1}{x-1} + rac{x-1}{x+1} - 2 $. Combine:

📌 f(x) = rac{(x+1)^2 + (x-1)^2}{(x-1)(x+1)} - 2 = rac{2x^2 + 2}{x^2 - 1} - 2 = 2\left( rac{x^2 + 1}{x^2 - 1} - 1

📌 ight) = 2\left( rac{2}{x^2 - 1}

📌 ight) = rac{4}{x^2 - 1}.

📌 Since $ x > 1 $, $ x^2 - 1 > 0 $, so $ f(x) > 0 $, and as $ x o 1^+ $, $ f(x) o \infty $, as $ x o \infty $, $ f(x) o 0^+ $. So $ R $ has no minimum, only an infimum of 0. But the problem likely intends for us to find the minimum under constraints â but none restrict. However, reconsider the original expression:

📌 rac{a}{b} + rac{b}{a} - 2 \geq 0,

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