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Home » Enter the Realm of Lord Drakkon — King of Darkness and Betrayal! - Get link 4share

Enter the Realm of Lord Drakkon — King of Darkness and Betrayal! - Get link 4share

📅 March 10, 2026 👤 scraface

Mar 10, 2026

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📌 R = rac{(p+q)^2 + (p-q)^2}{(p-q)(p+q)} - 2 = rac{p^2 + 2pq + q^2 + p^2 - 2pq + q^2}{p^2 - q^2} - 2 = rac{2p^2 + 2q^2}{p^2 - q^2} - 2.

📌 R = 2\left( rac{p^2 + q^2}{p^2 - q^2} - 1

📌 ight) = 2\left( rac{2q^2}{p^2 - q^2}

📌 ight) = rac{4q^2}{p^2 - q^2}.

📌 As $ p o q^+ $, $ R o \infty $. As $ p \gg q $, $ R o 0 $. So $ R > 0 $, but has no minimum? Wait â reconsider: the expression

📌 rac{p+q}{p-q} + rac{p-q}{p+q} \geq 2 \Rightarrow R = \left( rac{p+q}{p-q} + rac{p-q}{p+q}

📌 ight) - 2 \geq 2 - 2 = 0,

📌 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

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