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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 — perhaps we misread. Let’s 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: 📰 Youll Be Shocked By The Plasma Donation Criteria No One Tells You 3235728 📰 Is This Microsoft Software Engineers Secret Growth Hack Making Them Unstoppable 2818347 📰 Shocking Windows Hack Rename Files In Seconds With These Keyboard Shortcuts 9184575 📰 Duffel Bag Duffle 8614858 📰 Ragdool Archers Unleashed How These Unstoppable Sharpshooters Conquer Any Battle 398710 📰 Fsa Vs Hsa Account 7272862 📰 Business Mobile Phone Contracts 8863976 📰 Explodes The World Of Spider Man Gamesheres The Absolute Best Gamers Haven 388046 📰 Hundreds Of Free Games Availablefree Free Free Now 5671883 📰 Ucla Vs Maryland 4444696 📰 A Structural Engineer Designs A Tower With 42 Floors Each Floor Requiring 18 Steel Beams If She Uses A Backup Beam For Every 6 Regular Beams How Many Total Beams Regular Backups Are Needed 3581883 📰 Health Magazine 2350242 📰 Why 10 Million Mac Users Swear By Onedriveclick To See The Shocking Truth 7602015 📰 Cleaveland 4114462 📰 Top Rated Cable And Internet Providers 8008482