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📰 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 — perhaps we misread. Let’s compute minimum numerically. Let $ x = rac{p}{q} > 1 $, so 📰 Why Every Player Demands Super Mario Landsee What Makes It Unforgettable 253963 📰 The Shocking Truth About Pai Mei You Wont Believe What This Online Sens Has Done Next 5101621 📰 Bank Of America New Paltz 4315775 📰 Bank Of America Login In Page 5437898 📰 You Wont Believe Whos Raving About Almond Flour Crackers Zero Guilt Maximum Flavor 3441911 📰 Album Cover Generator 2000402 📰 Unsearchable Wasteline Omaha Whats Disappearings Silent Crisis Exposed 421854 📰 Finally Revealed The Easy Hack Everyone Wants You To Know About Making Money Fast 4931043 📰 How Old Is Lara Trump 6128432 📰 Chili Cookoff 571558 📰 Border Fire 1904502 📰 Women Famous Actors 3303030 📰 Stop Strugglingtask Bars Are The Secret Tool Every Task Manager Needs 8526523 📰 Google Disk Download 8650907 📰 Torch Your Path To Success In Animal Crossing Switch 2Heres How 9599867