with equality only if $ racp+qp-q = racp-qp+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 = racpq > 1 $, so - Abu Waleed Tea