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:

February 22, 2026

$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:$

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}.
rac{a}{b} + rac{b}{a} - 2 \geq 0,
with equality only if $ a = b $, i.e., $ rac{p+q}{p-q} = rac{p-q}{p+q} $, which implies $ (p+q)^2 = (p-q)^2 \Rightarrow pq = 0 $, impossible. So $ R > 0 $, but minimum not attained. But perhaps the hydrologist defines $ R $ only when flow is stable â yet no such condition
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