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christmas channels on xm sirius - Imagemakers

📅 May 1, 2026 👤 scraface

May 01, 2026

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📌 R = a + rac{1}{a} - 2.

📌 Define $ f(a) = a + rac{1}{a} - 2 $ for $ a > 1 $. By AM-GM, $ a + rac{1}{a} \geq 2 $, with equality at $ a = 1 $. But $ a > 1 $, so $ f(a) > 0 $. However, as $ a o 1^+ $, $ f(a) o 0^+ $. The minimum value is not attained, but we seek the infimum. Since $ f(a) $ increases for $ a > 1 $, the expression approaches 0 as $ p o q^+ $, but never reaches it. However, we analyze the original expression algebraically:

📌 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}

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