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rac12 \left[ \left( rac11 - rac13 - DNSFLEX

📅 March 11, 2026 👤 scraface

Mar 11, 2026

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📌 f(x) = \left(\frac{u}{\frac{u^2 - 1}{2}}\right)^2 = \left(\frac{2u}{u^2 - 1}\right)^2 = \frac{4u^2}{(u^2 - 1)^2}

📌 Now, since \(0 < x < \frac{\pi}{2}\), \(u = \sin x + \cos x = \sqrt{2} \sin\left(x + \frac{\pi}{4}\right)\), so \(u \in (1, \sqrt{2}]\) (maximum \(\sqrt{2}\) at \(x = \frac{\pi}{4}\)).

📌 Let \(v = u^2\), so \(v \in (1, 2]\). Then:

📌 f(x) = \frac{4v}{(v - 1)^2}

📌 Let \(g(v) = \frac{4v}{(v - 1)^2}\). Find maximum on \(v \in (1, 2]\).

📌 Take derivative:

📌 g'(v) = \frac{4(v - 1)^2 - 4v \cdot 2(v - 1)}{(v - 1)^4} = \frac{4(v - 1)[(v - 1) - 2v]}{(v - 1)^4} = \frac{4(-v - 1)}{(v - 1)^3}

📌 For \(v > 1\), \(g'(v) < 0\), so \(g(v)\) is decreasing on \((1, 2]\). Hence, maximum at \(v \to 1^+\)? But let’s check value at endpoint.

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