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Understanding the Context

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f(x) = \frac{1}{\frac{1}{4} \sin^2 2x} + 1 = \frac{4}{\sin^2 2x} + 1
Since $ 0 < x < \frac{\pi}{2} $, then $ 0 < 2x < \pi $, and $ \sin 2x > 0 $. The minimum of $ f(x) $ occurs when $ \sin^2 2x $ is maximized, i.e., when $ \sin^2 2x = 1 $. This happens when $ 2x = \frac{\pi}{2} \Rightarrow x = \frac{\pi}{4} $.
f\left(\frac{\pi}{4}\right) = \frac{4}{1} + 1 = 5

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