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We expand the expression:
g(x) = \left(\sqrt{2} \sin x + \frac{1}{\sqrt{2} \sin x}\right)^2 = 2 \sin^2 x + 2 + \frac{1}{2 \sin^2 x}
Let $ y = \sin^2 x $, where $ y > 0 $. Then:
g(x) = 2y + 2 + \frac{1}{2y}
We minimize $ 2y + \frac{1}{2y} + 2 $. Let:
h(y) = 2y + \frac{1}{2y}
Using the AM-GM inequality:
2y + \frac{1}{2y} \geq 2\sqrt{2y \cdot \frac{1}{2y}} = 2
Equality holds when $ 2y = \frac{1}{2y} \Rightarrow 4y^2 = 1 \Rightarrow y = \frac{1}{2} $.
g(x) = h(y) + 2 = 2 + 2 = 4