Therefore, the expression becomes:
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- \frac{1}{a^2} + \frac{1}{b^2} \geq \frac{2}{ab}
- Given \(a^2 + b^2 = 1\), the maximum of \(ab\) is \(\frac{1}{2}\) when \(a = b = \frac{\sqrt{2}}{2}\). Then:
- ab = \frac{1}{2} \quad \Rightarrow \quad \frac{1}{a^2} + \frac{1}{b^2} = \frac{(a^2 + b^2)^2}{a^2 b^2} = \frac{1}{(\frac{1}{2})^2} = 4
- + 4 = 9
- Verification at \(x = \frac{\pi}{4}\) shows \(\cos x = \sin x = \frac{\sqrt{2}}{2}\), yielding:
- Thus, the minimum value is \(\boxed{9}\).
