+ 4 + \frac{1}{a^2} + \frac{1}{b^2} = 5 + \frac{1}{a^2} + \frac{1}{b^2}
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- This expands to:
- a^2 + 2 + \frac{1}{a^2} + b^2 + 2 + \frac{1}{b^2}
- Since \(a^2 + b^2 = \cos^2 x + \sin^2 x = 1\), the expression simplifies to:
- We apply the AM-GM inequality to \(\frac{1}{a^2} + \frac{1}{b^2}\):
- \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:
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