Since \(x_1 > \sqrt{a}\), we will use induction to show that \(x_n > \sqrt{a}\) for all \(n\).
- **Base case**: For \(n=1\), we have \(x_1 > \sqrt{a}\) by assumption.
- **Inductive step**: Assume \(x_n > \sqrt{a}\) for some \(n\). We need to show that
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