Proof:
Prove by Principle of Mathematical Induction.
The case \( n=1, \sqrt{2}>1 \). This case is valid.
We assume the validity of the inequality for \( n=k \), and shall prove it for \( n=k+1 \).
The assumption \( a_{k}>2-\frac{1}{k} \) imply that
\[
a_{k+1}^{2}=2+a_{k}>2+2-\frac{1}{k}=4-\frac{1}{k}=\left(2-\frac{1}{k+1}\right)^{2}+\frac{4}{k+1}-\frac{1}{(k+1)^{2}}-\frac{1}{k},
\]
i.e., \( \quad a_{k+1}^{2}>\left(2-\frac{1}{k+1}\right)^{2}+\frac{3 k^{2}+k-1}{k(k+1)^{2}} \)
