30. Strengthen the proposition proof $a_{n}^{2}>2\left(\frac{a_{2}}{2}+\frac{a_{3}}{3}+\cdots+\frac{a_{n}}{n}\right)+\frac{1}{n}$.
(1) When $n=2$, $a_{n}^{2}=a_{2}^{2}=\frac{9}{4}$, and $2\left(\frac{a_{2}}{2}\right)^{2}+\frac{1}{2}=2$, the inequality holds.
(2) Assume that when $n=k$, the proposition holds, i.e., $a_{k}^{2}>2\left(\frac{a_{2}}{2}+\frac{a_{3}}{3}+\cdots+\frac{a_{k}}{k}\right)$. Then, when $n=k+1$, we have
$$\begin{aligned} a_{k+1}^{2}= & \left(a_{k}+\frac{1}{k+1}\right)^{2}=a_{k}^{2}+\frac{2 a_{k}}{k+1}+\frac{1}{(k+1)^{2}}> \\ & 2\left(\frac{a_{2}}{2}+\frac{a_{3}}{3}+\cdots+\frac{a_{k}}{k}\right)+\frac{1}{k}+\frac{2}{k+1}\left(a_{k}+\frac{1}{k+1}\right)-\frac{1}{(k+1)^{2}}= \\ & 2\left(\frac{a_{2}}{2}+\frac{a_{3}}{3}+\cdots+\frac{a_{k}}{k}+\frac{a_{k+1}}{k+1}\right)+\frac{(k+1)^{2}+k}{k(k+1)^{2}}> \\ & 2\left(\frac{a_{2}}{2}+\frac{a_{3}}{3}+\cdots+\frac{a_{k}}{k}+\frac{a_{k+1}}{k+1}\right)+\frac{k^{2}+k}{k(k+1)^{2}}= \\ & 2\left(\frac{a_{2}}{2}+\frac{a_{3}}{3}+\cdots+\frac{a_{k}}{k}+\frac{a_{k+1}}{k+1}\right)+\frac{1}{k+1} \end{aligned}$$

Thus, when $n=k+1$, the proposition holds. In summary, $a_{n}^{2}>2\left(\frac{a_{2}}{2}+\frac{a_{3}}{3}+\cdots+\frac{a_{n}}{n}\right)$ holds.