Question 35: Try to find the largest constant $c$, such that for any positive integer $n$ and any $n$ positive real numbers $a_{i}(1 \leq i \leq n)$, we have $\left(\sum_{i=1}^{n} a_{i}^{2}\right)\left(\sum_{1 \leq i, j \leq n} \max \{i, j\} a_{i} a_{j}\right) \geq c \cdot\left(\sum_{i=1}^{n} a_{i}\right)^{4}$.


The 35th question:
Solution: Note that:
$$
\sum_{1 \leq i, j \leq n} \max \{i, j\} a_{i} a_{j}=\sum_{i=1}^{n} i \cdot a_{i}^{2}+2 \sum_{1 \leq i<j \leq n} j a_{i} a_{j}=n\left(\sum_{i=1}^{n} a_{i}\right)^{2}-\sum_{k=1}^{n-1}\left(\sum_{i=1}^{k} a_{i}\right)^{2}
$$

Let \( S_{k}=\sum_{i=1}^{k} a_{i} (1 \leq k \leq n) \), and by homogeneity, let \( S_{n}=1, S_{0}=0 \), then \( 0=S_{0}<S_{1}<S_{2}<\ldots \leq S_{n}=1 \). According to the Cauchy-Schwarz inequality, we have:
$$
\begin{array}{l}
\left(\sum_{i=1}^{n} a_{i}^{2}\right)\left(\sum_{1 \leq i, j \leq n} \max \{i, j\} a_{i} a_{j}\right) \\
=\left(\sum_{i=1}^{n}\left(S_{i}-S_{i-1}\right)^{2}\right)\left(n-\sum_{k=1}^{n-1} S_{k}^{2}\right) \\
=\left(\sum_{i=0}^{n-1}\left(S_{i+1}-S_{i}\right)^{2}\right)\left(\sum_{i=0}^{n-1}\left(1-S_{i}^{2}\right)\right)
\end{array}
$$
$$
\begin{array}{l}
\geq\left(\sum_{i=0}^{n-1} \sqrt{1-S_{i}^{2}} \cdot\left(S_{i+1}-S_{i}\right)\right)^{2} \\
\geq\left(\int_{0}^{1} \sqrt{1-x^{2}} d x\right)^{2}=\frac{\pi^{2}}{16}=\frac{\pi^{2}}{16} \cdot\left(\sum_{i=1}^{n} a_{i}\right)^{4}
\end{array}
$$

On the other hand, according to the equality condition of the Cauchy-Schwarz inequality, we take \(\frac{\sqrt{1-s_{i}^{2}}}{s_{i+1}-S_{i}} (0 \leq i \leq n-1)\) as a positive constant (which also makes the slanted lines parallel in the figure), and when \( n \rightarrow+\infty \), then \(\frac{\left(\sum_{i=0}^{n-1}\left(S_{i+1}-s_{i}\right)^{2}\right)\left(\sum_{i=0}^{n-1}\left(1-s_{i}^{2}\right)\right)}{S_{n}^{4}} \rightarrow \frac{\pi^{2}}{16}\). In summary, the maximum value of \( c \) is \(\frac{\pi^{2}}{16}\).