The ninety-fourth problem: Let $\mathrm{n}$ be a given positive integer, and $\mathrm{p}(\mathrm{x})$ be a monic $\mathrm{n}$-degree polynomial with real coefficients. Find the minimum possible value of $\sum_{\mathrm{i}=1}^{\mathrm{n}+1} \frac{\mathrm{i}^{2}(\mathrm{p}(\mathrm{i} 2))^{2}}{(\mathrm{i}+1)^{2}(\mathrm{i}+2)^{2} \ldots(\mathrm{i}+\mathrm{n}+1)^{2}}$.
The 94th question:
Solution: According to the Lagrange interpolation formula, we have:
$$
p(x)=\sum_{i=1}^{n+1} \prod_{1 \leq j \leq n+1}^{j \neq i} \frac{x-j^{2}}{i^{2}-j^{2}} \cdot p\left(i^{2}\right)
$$

By comparing the leading coefficient of the above formula, we get:
$$
\begin{array}{l}
1=\sum_{i=1}^{n+1} \prod_{1 \leq j \leq n+1}^{j \neq i} \frac{p\left(i^{2}\right)}{i^{2}-j^{2}} \\
=\sum_{i=1}^{n+1} \frac{(-1)^{n+1-i} p\left(i^{2}\right)}{(i-1)!\cdot(i+1)(i+2) \ldots(2 i-1) \cdot(n+1-i)!\cdot(2 i+1)(2 i+2) \ldots(n+1+i)} \\
=\frac{1}{n!} \sum_{i=1}^{n+1} \frac{2 i \cdot C_{n}^{i-1}(-1)^{n+1-i-i} p\left(i^{2}\right)}{(i+1)(i+2) \ldots(n+1+i)}
\end{array}
$$

Rearranging, we get:
$$
\sum_{i=1}^{n+1} \frac{i \cdot C_{n}^{i-1}(-1)^{n+1-i} p\left(i^{2}\right)}{(i+1)(i+2) \ldots(n+1+i)}=\frac{n!}{2}
$$

According to the Cauchy-Schwarz inequality and (*), we have:
$$
\begin{array}{l}
\left(\sum_{i=1}^{n+1} \frac{i^{2}\left(p\left(i^{2}\right)\right)^{2}}{(i+1)^{2}(i+2)^{2} \ldots(i+n+1)^{2}}\right) \cdot\left(\sum_{i=1}^{n+1}\left(C_{n}^{i-1}(-1)^{n+1-i}\right)^{2}\right) \geq\left(\sum_{i=1}^{n+1} \frac{i \cdot C_{n}^{i-1}(-1)^{n+1-1} p\left(i^{2}\right)}{(i+1)(i+2) \ldots(n+1+i)}\right)^{2} \\
\Rightarrow\left(\sum_{i=1}^{n+1} \frac{i^{2}\left(p\left(i^{2}\right)\right)^{2}}{(i+1)^{2}(i+2)^{2} \ldots(i+n+1)^{2}}\right) \cdot\left(\sum_{i=1}^{n+1}\left(C_{n}^{i-1}\right)^{2}\right) \geq\left(\frac{n!}{2}\right)^{2}
\end{array}
$$

Since $\sum_{\mathrm{i}=1}^{\mathrm{n}+1}\left(C_{n}^{\mathrm{i}-1}\right)^{2}=\sum_{i=0}^{n}\left(C_{n}^{i}\right)^{2}=C_{2 n}^{n}$, substituting into (2) gives:
$$
\sum_{i=1}^{n+1} \frac{i^{2}\left(p\left(i^{2}\right)\right)^{2}}{(i+1)^{2}(i+2)^{2} \ldots(i+n+1)^{2}} \geq \frac{(n!)^{4}}{4 \cdot(2 n)!}
$$

According to the equality condition of the Cauchy-Schwarz inequality, the above inequality can achieve equality.