54. Let $x_{1}, x_{2}, \cdots, x_{5}$ be real numbers, and $\sum_{i=1}^{5} x_{i}=0$, let $x_{6}=x_{1}$, find the smallest real number $k$ such that the inequality $\sum_{i=1}^{5} x_{i} x_{i+1} \leqslant k\left(\sum_{i=1}^{5} a_{i}^{2}\right) .(2001$ Czech and Slovak Olympiad problem)
54. Let $x_{6}=x_{1}, x_{7}=x_{2}, 0=\left(\sum_{i=1}^{5} x_{i}\right)^{2}=\sum_{i=1}^{5} x_{i}^{2}+2 \sum_{i=1}^{5} x_{i} x_{i+1}+2 \sum_{i=1}^{5} x_{i} x_{i+2}$. For any real number $\lambda$,
$$\begin{aligned}
0 \leqslant & \sum_{i=2}^{6}\left(x_{i+1}+x_{i-1}-\lambda x_{i}\right)^{2}= \\
& \left(2+\lambda^{2}\right) \sum_{i=1}^{5} x_{i}^{2}-2 \sum_{i=1}^{5} x_{i} x_{i+2}-4 \lambda \sum_{i=1}^{5} x_{i} x_{i+1}= \\
& \left(1+\lambda^{2}\right) \sum_{i=1}^{5} x_{i}^{2}-(2+4 \lambda) \sum_{i=1}^{5} x_{i} x_{i+1}
\end{aligned}$$

Therefore, $\sum_{i=1}^{5} x_{i} x_{i+1} \leqslant \inf _{\lambda>-\frac{1}{2}} \frac{1+\lambda^{2}}{2+4 \lambda} \sum_{i=1}^{5} x_{i}^{2}=\frac{\sqrt{5}-1}{4} \sum_{i=1}^{5} x_{i}^{2}$. That is, the minimum value of $k$ is
$$\frac{\sqrt{5}-1}{4}$$