70. Find the maximum value of the expression $k=k(n)$, such that for any real numbers $x_{1}, x_{2}, \cdots, x_{n}$, the following inequality always holds. $x_{1}^{2}+\left(x_{1}+x_{2}\right)^{2}+\cdots+\left(x_{1}+x_{2}+\cdots+x_{n}\right)^{2} \geq k\left(x_{1}^{2}+x_{2}^{2}+\cdots+x_{n}^{2}\right)$ (Le Hong Quy)
Proof: Let $a_{1}, a_{2}, \cdots, a_{n}$ be positive real numbers, then
$$\begin{array}{l}
a_{1} y_{1}^{2}+\frac{1}{a_{1}} \cdot y_{2}^{2}+2 y_{1} y_{2} \geq 0 \\
a_{2} y_{2}^{2}+\frac{1}{a_{2}} \cdot y_{3}^{2}+2 y_{2} y_{3} \geq 0 \\
\ldots \ldots \ldots \ldots \ldots \ldots \\
a_{n-1} y_{n-1}^{2}+\frac{1}{a_{n}} \cdot y_{n}^{2}+2 y_{n-1} y_{n} \geq 0
\end{array}$$

Adding these inequalities, we have
$$a_{1} y_{1}^{2}+\left(\frac{1}{a_{1}}+a_{2}\right) y_{2}^{2}+\cdots+\left(\frac{1}{a_{n-2}}+a_{n-1}\right) y_{n-1}^{2}+\frac{1}{a_{n-1}} \cdot y_{n}^{2}+2 \sum_{i=1}^{n-1} y_{i} y_{i+1} \geq 0 \quad(*)$$

We will choose $n$ numbers $a_{1}, a_{2}, \cdots, a_{n}$, such that
$$a_{1}=\frac{1}{a_{1}}+a_{2}=\cdots=\frac{1}{a_{n-1}}-1$$

After calculation, we find $a_{k}=\frac{\sin (k+1) \alpha}{\sin (k \alpha)}$, where $\alpha=\frac{2 \pi}{2 n+1}$. Substituting them into (*), we have
$$2 \cos \alpha\left(\sum_{k=1}^{n} y_{k}^{2}\right)+2\left(\sum_{k=1}^{n-1} y_{k} y_{k+1}\right)+y_{n}^{2} \geq 0 \Leftrightarrow 2(1+\cos \alpha)\left(\sum_{k=1}^{n} y_{k}^{2}\right) \geq y_{1}^{2}+\sum_{k=1}^{n-1}\left(y_{k}-y_{k+1}\right)^{2}$$

Let $y_{k}=x_{1}+x_{2}+\cdots+x_{k} \quad k \in\{1,2, \cdots, n\}$, we have
$$4 \cos ^{2} \frac{\alpha}{2}\left[x_{1}^{2}+\left(x_{1}+x_{2}\right)^{2}+\cdots+\left(x_{1}+x_{2}+\cdots+x_{n}\right)^{2}\right] \geq x_{1}^{2}+x_{2}^{2}+\ldots+x_{n}^{2}$$
The best (maximum) value of $k$ is $\frac{1}{4 \cos ^{2} \frac{\pi}{2 n+1}}$.
Equality holds when $x_{k}=(-1)^{k}\left[\sin \frac{2 k \pi}{2 n+1}+\sin \frac{2(k-1) \pi}{2 n+1}\right]$
Note: In Example 6.2.4, we proved (using the Cauchy-Schwarz inequality)
$$x_{1}^{2}+\left(x_{1}+x_{2}\right)^{2}+\cdots+\left(x_{1}+x_{2}+\cdots+x_{n}\right)^{2} \leq \frac{1}{4 \sin ^{2} \frac{\pi}{2(2 n+1)}}\left(x_{1}^{2}+x_{2}^{2}+\ldots+x_{n}^{2}\right)$$

What a strange and interesting coincidence, the same problem, one seeks the maximum, the other the minimum, one based on the AM-GM inequality, and the other on the Cauchy-Schwarz inequality, but the results are similar in appearance, both involving $\frac{2 \pi}{2 n+1}$.