Problem 70. Find the maximum value of $k=k(n)$ for which the following inequality is true for all real numbers $x_{1}, x_{2}, \ldots, x_{n}$
$$x_{1}^{2}+\left(x_{1}+x_{2}\right)^{2}+\ldots+\left(x_{1}+x_{2}+\ldots+x_{n}\right)^{2} \geq k\left(x_{1}+x_{2}+\ldots+x_{n}\right)^{2}$$
SOLUTION. Let $a_{1}, a_{2}, \ldots, a_{n}$ be positive real numbers, then
$$\begin{array}{c}
a_{1} y_{1}^{2}+\frac{1}{a_{1}} \cdot y_{2}^{2}+2 y_{1} y_{2} \geq 0 \\
a_{2} y_{2}^{2}+\frac{1}{a_{2}} \cdot y_{3}^{2}+2 y_{2} y_{3} \geq 0 \\
\cdots \\
\cdots \\
a_{n-1} y_{n-1}^{2}+\frac{1}{a_{n}} \cdot y_{n}^{2}+2 y_{n-1} y_{n} \geq 0
\end{array}$$

Adding up these results, we obtain
$$a_{1} y_{1}^{2}+\left(\frac{1}{a_{1}}+a_{2}\right) y_{2}^{2}+\ldots+\left(\frac{1}{a_{n-2}}+a_{n-1}\right) y_{n-1}^{2}+\frac{1}{a_{n-1}} \cdot y_{n}^{2}+2 \sum_{i=1}^{n-1} y_{i} y_{i+1} \geq 0$$

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

With some calculations, we find out $a_{k}=\frac{\sin (k+1) \alpha}{\sin (k \alpha)}$ where $\alpha=\frac{2 \pi}{2 n+1}$. Inserting these in $(\star)$, we conclude
$$\begin{aligned}
& 2 \cos \alpha\left(\sum_{k=1}^{n} y_{k}^{2}\right)+2\left(\sum_{k=1}^{n-1} y_{k} y_{k+1}\right)+y_{n}^{2} \geq 0 \\
\Leftrightarrow & 2(1+\cos \alpha)\left(\sum_{k=1}^{n} y_{k}^{2}\right) \geq y_{1}^{2}+\sum_{k=1}^{n-1}\left(y_{k}-y_{k+1}\right)^{2}
\end{aligned}$$

Setting $y_{k}=x_{1}+x_{2}+\ldots+x_{k} \forall k \in\{1,2, \ldots, n\}$, we obtain
$$4 \cos ^{2} \frac{\alpha}{2}\left(x_{1}^{2}+\left(x_{1}+x_{2}\right)^{2}+\ldots+\left(x_{1}+x_{2}+\ldots+x_{n}\right)^{2}\right) \geq x_{1}^{2}+x_{2}^{2}+\ldots+x_{n}^{2}$$

The best (greatest) value of $k$ is $\frac{1}{4 \cos ^{2} \frac{\pi}{2 n+1}}$ with equality for
$$x_{k}=(-1)^{k}\left(\sin \frac{2 k \pi}{2 n+1}+\sin \frac{2(k-1) \pi}{2 n+1}\right)$$