Determine the largest constant $C$ such that for all real numbers $x_{1}, x_{2}, \ldots, x_{6}$ the inequality

$$
\left(x_{1}+x_{2}+\cdots+x_{6}\right)^{2} \geq C \cdot\left(x_{1}\left(x_{2}+x_{3}\right)+x_{2}\left(x_{3}+x_{4}\right)+\cdots+x_{6}\left(x_{1}+x_{2}\right)\right)
$$

holds.

Determine for this $C$ all $x_{1}, x_{2}, \ldots, x_{6}$ for which equality holds.

(Walther Janous)
With $x_{1}=x_{2}=\cdots=x_{6}=1$, we get $36 \geq 12 C$, hence $C \leq 3$.

We now prove that $C_{\max }=3$.

Thus, the inequality in question reads

$$
\left(x_{1}+x_{2}+\cdots+x_{6}\right)^{2} \geq 3 \cdot\left(x_{1}\left(x_{2}+x_{3}\right)+x_{2}\left(x_{3}+x_{4}\right)+\cdots+x_{6}\left(x_{1}+x_{2}\right)\right),
$$

i.e., (after some transformations)

$$
\begin{aligned}
x_{1}^{2}+\cdots+x_{6}^{2}+2 x_{1} x_{4}+2 x_{2} x_{5}+2 x_{3} x_{6} \geq & x_{1} x_{2}+x_{1} x_{3}+x_{2} x_{3}+x_{2} x_{4} \\
& +x_{3} x_{4}+x_{3} x_{5}+x_{4} x_{5}+x_{4} x_{6}+x_{5} x_{6}+x_{1} x_{5}+x_{1} x_{6}+x_{2} x_{6}
\end{aligned}
$$

However, this inequality can be expressed in the form

$$
\begin{aligned}
\left(x_{1}+x_{4}\right)^{2}+\left(x_{2}+x_{5}\right)^{2}+\left(x_{3}+x_{6}\right)^{2} & \\
& \geq\left(x_{1}+x_{4}\right)\left(x_{2}+x_{5}\right)+\left(x_{2}+x_{5}\right)\left(x_{3}+x_{6}\right)+\left(x_{3}+x_{6}\right)\left(x_{1}+x_{4}\right)
\end{aligned}
$$

Therefore, its validity follows from the general statement

$$
X^{2}+Y^{2}+Z^{2} \geq X Y+Y Z+Z X
$$

with equality holding exactly for $X=Y=Z$. (Here, $X, Y$, and $Z$ are arbitrary real numbers.)

In the inequality in question, equality holds exactly for all 6-tuples $\left(x_{1}, \ldots, x_{6}\right)$ of real numbers with $x_{1}+x_{4}=x_{2}+x_{5}=x_{3}+x_{6}$.

(Walther Janous)