Konigsberg wrote: Let $x_1$, $x_2$, ..., $x_5$ be nonnegative real numbers such that $x_1+x_2+x_3+x_4+x_5=5$. Determine the maximum value of $x_1x_2+x_2x_3+x_3x_4+x_4x_5$. If $x_3\leq x_4$ we see that $x_1x_2+x_2x_3+x_3x_4+x_4x_5\leq(x_1+x_4)(x_2+x_3+x_5)\leq$ $\leq\left(\frac{x_1+x_2+x_3+x_4+x_5}{2}\right)^2=\frac{25}{4}$. If $x_4\leq x_3$ we obtain: $x_1x_2+x_2x_3+x_3x_4+x_4x_5\leq(x_1+x_3)(x_2+x_4+x_5)\leq$ $\leq\left(\frac{x_1+x_2+x_3+x_4+x_5}{2}\right)^2=\frac{25}{4}$. The equality occurs for $x_1=x_2=\frac{5}{2}$ and $x_3=x_4=x_5=0$.

Without loss of generality I will assume that $x_1$ is the largest of all (because my proof will follow exactly the same steps in the other cases).Also denote the LHS by $f(x_1,x_2,x_3,x_4)$.Then note that $f(x_1,x_2+x_4,x_3,0) > f(x_1,x_2,x_3,x_4)$ so the LHS cannot attain its maximum until $x_5$ is zero.Similarly we can reduce the three-variables to two variables consideration.At this stage we have $x_1+x_2=5$ and consequently $x_1x_2 \le \frac{25}{4}$ by A.M-G.M inequality.Equality occurs when two of the five variables are $2.5$ and the rest are $0$.