Let $a, b, c, d, e, f, g$ be non-negative numbers whose sum is 1. Select the largest of the sums $a+b+c, b+c+d, c+d+e, d+e+f$, $e+f+g$. What is the minimum value of the thus obtained value?
Since

$$
\begin{aligned}
& (a+b+c)+(c+d+e)+(e+f+g)= \\
& =a+b+c+d+e+f+g)+c+e \geq 1
\end{aligned}
$$

it follows that among the numbers $a+b+c, c+d+e, e+f+g$, at least one is at least $1/3$. Therefore, the largest value of the given 5 three-term sums is always at least $1/3$.

On the other hand, when $a=d=g=1/3, b=c=e=f=0$, each of the three-term sums is $1/3$. Consequently, the sought minimum is $1/3$.

Remark. A similar statement can be proven as follows: If the sum of non-negative numbers $a_{1}, a_{2}, \ldots, a_{n}$ is 1, then the maximum of the sums $\left(a_{i+1}+a_{i+2}+\ldots+a_{i+k}\right)$ $(i=0,1, \ldots, n-k)$ has a minimum of the reciprocal of $1+\left[\frac{n-1}{k}\right]$.