8. The sum of eight numbers is $\frac{4}{5}$. It turns out that the sum of any seven of these eight numbers is non-negative. What is the smallest value that one of these numbers can take?
Answer: $-4.8$.

Solution. Let the numbers be denoted by $a_{1}, a_{2}, \ldots, a_{8}$. According to the condition, $a_{1}+a_{2}+\ldots+a_{8}=\frac{4}{5}$. Note that the sum of all numbers except one is non-negative, so each number does not exceed $\frac{4}{5}$ (and conversely, if each number does not exceed $\frac{4}{5}$, the sum without any one of the numbers is non-negative). Therefore, the sum of any seven numbers does not exceed $7 \cdot \frac{4}{5}$, and then the remaining number must be at least $-6 \cdot \frac{4}{5}$.

On the other hand, if one of the numbers is $-\frac{24}{5}$, and each of the remaining numbers is $\frac{4}{5}$, then all conditions are satisfied. Thus, the smallest possible value of one of the numbers is $-\frac{24}{5}=-4.8$.

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