Problem 1395

Let $M$ be the set of mathematicians, $n=|M|, F(m)$ denote the set of friends of mathematician $m$, and $f(m)$ denote the number of friends of mathematician $m$ $(f(m)=|F(m)|)$. The proposition is equivalent to proving: there must be an $m_{0}$ such that
$$\frac{1}{f\left(m_{0}\right)} \sum_{m \in F\left(m_{0}\right)} f(m) \geqslant \frac{1}{n} \sum_{m \in M} f(m)$$

We will use proof by contradiction to prove this proposition. If there is no such mathematician $m_{0}$, then for any $m_{0}$, we have
$$n \cdot \sum_{m \in F\left(m_{0}\right)} f(m) < f\left(m_{0}\right) \sum_{m \in M} f(m)$$

Summing over all $m_{0}$, we get
$$n \cdot \sum_{m_{0}} \sum_{m \in F\left(m_{0}\right)} f(m) = n \sum_{m} \sum_{m \in F\left(m_{0}\right)} f(m) = n \sum_{m \in M} f^{2}(m) < \left(\sum_{m \in M} f(m)\right)^{2}$$

This contradicts the Cauchy-Schwarz inequality, hence the proposition is true.