Problem 1587

1. Graph Representation and Assumptions:
- Denote the group of mathematicians by a graph $G$ with vertices $v_1, v_2, \ldots, v_n$.
- Assume $\deg(v_i) \geq 1$ for each $i, 1 \leq i \leq n$. This assumption simplifies the problem without loss of generality.

2. Contradiction Setup:
- Suppose the contrary, that for every mathematician $v_i$, the average number of friends of their friends is less than the average number of friends of all mathematicians.
- Let $k_i = \deg(v_i)$ be the number of friends of $v_i$.
- Let $a_i$ be the total number of friends of friends of $v_i$. If $w_1, w_2, \ldots, w_{k_i}$ are all friends of $v_i$, then $a_i = \sum_{j=1}^{k_i} \deg(w_j)$.

3. Good Subgraph Definition:
- Define a subgraph $x$ of $G$ as $good$ if it has 3 vertices $v_i, v_j, v_k$ such that $\deg(v_i) = \deg(v_j) = 1$ and $\deg(v_k) = 2$ in $x$.
- Let $Y$ be the total number of good subgraphs of $G$.

4. Total Number of Friends:
- Let $t$ be the total number of friends of all mathematicians, $t = \sum_{i=1}^{n} \deg(v_i)$.

5. Average Friends Calculation:
- Since $\frac{a_i}{k_i} < \frac{t}{n}$ for each $i, 1 \leq i \leq n$, we have:
$$A = \sum_{i=1}^{n} a_i < \frac{t}{n} \sum_{i=1}^{n} k_i = \frac{t^2}{n}$$

6. Counting Good Subgraphs:
- Select a $v_i$, $1 \leq i \leq n$. Let $g_i$ be the number of good subgraphs of $G$ where $v_i$ is a vertex with $\deg(v_i) = 1$.
- Then $v_i$ counts exactly $\deg(v_i) + g_i$ times in $A$ (once for each friend and once for each good subgraph it is part of).

7. Summing Up:
- Thus, $A = \sum_{i=1}^{n} a_i = \sum_{i=1}^{n} \deg(v_i) + g_i$.
- Since each good subgraph is counted exactly twice in $\sum_{i=1}^{n} g_i$, we have:
$$\sum_{i=1}^{n} g_i = 2Y = 2 \sum_{i=1}^{n} \binom{\deg(v_i)}{2}$$
- Therefore:
$$A = \sum_{i=1}^{n} a_i = \left( \sum_{i=1}^{n} \deg(v_i) \right) + 2 \sum_{i=1}^{n} \binom{\deg(v_i)}{2}$$
$$= \left( \sum_{i=1}^{n} \deg(v_i) \right) + 2 \sum_{i=1}^{n} \frac{\deg(v_i)(\deg(v_i) - 1)}{2}$$
$$= \sum_{i=1}^{n} (\deg(v_i))^2$$

8. Contradiction:
- Thus, $A = \sum_{i=1}^{n} (\deg(v_i))^2 < \frac{t^2}{n} = \left( \sum_{i=1}^{n} \deg(v_i) \right)^2 \frac{1}{n}$.
- By the Root Mean Square-Arithmetic Mean Inequality:
$$\sqrt{\frac{\sum_{i=1}^{n} (\deg(v_i))^2}{n}} \geq \frac{\sum_{i=1}^{n} \deg(v_i)}{n}$$
$$\frac{\sum_{i=1}^{n} (\deg(v_i))^2}{n} \geq \frac{\left( \sum_{i=1}^{n} \deg(v_i) \right)^2}{n^2}$$
- This contradicts the earlier inequality, thus proving the original statement.

$\blacksquare$