I know of simpler proofs, but I am concerned with the validity of the following probabilistic approach.
Let $a_i$ be the number of friends that person $i$ has. Assume for contradiction that all $N$ people in the have more friends than the average number of friends their friends have. Pick a person at random, weighted by how many friends they have in the room. Let $X$ be the number of friends your pick has. Let $Y$ be your pick's friends' average number of friends. By our opening assumption, $P(X > Y) = 1$ and therefore $E[X] > E[Y]$. However, $$E[X] = \frac{a_1}{a_1+\cdots+a_n}a_1 + \cdots + \frac{a_n}{a_1+\cdots+a_n}a_n = \frac{\sum_{i=1}^{N}{a_i^2}}{\sum_{i=1}^{N}{a_i}}$$ And also, $$E[Y] = \frac{\text{total friends of friends}}{\text{total friends}}=\frac{\sum_{i=1}^{N}{a_i^2}}{\sum_{i=1}^N {a_i}}=E[X]$$ This contradicts $E[X] > E[Y]$. (In the expression for $E[Y]$ the numerator and denominator are both overcounting. For example, total friends is counting every person $a_i$ times. The numerator counts person $i$'s friends $a_i$ times.)
Is this correct? I'm not entirely sure my argument is correct, especially the way I deal with $E[Y]$. Any insight appreciated!
You're right that $E[Y]=E[X]$, but I don't quite follow how you calculated $E[Y]$. I'd phrase the probabilistic argument as follows:
Your weighted selection of a person can be expressed as uniformly picking a friendship and then uniformly picking one of its two friends. If we now uniformly select a friend of this person, we're uniformly selecting a friendship incident at this person, and this gets us back to a uniform distribution over all friendships. Then uniformly selecting one of the two friends of this friendship takes us to the original person or to a uniformly selected one of their friends with equal probability, which shows $\frac12E[X]+\frac12E[Y]=E[X]$.