Consider a group of people (as a connected graph where each vertex represents a person and the ones that are connected to that vertex are her friends). If a person gets an amount of wage which is above the average wage level of her friends then that person is happy. Show that for everybody to be happy, each person gets the same amount of wage.
Well, I have thought about Friendship Paradox. However, in this case, we don't deal with degrees but wages. How am I supposed to think?
Proceed by induction on the number of vertices. The base case where $n=1$ is trivial.
For the inductive step, we suppose the hypothesis holds for any connected graph with $n$ vertices. Let $G$ be a connected graph with $n+1$ vertices, and suppose everyone is happy. Choose a vertex $v_0 \in G$ and note that the graph $G-\{v_0\}$ is the graph generated by eliminating the vertex $v_0$ and the edges involving $v_0$. The inductive hypothesis implies that everyone in the component graphs $C_1, \ldots, C_k$ of $G-\{v_0\}$ has the same wage (that is, everyone within each component has the same wage, we still need to show that everyone in one component has the same wage with everyone in a different component, and also that $v_0$ has a wage that agrees).
It suffices at this point to prove that every neighboring vertex of $v_0$ has the same wage as $v_0$. Why and how that is I'll leave to you as an exercise. Hint: It's a pretty straightforward argument of showing equality using the antisymmetry property of $\leq$, and then using the transitivity of $=$.
EDIT: Ok, maybe the argument I'm leading you to is not as straightforward as I'm implying, and pretty sure you need one other "trick", so I'll give this trick: Proceed by induction on the number of components $C_1, \ldots, C_k$. Both the "base case" and the "inductive step" are pretty immediate, and I'll leave to the reader. But I'll give a hint in each case for good measure should you get stuck.
Base Case Hint: Note that the fact that every vertex in each component has the same wage means in this particular case that every neighbor of $v_0$ has the same wage.
Inductive Step Hint: Interestingly enough, you don't actually need the "inductive hypothesis" on $k$ (and really it's more accurate to call this argument "cases on $k$" rather than "induction on $k$"), One proceeds by applying the inductive hypothesis we have for $n$ on $C_j \cup \{v_0\}$ ($1 \leq j \leq k$).