I am trying to solve the following exercise:
Given a simple connected graph $G$, we define the barycentricity of a vertex $x \in V(G)$ as $s(x)=\sum_{v \in V(G)} dist (x,v)$.
Let $T$ be a tree and $x\in V(T)$ a vertex that is not a leaf. Then prove that for every $y,z \in N(x)$ ($N(x)$: the neighborhood of x): $$2s(x)\lt s(y)+s(z)$$
I'd like to get some intuition for the barycentricity of a graph, cause i can't really find a lot of examples with it. I started by thinking that $s(x)$ depends on the distance of $x$ from the center (which is 1 or 2 vertices for trees), but then i found some counterexamples to this.
Taking cases, when y,z have a greater distance from the center than x i think i can prove that $s(x) \lt s(y)$ and $s(x)\lt s(z)$. What tells me that when y or z are closer to the center then $2s(x)\lt s(y)+s(z)$?
I guess that $y$ and $z$ should be distinct. Otherwise if $x$ and $y$ are neighbors which both are not leaves than we would have both $s(x)<s(y)$ and $s(y)<s(x)$, which is a contradiction. Now fix $x$ as a root of the tree. Let $T_1,\dots, T_l$ are subtrees of $T$ rooted on children of $x$. That is $V(T)=\{x\}\bigcup V(T_i)$. Let $y$ and $z$ be two distinct neighbors of $x$. They both of them are children of $x$. Assume that $y\in T_i$ and $z\in T_j$. Since $y$ and $z$ are distinct, $i\ne j$. Since $T$ has is a tree, between each two vertices $v$ and $u$ of $T$ there exists a unique simple path. This path realizes distance between $v$ and $u$. This property allows us easily calculate that $$s(y)=s(x)+\sum_{k\ne i} |V(T_k)|-|V(T_i)|+1=|V(T)|-2|V(T_i)|$$ and, similarly, $$s(z)=s(x)+|V(T)|-2|V(T_j)|.$$ Adding these two equalities we obtain
$$s(y)+s(z)=2s(x)+ 2|V(T)|-|V(T_i)|-|V(T_j)|> 2s(x)+|V(T)|>2s(x).$$