Let $G$ be a triangle-free graph. We may assume that $G$ has at least three vertices and at least one edge; otherwise, there is nothing to prove. Consider the set $P$ of all functions $c\colon V(G)\to\mathbb{R}_+$ such that $\sum_{v\in V(G)} c(v)=1$ . Define the total weight $W(c)$ of such a function by$$ W(c) = \sum_{uv\in E(G)} c(u)\cdot c(v).$$ By declaring that $c\le c^*$ if and only if $W(c)\le W(c^*)$ we make $P$ into a poset.

Consider the function $c_0\in P$ which takes the constant value $\frac{1}{|V(G)|}$ on each vertex. The total weight of this function is$$ W(c_0) = \sum_{uv\in E(G)} \frac{1}{|V(G)|}\cdot\frac{1}{|V(G)|}=\frac{|E(G)|}{|V(G)|^2},$$ which is positive because $G$ has an edge. So if $c\ge c_0$ in $P$ , then $c$ has support on an induced subgraph of $G$ with at least one edge.

We claim that a maximal element of $P$ above $c_0$ is supported on a copy of $K_2$ inside $G$ . To see this, suppose $c\ge c_0$ in $P$ . If $c$ has support on a subgraph larger than $K_2$ , then there are nonadjacent vertices $u$ and $v$ such that $c(u)$ and $c(v)$ are both positive. Without loss of generality, suppose that \begin{equation*} \sum_{uw\in E(G)} c(w) \ge \sum_{vw\in E(G)} c(w).\tag{*} \end{equation*}Now we push the function off $v$ . To do this, define a function $c^*\colon V(G)\to\mathbb{R}_+$ by