Does a strict 2-balanced graph must to be strict balanced? Let $H$ be a finite simple graph, with $v_H\ge3$ vertices and $e_H\ge3$ edges. Say $H$ is strict 2-balanced if $\frac{e_H-1}{v_H-2}\gt \frac{e_K-1}{v_K-2}$ for all proper subgraphs $K$ with $v_K\ge3$, and say $H$ is strict balanced if $\frac{e_H}{v_H}\gt \frac{e_K}{v_K}$ for all proper subgraphs $K$ with $v_K\ge1$.
Does a strict 2-balanced graph must to be strict balanced? (May add conditions such that the density of the graph is not so small, bigger than 2 etc.)
 A: This is true. Indeed, provided $H$ has a connected component with at least two edges, then just being $2$-balanced, that is, knowing that $\frac{e_H-1}{v_H-2} \ge \frac{e_K-1}{v_K-2}$ for all $K \subset H$ is enough to imply that $H$ is strictly balanced. To prove this, suppose that $H$ is not strictly balanced and let $K \subset H$ be a subgraph with $\frac{e_K}{v_K} \ge\frac{e_H}{v_H}$. That is, $e_K v_H \geq e_H v_K$. Since $H$ is $2$-balanced, we have $\frac{e_H-1}{v_H-2} \ge \frac{e_K-1}{v_K-2}$. Multiplying out gives
$$e_H v_K - 2e_H - v_K + 2 \geq e_K v_H - 2e_K - v_H + 2.$$
Substituting in $e_K v_H \geq e_H v_K$ on the right-hand side, we get that
$$e_H v_K - 2e_H - v_K + 2 \geq e_H v_K - 2e_K - v_H + 2.$$
Cancelling the like terms gives $-2e_H - v_K \geq -2e_K - v_H$, which in turn implies that $2(e_K - 1) - (v_K - 2) \geq 2(e_H - 1) - (v_H - 2)$ and this can be rewritten as
$$\left(2\frac{e_K-1}{v_K-2} - 1\right)(v_K - 2) \geq \left(2\frac{e_H-1}{v_H-2} - 1\right)(v_H - 2).$$
However, this is a contradiction, since $2\frac{e_H-1}{v_H-2} - 1 \geq 2\frac{e_K-1}{v_K-2} - 1$ by assumption, $\frac{e_H-1}{v_H-2} > 1/2$ for all $H$ with a connected component with at least $2$ edges and $v_H - 2 > v_K - 2$. The remaining case, where $H$ is a matching is not strictly $2$-balanced unless $H$ is a single edge, so the required result also holds in this case.
