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.)



Does a strict 2-balanced graph must to be strict balanced?
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.


