An inequality involving an expected value in a paper about accessibility percolation In a paper called "Increasing paths in regular trees", by Roberts and Zhao, the authors complete results from another paper, by Nowak and Krug, in which they study accessibility percolation on $n$-ary trees.
They define a random variable $N_\epsilon$ and show that its expected value satisfies $$\mathbb{E}[N_\epsilon] \geq \dfrac{(\alpha(1-\epsilon)e)^h}{3h^{3/2}},$$ where $\alpha,h > 0$ and $\epsilon \in [0,1)$.
Then, assuming that $\alpha(1-\epsilon)e > 1$, they choose $\delta \in (0, \log(\alpha(1-\epsilon)e)$ and claim that $\mathbb{E}[N_\epsilon]/2 \geq e^{\delta h}$ for all sufficiently large $h$.
I don't understand why this last inequality holds true. For instante, if $0 < \delta < \log(\alpha(1-\epsilon)e)$, then $e^\delta < \alpha(1-\epsilon)e$, whence $e^{\delta h} < (\alpha(1-\epsilon)e)^h$, so $$\mathbb{E}[N_\epsilon] \geq \dfrac{e^{\delta h}}{3h^{3/2}}.$$ In order for the claim to hold, shouldn't we need to have $$\dfrac{e^{\delta h}}{3h^{3/2}} \geq 2 e^{\delta h},$$ that is, $$3h^{3/2} \leq 2$$ for sufficiently large $h$? What am I missing here?

Simply saying that $e^{\delta h} < (\alpha(1-\epsilon)e)^h$ as you do forgets crucial information about the growth rate of both sides as $h \to \infty$.
We can write
$$
   \frac{\mathbb E[N_\epsilon]}{e^{\delta h}} \ge \frac1{h^{3/2}} \left(\frac{\alpha(1-\epsilon)e}{e^{\delta}}\right)^h
$$
and observe that $\frac{\alpha(1-\epsilon)e}{e^{\delta}} > 1$. Therefore $\left(\frac{\alpha(1-\epsilon)e}{e^{\delta}}\right)^h$ increases exponentially, while $h^{3/2}$ only increases polynomially. Therefore as $h \to \infty$, the right-hand side also increases to $\infty$. Therefore, for all sufficiently large $h$, $\frac{\mathbb E[N_\epsilon]}{e^{\delta h}} \ge 2$.
