Let $(e_1,\lambda_1)$ denote the first eigenpair of the Dirichlet Laplacian on a bounded open set $\Omega$ with smooth boundary such that $\max_{\overline\Omega}e_1=1$. We can show that there is a $c>0$ with $\partial_\nu e_1\le-c$. Moreover, by the minimum principle for superharmonic functions, $\min_{\overline\Omega}e_1=\min_{\partial\Omega}e_1$.
But why can we conclude that $0<\inf_{x\in\Omega}\frac{e_1(x)}{\operatorname{dist}(x,\:\partial\Omega)}<\sup_{x\in\Omega}\frac{e_1(x)}{\operatorname{dist}(x,\:\partial\Omega)}<\infty$?
Let $K$ be any compact subset of $\Omega$. Since $e_1 >0$ on $\Omega$ and that $d(\cdot, \partial \Omega)$ is bounded below by a positive constant (depending on $K$) on $K$, we have
$$0<\inf_{x\in K}\frac{e_1(x)}{\operatorname{dist}(x,\:\partial\Omega)}<\sup_{x\in K}\frac{e_1(x)}{\operatorname{dist}(x,\:\partial\Omega)}<\infty.$$
Now we consider those points closed to the boundary. By the tubular neighborhood theorem, there is $\epsilon >0$ so that
$$ \Phi : \partial \Omega \times (-\epsilon, \epsilon) \to \mathbb R^n, \Phi (x, t) = x - t\nu$$
is a diffeomorphism onto it's image, and $\bar \Omega \cap \operatorname{Im}\Phi = \Phi (\partial\Omega \times [0,\epsilon))$ ($\nu$ is the outward unit normal here). Moreover, we have, for all $x\in \partial \Omega$ and $t >0$,
$$\tag{1} d( \Phi (x, t), \Omega) = d(\Phi (x, t), x) = t.$$
Since $\partial_v e_1\le -c$ on $\partial \Omega$ and that $\partial _v e_1$ is continuous, there is $\epsilon >0$ small enough so that $$ -C \le \partial _v e_1 (p) \le -c/2 <0$$ for all $p \in \Phi (\partial \Omega \times [0,\epsilon))$. In particular, since $e_1 = 0$ on $\partial \Omega$,
$$ e_1( x- t\nu) = e_1( x- t\nu) - e_1(x) = -\int_0^t \partial_v e_1 (x-s\nu) ds.$$
Together with (1), for all $p \in \Phi (\partial \Omega\times [0,\epsilon))$,
$$ C d(p, \partial \Omega) \ge e_1(p) \ge (c/2) d(p, \partial \Omega), $$
together with the choice $K = \Omega\setminus \operatorname{Im} \Phi$, we have the desired inequalities.