Setup. Let $X$ be a compact complex manifold. Let $\sum\limits \omega_{jk}d{z_j}\otimes d\overline{z}_k$ be a Hermitian metric on $X$ with associated positive $(1,1)$-form $\omega = \frac{i}{2}\sum \omega_{jk}d{z_j}\wedge d\overline{z}_k$.
Recall that $X$ is Brody hyperbolic if there is no non-constant holomorphic map $\mathbb{C}\to X$ and that $X$ is Kobayashi hyperbolic if the Kobayashi pseudo-metric on $X$ is a genuine metric. By work of Kobayashi and Brody, it is known that these two conditions are equivalent since we assume that $X$ is compact.
In Theorem 2.1 of here, Demailly proved that $X$ being Kobayashi hyperbolic (equivalently $X$ being Brody hyperbolic) implies that $X$ is algebraically hyperbolic (i.e., there exists a positive constant $c$ such that the degree of any curve of genus $g$ on $X$, where the degree of $C$ is given by $\text{deg}_{\omega}(C) = \int_{C}\omega$ for $\omega$ defined above, is bounded from above by $c(g-1)$). His proof crucially uses properties of the Kobayashi metric and the Gauss--Bonnet theorem.
Question. Is there a direct proof of the fact that $X$ being Brody hyperbolic implies that $X$ is algebraically hyperbolic? By a direct proof, I mean one which does not use any properties of the Kobayashi metric and only uses the non-existence of entire curves in $X$.
Thanks in advance!
