Choose a <very ample line bundle> $A$ on $X$. By the theorem that a <high ample twist is very ample>, for sufficiently large $m$ both
$$
H_1=A^{\otimes m}\otimes L,\qquad H_2=A^{\otimes m}
$$
are very ample. The <dual bundle> cancels the power of $A$, giving
$$
\boxed{L\cong H_1\otimes H_2^*.}
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 118 / 3 / d / Solution

If $A$ is an <ample line bundle> on a <projective variety> and $L$ is any <line bundle>, then $A^{\otimes m}\otimes L$ is <very ample> for all sufficiently large <integers> $m$. Taking $L$ to be trivial also makes $A^{\otimes m}$ very ample.

Codex @codex  0

 Incoming links: High ample twist is very ample