If $f$ is $\beta$-smooth and non-negative, then $|f'(x)|^2\le 2\beta f(x)$?

Question

I am reading a paper, and I found this conclusion from a proof.

I am wondering why we can conclude that if a function $f$ is $\beta$-smooth and non-negative, then $|f'(x)|^2\le 2\beta f(x)$.

A simple definition of $\beta$-smooth is $|f''(x)| \le \beta$.