I'm currently studying the fact that $\chi^2$ is sub-exponential, which seems similar to this site, and what I'm confused looks similar to what the author of the link was confused as well.
Show that $\frac{e^{-t}}{\sqrt{1-2t}} \le e^{2t^2}$ for all $t$ with $\vert t \vert \le 1/4$
What I need is the exact value of $\alpha$ and $\nu$, but the link seems to show the numerical result; that is, since the result shows that $t < 0.34$ meets the inequality $\frac{e^{-t}}{\sqrt{1-2t}} \le e^{2t^2}$, $|t| \le 1/4$ satisfies it. But, I want a more sophisticated proof of this inequality using some kind of closed forms such as Taylor's expansion.
If such closed form exists, any hint regarding this proof would be grateful. Thank you.